tag:blogger.com,1999:blog-30042132096918179782024-03-17T20:04:01.765-07:00Physics Problems & SolutionsUnknownnoreply@blogger.comBlogger417125tag:blogger.com,1999:blog-3004213209691817978.post-12527309024401096922016-03-16T09:54:00.000-07:002024-01-25T14:40:16.270-08:00Constants<div dir="ltr" style="text-align: left;" trbidi="on">
<div style="text-align: center;">
TABLE OF INFORMATION<br />
(Printed in the test booklet)</div>
<br />
<i>m</i><sub><i>e</i></sub><i> </i>= 9.11 × 10<sup>−31 </sup>kg<br />
<i>e </i>= 1.60 ×<sup> </sup>10<sup>−</sup><sup>19 </sup><span style="text-align: justify;">coulomb</span><br />
<i>N</i><sub>0 </sub>= 6.02 ×<sup> </sup>10<sup>23 </sup><span style="text-align: justify;">per mole</span><br />
<i>R </i>= 8.31 joules/(mole K)<br />
<i>k </i>= 1.38 ×<sup> </sup>10<sup>−</sup><sup>23</sup><sup> </sup>joule/K = 1.38 ×<sup> </sup><sup> </sup>10<sup>−16</sup><sup> </sup>erg/K<br />
<i>c </i>= 3 ×<sup> </sup>10<sup>8</sup><sup> </sup><span style="text-align: justify;">m/s</span><br />
<i>h </i>= 6.63 ×<sup> </sup>10<sup>−</sup><sup>34 </sup><span style="text-align: justify;">Joule second </span>= 4.1 ×<sup> </sup>10<sup>−</sup><sup>15 </sup><span style="text-align: justify;">eV.second</span><br />
<i>ħ</i><i> </i>= <i>h</i>/2<i>π</i><br />
<i>ɛ</i><sub>0 </sub>= 8.85 ×<sup> </sup>10<sup>−12</sup><sup> </sup><span style="text-align: justify;">coulomb</span><sup>2</sup><span style="text-align: justify;">/(newton meter</span><sup>2</sup><span style="text-align: justify;">) </span><br />
<div style="text-align: justify;">
<i>μ</i><sub>0 </sub>= 4<i>π</i> ×<sup> </sup>10<sup>−7</sup><sup> </sup>weber <span style="text-align: justify;">/(ampere meter</span><span style="text-align: justify;">) </span></div>
<div style="text-align: justify;">
<span style="text-align: justify;"><i style="text-align: left;">G </i><span style="text-align: left;">= 6.67 ×</span><sup style="text-align: left;"> </sup>10<sup>−11</sup><sup style="text-align: left;"> </sup><span style="text-align: justify;"><span style="text-align: justify;">meter</span><sup style="text-align: left;">3</sup>/</span></span><span style="text-align: justify;">(kilogram second</span><sup style="text-align: left;">2</sup><span style="text-align: justify;">) </span></div>
<div style="text-align: justify;">
<span style="text-align: justify;"><span style="text-align: justify;"><i style="text-align: left;">g </i><span style="text-align: left;">= 9.80 m/s</span><span style="text-align: justify;"><sup style="text-align: left;">2</sup></span></span></span></div>
<div style="text-align: justify;">
<span style="text-align: justify;"><span style="text-align: justify;"><span style="text-align: justify;"><span style="text-align: left;">1 atm = 1.0 ×</span><sup style="text-align: left;"> </sup><span style="text-align: left;">10</span><sup style="text-align: left;">5</sup><sup style="text-align: left;"> </sup><span style="text-align: justify;">newton</span><span style="text-align: justify;">/meter</span><sup style="text-align: left;">2</sup><span style="text-align: justify;"> </span></span></span></span><span style="text-align: left;">= 1.0 ×</span><sup style="text-align: left;"> </sup><span style="text-align: left;">10</span><sup style="text-align: left;">5</sup><sup style="text-align: left;"> </sup>pascals (Pa)</div>
<div style="text-align: justify;">
1 Å = <span style="text-align: left;">1 </span><span style="text-align: left;">×</span><span style="text-align: left;"> </span>10<sup>−10</sup><sup style="text-align: left;"> </sup><span style="text-align: justify;">meter</span><br />
<span style="text-align: justify;">1 weber/<span style="text-align: justify;">meter</span><sup style="text-align: left;">2</sup> = 1 tesla = </span><span style="text-align: left;">10</span><sup style="text-align: left;">4</sup><sup style="text-align: left;"> </sup>gauss<br />
<br />
<br />
Moments of inertia about center of mass<br />
<br />
Rod : <sup style="text-align: left;">1</sup><span style="text-align: left;">/</span><sub style="text-align: left;">12</sub><i>Ml</i><sup style="text-align: left;">2 </sup><br />
Disc : <sup style="text-align: left;">1</sup><span style="text-align: left;">/</span><sub style="text-align: left;">2</sub><i>MR</i><sup style="text-align: left;">2</sup><br />
Sphere: <sup style="text-align: left;">2</sup><span style="text-align: left;">/</span><sub style="text-align: left;">5</sub><i>MR</i><sup style="text-align: left;">2</sup></div>
</div>
Unknownnoreply@blogger.com0tag:blogger.com,1999:blog-3004213209691817978.post-44916175283779690982016-03-05T18:36:00.001-08:002023-12-23T17:41:24.320-08:00Electromagnetism - Capacitor<div dir="ltr" style="text-align: left;" trbidi="on">
<div dir="ltr" style="text-align: left;" trbidi="on">
<div style="text-align: center;"><br /></div>
<div dir="ltr" style="text-align: left;" trbidi="on">
<div style="text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEia-MUjkXNHqb8FT9oUPZfHiyp4MDjpMdioVxUOWm_-Jwfz4CgutUfI4sDq2imIqCh0AkubfgE55zFzchmxW-5VdkdgCyRT2ezJz0lRoXYqm3H_m98XUV13GkH6uL5w3Db8hbo4-6fOleOmhE6HjKp6wm6LIC0VE8nUGwMHKNOCyPDlLRDIvH1JTzULaGsP/s584/GR9677-01a.JPG" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="442" data-original-width="584" height="192" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEia-MUjkXNHqb8FT9oUPZfHiyp4MDjpMdioVxUOWm_-Jwfz4CgutUfI4sDq2imIqCh0AkubfgE55zFzchmxW-5VdkdgCyRT2ezJz0lRoXYqm3H_m98XUV13GkH6uL5w3Db8hbo4-6fOleOmhE6HjKp6wm6LIC0VE8nUGwMHKNOCyPDlLRDIvH1JTzULaGsP/w254-h192/GR9677-01a.JPG" width="254" /></a></div></div></div><br /><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhJyQjbgRADA2NwSLMeVoroovDsO14FbsnrvTUnO0Upl9RGmqIwnxJt-UZwYZ2eOZ2fh3QS5IuoJ5NPDFMcwmbcAZLY4RYCalbIiKn5XEKKA-QF3wQnLr62Y5x_b7JxkY4xohVFev6NBWVDwyY-MKFUcSawXePYy5umWYKDqhnYN_rUf867YyTqqXE-k4Ou/s778/GR9677-01b.JPG" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="568" data-original-width="778" height="234" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhJyQjbgRADA2NwSLMeVoroovDsO14FbsnrvTUnO0Upl9RGmqIwnxJt-UZwYZ2eOZ2fh3QS5IuoJ5NPDFMcwmbcAZLY4RYCalbIiKn5XEKKA-QF3wQnLr62Y5x_b7JxkY4xohVFev6NBWVDwyY-MKFUcSawXePYy5umWYKDqhnYN_rUf867YyTqqXE-k4Ou/s320/GR9677-01b.JPG" width="320" /></a></div><br /><div style="text-align: justify;"><br /></div><div style="text-align: justify;">
The capacitor shown in Figure 1 above is charged by connecting switch <i>S</i> to contact <i>a</i>. If switch <i>S</i> is thrown to contact <i>b</i> at time <i>t</i> = 0, which of the curve in Figure 2 above represents the magnitude of the current through the resistor <i>R </i>as a function of time?</div>
<br />
<a href="https://www.blogger.com/blogger.g?blogID=3004213209691817978" style="clear: right; float: right; margin-bottom: 1em; margin-left: 1em;"></a>A. A<br />
B. B<br />
C. C<br />
D. D<br />
E. E
<br />
<div style="text-align: right;">
<b><b><span style="color: red; font-weight: bold;">(GR9677 #01)</span></b></b></div>
<b><b><span style="font-weight: bold;">Solution</span>:</b></b><br />
<br />
The capacitor is charged by connecting switch <i style="text-align: justify;">S</i> to contact <i>a</i>, so the current after connecting switch to contact <i>b</i> (<i>t</i> = 0) must start with
<i>I</i> = <i>V/R</i>.<br />
<br />
Only plot A and B are right.<br />
<br />
The current can not be the same all the time and since the voltage of a capacitor follows an exponential decay:<br />
<i><br /></i>
<i>V</i>(<i>t</i>) = <i>V</i><sub>0</sub>e<sup><i>−t/RC </i></sup><br />
<i>I</i>(<i>t</i>) = <i>V</i>(<i>t</i>)<i>/R</i> = <i>V</i><sub>0</sub>e<sup><i>−t/RC</i></sup><i>/</i> <i>R</i><br />
<br />
Only plot B is right.<br />
<b><b><br /></b></b>
<b><b><span style="font-weight: bold;">Answer</span>: B</b></b>
<b>
<b>
</b>
</b>
</div>
Unknownnoreply@blogger.com0tag:blogger.com,1999:blog-3004213209691817978.post-65530497818228165012016-03-05T18:35:00.002-08:002023-12-23T17:43:07.635-08:00Electromagnetism - Faraday’s law<div dir="ltr" style="text-align: left;" trbidi="on">
<div style="text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjbVJneG3ENHqk3NEsfdioc35r_8CQtepecOg2SWh_SleCm_hblHkbd4CvEYISdfTaB_cFPnfqshLAVnKatpOqs_FQ927jMxqjgdbejPZC8KoBv1-mmDFmd4fOmVUlTsWBtf2WIQgbpQiT3iReRA2oD7ojpKVKLImrWUo3-mK49fNo7oQtHzOgH5WhzDm1z/s566/GR9677-02.JPG" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="547" data-original-width="566" height="309" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjbVJneG3ENHqk3NEsfdioc35r_8CQtepecOg2SWh_SleCm_hblHkbd4CvEYISdfTaB_cFPnfqshLAVnKatpOqs_FQ927jMxqjgdbejPZC8KoBv1-mmDFmd4fOmVUlTsWBtf2WIQgbpQiT3iReRA2oD7ojpKVKLImrWUo3-mK49fNo7oQtHzOgH5WhzDm1z/s320/GR9677-02.JPG" width="320" /></a></div><span style="text-align: justify;"><div dir="ltr" style="text-align: left;" trbidi="on"><br /></div>The circuit shown is in a uniform magnetic field that is into the page and is decreasing in magnitude at rate of 150 tesla/second. The ammeter reads</span><br />
<br />
A. 0.15 A <br />
B. 0.35 A <br />
C. 0.50 A <br />
D. 0.65 A <br />
E. 0.80 A
<br />
<div style="text-align: right;">
<b><span style="color: red;"><b>(GR9677 #02)
</b></span></b></div>
<b>
</b>
<br />
<div class="separator" style="clear: both; text-align: center;">
</div>
<b>Solution:</b>
<br />
<br />
<i>V </i>− <i>IR </i>− <i>ɛ</i> = 0<br />
<i>I </i>= (<i>V</i> − <i>ɛ</i>)/<i>R</i><br />
<br />
<i>ɛ</i> = − <i>d</i>Φ/<i>dt</i> = −<i>A</i><i>dB</i><i>/dt</i><br />
<br />
Given:<br />
<i>dB/dt</i> = −150 t/s (minus sign because it’s decreasing)<br />
<i>A</i> = (0.1 m)<sup>2 </sup>= 0.01 m<sup>2</sup><br />
<i>R</i> = 10 Ω<br />
<i>V</i> = 5 V<br />
<br />
<i>ɛ</i> = − (0.01)(−150) = 1.5 V<br />
<i>I </i>= (5 − 1.5)/10<i> </i>= 3.5/10 = 0.35 A<br />
<br />
<b>Answer: B</b><br />
<br /></div>
Unknownnoreply@blogger.com0tag:blogger.com,1999:blog-3004213209691817978.post-64019662788958139182016-03-05T18:34:00.001-08:002023-12-23T17:44:44.403-08:00Electromagnetism - Electric Potential<div dir="ltr" style="text-align: left;" trbidi="on">
<div style="text-align: center;">
<div style="text-align: left;">
<u>Question 3-4</u>: refer to a thin, nonconducting ring of radius <i>R</i>, as shown below, which has a charge Q uniformly spread out on it.</div>
</div>
<div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhXQnczggJqOv-zhLwNhVLcBGOHUomNnr7UXRTUqU0BW7OeR6S-RG-BoaJ8ImUck4jglP2zEdU2x1I4-GKLLsF-q_L2TorQVJf9BqD94zCqT9BXtuPaP-pHNPYS_dXo3pimI-2vtQ68IWhu-zFLI1-WSRw0f5hnz0ndpE2qc5jhZ57KGxqD7RGTWyLSpFD9/s1116/GR9677-03.JPG" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="589" data-original-width="1116" height="169" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhXQnczggJqOv-zhLwNhVLcBGOHUomNnr7UXRTUqU0BW7OeR6S-RG-BoaJ8ImUck4jglP2zEdU2x1I4-GKLLsF-q_L2TorQVJf9BqD94zCqT9BXtuPaP-pHNPYS_dXo3pimI-2vtQ68IWhu-zFLI1-WSRw0f5hnz0ndpE2qc5jhZ57KGxqD7RGTWyLSpFD9/s320/GR9677-03.JPG" width="320" /></a></div><br /><div style="text-align: justify;">The electric potential at a point P, which is located on the axis of symmetry a distance <i>x</i> from the center of the ring, is given by</div>
<br />
A. <i>Q / </i>(4<i style="text-align: center;">π</i><i style="text-align: center;">ɛ</i><sub style="text-align: center;">0</sub><i>x</i>)
<br />
B. <i>Q / </i>[4<i style="text-align: center;">π</i><i style="text-align: center;">ɛ</i><sub style="text-align: center;">0</sub>(<i>R</i><sup>2 </sup>+ <i>x</i><sup>2</sup>)<sup>1/2</sup>]<br />
C. <i>Q</i><i>x</i><i> / </i>[4<i style="text-align: center;">π</i><i style="text-align: center;">ɛ</i><sub style="text-align: center;">0</sub>(<i>R</i><sup>2 </sup>+ <i>x</i><sup>2</sup>)]<br />
D. <i>Q</i><i>x</i><i> / </i>[4<i style="text-align: center;">π</i><i style="text-align: center;">ɛ</i><sub style="text-align: center;">0</sub>(<i>R</i><sup>2 </sup>+ <i>x</i><sup>2</sup>)<sup>3/2</sup>]<br />
E. <i>Q</i><i>R</i><i> / </i>[4<i style="text-align: center;">π</i><i style="text-align: center;">ɛ</i><sub style="text-align: center;">0</sub>(<i>R</i><sup>2 </sup>+ <i>x</i><sup>2</sup>)]<br />
<div style="text-align: right;">
<b><span style="color: red;"><b>(GR9677 #03)
</b></span></b></div>
<b>
</b>
<br />
<div class="separator" style="clear: both; text-align: center;">
</div>
<b>Solution:</b>
<br />
<br />
Electric Potential, <i>V </i>= <i>kQ/</i><i>r</i><i> </i><br />
with <i>k </i>= 1/4<i style="text-align: center;">π</i><i style="text-align: center;">ɛ</i><sub style="text-align: center;">0</sub><br />
<br />
The distance <i>r</i> of P from the charged ring is <i>r</i><sup>2</sup> = <i>R</i><sup>2 </sup>+ <i>x</i><sup>2</sup><br />
<i><br /></i>
<i>V </i>= <i>Q / </i>[4<i style="text-align: center;">π</i><i style="text-align: center;">ɛ</i><sub style="text-align: center;">0</sub>(<i>R</i><sup>2 </sup>+ <i>x</i><sup>2</sup>)<sup>1/2</sup>]<br />
<br />
<b>Answer: B</b></div>
Unknownnoreply@blogger.com0tag:blogger.com,1999:blog-3004213209691817978.post-70471795192756286522016-03-05T18:33:00.000-08:002023-12-23T18:02:35.127-08:00Electromagnetism - Oscillation<div dir="ltr" style="text-align: left;" trbidi="on">
<div style="text-align: justify;">
<div style="text-align: center;">
<div style="text-align: left;">
<u>Question 3-4</u>: refer to a thin, nonconducting ring of radius <i>R</i>, as shown below, which has a charge <i>Q</i> uniformly spread out on it.</div><div style="text-align: left;"><br /></div>
</div><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgiN_xOoM69Io8YrS-b8BKAS1lAGQfDs3Ub5fGCg1Zb65xFQJo2YY39kNeBw5v2zl7HKWill78bZ2XWnRxWvhAokowZP6xbDcaotP_vjpDX0ubK6DkfgEtOhd6rXIXU1JWbAlLZfIhxCD5HWCc7mPhngytIAo2LUeZKix5Qk8UeaBBCsv8HKGbhWQIAFq65/s1116/GR9677-03.JPG" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="589" data-original-width="1116" height="169" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgiN_xOoM69Io8YrS-b8BKAS1lAGQfDs3Ub5fGCg1Zb65xFQJo2YY39kNeBw5v2zl7HKWill78bZ2XWnRxWvhAokowZP6xbDcaotP_vjpDX0ubK6DkfgEtOhd6rXIXU1JWbAlLZfIhxCD5HWCc7mPhngytIAo2LUeZKix5Qk8UeaBBCsv8HKGbhWQIAFq65/s320/GR9677-03.JPG" width="320" /></a></div><div style="text-align: justify;"><br /></div><div style="text-align: justify;">A small particle of mass <i>m</i> and charge –<i>q </i>is placed at point P and released. If <i>R</i> <span style="text-align: left;">≫ </span><i>x</i>, the particle will undergo oscillations along the axis of symmetry with an angular frequency that is equal to:</div></div>
<div dir="ltr" style="text-align: left;" trbidi="on"><br /></div><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjcU8_lBEQrt9c6Fw0EX5RAIDQhIGkQMB3C-eNmGz0rx9iyNbJguniaS46lZY-MYNQJPzo1Ci9iwVTcfSUSq4LDevBRcbJyex5jSEhdatgNGm5_skYBnYHBd-A5hK2JVlKGUubNzL1NGq0kFDI69KVHOzntAt1sQEpf20UuIV63DeaHuiXtPeNzUif1E58Q/s622/GR9677-04%20answer.JPG" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="622" data-original-width="403" height="320" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjcU8_lBEQrt9c6Fw0EX5RAIDQhIGkQMB3C-eNmGz0rx9iyNbJguniaS46lZY-MYNQJPzo1Ci9iwVTcfSUSq4LDevBRcbJyex5jSEhdatgNGm5_skYBnYHBd-A5hK2JVlKGUubNzL1NGq0kFDI69KVHOzntAt1sQEpf20UuIV63DeaHuiXtPeNzUif1E58Q/s320/GR9677-04%20answer.JPG" width="207" /></a></div><br /><div style="text-align: right;"><b><span style="color: red;"><b><br />(GR9677 #04)
</b></span></b></div>
<b>
</b>
<br />
<div class="separator" style="clear: both; text-align: center;">
</div>
<b>Solution:</b>
<br />
<br />
<i style="text-align: center;">F</i><sub style="text-align: center;">electric </sub>= <i>k</i><i>q</i><i>Q</i>/<i>r</i>²<br />
<i style="text-align: center;">F</i><sub style="text-align: center;">centripetal</sub> = <i>mv</i>²<i>/r</i> = <i>m</i><i>ω</i>²<i>r</i><br />
<i><br /></i>
<i style="text-align: center;">F</i><sub style="text-align: center;">e </sub>= <i style="text-align: center;">F</i><sub style="text-align: center;">c</sub><br />
<i>k</i><i>q</i><i>Q</i>/<i>r</i>² = <i>m</i><i>ω</i>²<i>r</i><br />
<i>ω</i>² = <i>kqQ</i>/<i>m</i><i>r</i>³<br />
<br />
with<br />
<i>k </i>= 1/4<i>π</i><i style="text-align: center;">ɛ</i><sub style="text-align: center;">0</sub><br />
<i>r</i>² <i> = </i><i>R</i>² + <i>x</i>²<br />
<i style="text-align: justify;">R</i><span style="text-align: justify;"> </span><span style="text-align: left;">≫ </span><i style="text-align: justify;">x</i><br />
<i>r</i>² ∼<i> </i><i>R</i>² → <i>r</i>³ ∼<i> </i><i>R</i>³<br />
<br />
<i>ω</i> = √(<sup><i>qQ</i></sup>/<sub>4πɛ<sub>0</sub><i>mR</i><sup>3</sup></sub>)</div><div dir="ltr" style="text-align: left;" trbidi="on"><p>
<b>Answer: A<br /><br /><span style="color: red;">Notes:</span> </b>see problem <a href="http://physics-problems-solutions.blogspot.com/2014/10/electromagnetism-coulombs-law.html" target="_blank">GR9277 #65</a></p></div>
Unknownnoreply@blogger.com0tag:blogger.com,1999:blog-3004213209691817978.post-6283593081395911562016-03-05T18:24:00.004-08:002023-12-23T18:05:28.493-08:00Classical Mechanics - Circular Motion<div dir="ltr" style="text-align: left;" trbidi="on">
<div style="text-align: center;"><br /></div>
<div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEibd0WCx7vbScfFuDCLfUMYgkdW8PwZKauMLcyjBogH7k1bkw0fufZ86Cq45i8GwiIAisU87hm3ELWHHgKCl63r1pydw0yhvNLIQ24enOGY10alu-dHt_Se5h0vfujRB_toMhyphenhyphenAAvbDB8Yo0dyxxVgKPu-EcgcHOaKFxjlM-JXWo99HlE7vIKxR8lo_UqLp/s698/GR9677-05.JPG" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="435" data-original-width="698" height="199" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEibd0WCx7vbScfFuDCLfUMYgkdW8PwZKauMLcyjBogH7k1bkw0fufZ86Cq45i8GwiIAisU87hm3ELWHHgKCl63r1pydw0yhvNLIQ24enOGY10alu-dHt_Se5h0vfujRB_toMhyphenhyphenAAvbDB8Yo0dyxxVgKPu-EcgcHOaKFxjlM-JXWo99HlE7vIKxR8lo_UqLp/s320/GR9677-05.JPG" width="320" /></a></div><div style="text-align: justify;"><br /></div><div style="text-align: justify;">
A car travels with constant speed on a circular road on level ground. In the diagram above, <i style="text-align: left;">F</i><sub style="text-align: left;">air</sub> is the force of air resistance on the car. Which of the other force shown best represents the horizontal force of the road on the car's tires?</div>
<br />
A. <i>F</i><sub>A</sub><span style="text-align: justify;"> </span><br />
B. <i>F</i><sub>B</sub><br />
C. <i>F</i><sub>C</sub><br />
D. <i>F</i><sub>D</sub><br />
E. <i>F</i><sub>E</sub><br />
<div style="text-align: right;">
<b><span style="color: red;"><b>(GR9677 #05)
</b></span></b></div>
<b>
</b>
<br />
<div class="separator" style="clear: both; text-align: center;">
</div>
<b>Solution:</b>
<br />
<br />
<i>F</i><sub>A</sub> = Centripetal force<br />
<i>F</i><sub>C </sub>= Frictional force of the road, equals to force exerted by tires exert in the backward direction so that the car moves in the forward direction (Newton's third law Action-Reaction).<br />
<br />
The horizontal force on the car’s tires, <i>F</i><sub>A</sub> + <i>F</i><sub>C </sub>= <i>F</i><sub>B</sub><br />
<br />
<b>Answer: B</b></div>
Unknownnoreply@blogger.com0tag:blogger.com,1999:blog-3004213209691817978.post-59215282492173139782016-03-05T18:24:00.002-08:002023-12-23T18:13:40.119-08:00Classical Mechanics - Friction Force<div dir="ltr" style="text-align: left;" trbidi="on">
<div style="text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhdVPGFlRXKRdUNHUEtIkTBNxGulRefYREf6XozZIh1tXQx7nuwxsN4PrAiXWdyxVBojjW7keSP3N2XJXgXk4eDQkvY_Df-VGTka24q1M1B1d_we3kPTeQNwIcfb1e0DK9R3dsRH-WYUp7GQZHQI7CREiuUZQQH7CxRxDGVRdXo30YpyX8bAkamoMLAS2Eg/s380/GR9677-06.JPG" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="288" data-original-width="380" height="243" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhdVPGFlRXKRdUNHUEtIkTBNxGulRefYREf6XozZIh1tXQx7nuwxsN4PrAiXWdyxVBojjW7keSP3N2XJXgXk4eDQkvY_Df-VGTka24q1M1B1d_we3kPTeQNwIcfb1e0DK9R3dsRH-WYUp7GQZHQI7CREiuUZQQH7CxRxDGVRdXo30YpyX8bAkamoMLAS2Eg/s320/GR9677-06.JPG" width="320" /></a></div><br /><div style="text-align: justify;"><br /></div><div style="text-align: justify;">
A block of mass <i>m </i>sliding down an incline at constant speed is initially at height <i>h </i>above the ground as shown in the figure. The coefficient of kinetic friction between the mass and the incline is <i>µ</i>. If the mass continues to side down the incline at a constant speed, how much energy is dissipated by friction by the time the mass reached the bottom of the incline?
</div>
<br />
A. <i>mgh/µ
</i><br />
B. <i>mgh
</i><br />
C. <i>µmgh/</i>sin <i>θ </i><br />
D. <i>mgh</i> sin <i>θ </i><br />
E. 0<br />
<div style="text-align: right;">
<b><span style="color: red;"><b>(GR9677 #06)
</b></span></b></div>
<b>
</b>
<br />
<div class="separator" style="clear: both; text-align: center;">
</div>
<b>Solution:</b>
<br />
<br />
<i>mg</i> sin <i>θ </i>− <i>F</i><sub><i>r </i></sub>= <i>ma</i><br />
<span style="text-align: justify;">At a constant speed, <i>a</i> = 0</span><br />
<span style="text-align: justify;"><i style="text-align: left;">F</i><sub style="text-align: left;"><i>r </i></sub><span style="text-align: left;">= </span><i style="text-align: left;">mg</i><span style="text-align: left;"> sin </span><i style="text-align: left;">θ </i></span><br />
<span style="text-align: justify;"><br /></span>
<span style="text-align: justify;">Energy dissipated = Work done by </span>the frictional force<br />
<i>W</i> = <i>F</i><sub><i>r </i></sub>⋅ <i>s</i><br />
<i><br /></i>
<i>s = </i>length of the inclined surface = <i>h</i>/sin <i>θ</i><br />
<i><br /></i>
<i>W</i> = <i>mg</i> sin <i>θ </i>(<i>h</i>/sin <i>θ</i>) = <i>mgh</i><br />
<br />
<b>Answer: B</b><br />
<br /></div>
Unknownnoreply@blogger.com0tag:blogger.com,1999:blog-3004213209691817978.post-31462223668102399552016-03-05T17:49:00.000-08:002023-12-23T18:12:54.135-08:00Classical Mechanics - Linear Momentum<div dir="ltr" style="text-align: left;" trbidi="on">
<div style="text-align: center;"><br /></div><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEioyAdm2lD1JAGLzaJeyojmFXxng42o-Z8TJoX_iIrt6FsNks1JsayY4m7n_NMLzLvV_xARWcVkiPY001DEBK9sjQlDSRJH8NIDec7CqVRTnKTLbj74WeLo-tAuwKOACWe0-Cm9PqX4PiE5FNL8gQuqmeCnNb0zcYIwtIvgba-bQDZ_cCM039JgsZnCxt0E/s458/GR9677-07.JPG" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="367" data-original-width="458" height="256" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEioyAdm2lD1JAGLzaJeyojmFXxng42o-Z8TJoX_iIrt6FsNks1JsayY4m7n_NMLzLvV_xARWcVkiPY001DEBK9sjQlDSRJH8NIDec7CqVRTnKTLbj74WeLo-tAuwKOACWe0-Cm9PqX4PiE5FNL8gQuqmeCnNb0zcYIwtIvgba-bQDZ_cCM039JgsZnCxt0E/s320/GR9677-07.JPG" width="320" /></a></div><br /><div style="text-align: center;"><span style="text-align: justify;"><br /></span></div>
<p style="text-align: justify;">
As shown in the picture, a ball of mass <i>m</i> suspended on the end of a wire, is released from height <i>h</i> and collides elastically, when it is at its lowest point, with a block of mass 2<i>m</i> at rest on a frictionless surface. After the collision, the ball rises to a final height equal to
</p><p>
A. 1/9 <i style="text-align: justify;">h</i><br />
B. 1/8 <i style="text-align: justify;">h</i><br />
C. 1/3 <i style="text-align: justify;">h</i><br />
D. 1/2 <i style="text-align: justify;">h</i><br />
E. 2/3 <i style="text-align: justify;">h</i><br />
</p><div style="text-align: right;">
<b><span style="color: red;"><b>(GR9677 #07)
</b></span></b></div>
<b>
</b>
<br />
<div class="separator" style="clear: both; text-align: center;">
</div>
<b>Solution:</b><br />
<br />
Conservation of momentum of the system:<br />
<br />
<i>m</i><sub><i>a</i></sub><i>v</i><sub><i>a </i></sub>+ <i>m</i><sub><i>b</i></sub><i>v</i><sub><i>b</i></sub> = <i>m</i><sub><i>a</i></sub><i>v</i><sub><i>a</i></sub><i>' </i>+ <i>m</i><sub><i>b</i></sub><i>v</i><sub><i>b</i></sub><i>' </i><br />
<br />
Given:<br />
<i>m</i><sub><i>a </i></sub>= <i>m</i><br />
<i>m</i><sub><i>b </i></sub>= 2<i>m</i><br />
<i>v</i><sub><i>b</i></sub> = 0<br />
<i><br /></i>
<i>m</i><i>v</i><sub><i>a </i></sub>+ 0 = <i>m</i><i>v</i><sub><i>a</i></sub><i>' </i>+ 2<i>m</i><i>v</i><sub><i>b</i></sub><i>' </i><sub><i> </i></sub><br />
<i>v</i><sub><i>a </i></sub> = <i>v</i><sub><i>a</i></sub><i>' </i>+ 2<i>v</i><sub><i>b</i></sub><i>' </i><sub><i> </i></sub><sub><i> </i></sub>(Eq.1)<br />
<div>
<br /></div>
Conservation of kinetic energy of the system:<br />
<br />
½ <i>m</i><sub><i>a</i></sub><i>v</i><sub><i>a</i></sub>² + ½ <i>m</i><sub><i>b</i></sub><i>v</i><sub><i>b</i></sub>² = ½ <i>m</i><sub><i>a</i></sub><i>v</i><sub><i>a</i></sub><i>'</i>² + ½ <i>m</i><sub><i>b</i></sub><i>v</i><sub><i>b</i></sub><i>'</i>²<br />
<i>m</i><i>v</i><sub><i>a</i></sub>² + 0 = <i>m</i><i>v</i><sub><i>a</i></sub><i>'</i>² + 2<i>m</i><i>v</i><sub><i>b</i></sub><i>'</i>²<br />
<i>v</i><sub><i>a</i></sub>² = <i>v</i><sub><i>a</i></sub><i>'</i>² + 2<i>v</i><sub><i>b</i></sub><i>'</i>² (Eq.2)<br />
<br />
(Eq.1) → (Eq.2)<br />
(<i>v</i><sub><i>a</i></sub><i>' </i>+ 2<i>v</i><sub><i>b</i></sub><i>'</i>)² = <i>v</i><sub><i>a</i></sub><i>'</i>² + 2<i>v</i><sub><i>b</i></sub><i>'</i>²<br />
<i>v</i><sub><i>a</i></sub><i>'</i>² + 4<i>v</i><sub><i>a</i></sub><i>'</i><i>v</i><sub><i>b</i></sub><i>' </i>+ 4<i>v</i><sub><i>b</i></sub><i>'</i>² = <i>v</i><sub><i>a</i></sub><i>'</i>² + 2<i>v</i><sub><i>b</i></sub><i>'</i>²<br />
4<i>v</i><sub><i>a</i></sub><i>'</i><i>v</i><sub><i>b</i></sub><i>' </i><sub><i> </i></sub>= 2<i>v</i><sub><i>b</i></sub><i>'</i>² − 4<i>v</i><sub><i>b</i></sub><i>'</i>²<br />
2<i>v</i><sub><i>a</i></sub><i>'</i><sub><i> </i></sub>= − <i>v</i><sub><i>b</i></sub><i>' </i><br />
<i>v</i><sub><i>b</i></sub><i>' </i><sub><i> </i></sub>= −2<i>v</i><sub><i>a</i></sub><i>'</i><i> </i>(Eq.3)<br />
<br />
(Eq.3) → <sub><i> </i></sub>(Eq.1)<br />
<i>v</i><sub><i>a </i></sub> = <i>v</i><sub><i>a</i></sub><i>' </i>+ 2<i>v</i><sub><i>b</i></sub><i>' </i><br />
<i>v</i><sub><i>a </i></sub> = <i>v</i><sub><i>a</i></sub><i>' </i>+ 2(−2<i>v</i><sub><i>a</i></sub><i>'</i>)<br />
<i>v</i><sub><i>a </i></sub> = − 3<i>v</i><sub><i>a</i></sub><i>' </i>(Eq. 4)<br />
<i><br /></i>
For the pendulum, conservation of energy:<br />
<br />
at the moment of the collision<br />
<i>U<span style="font-size: 13.3333px;"> </span></i>= <i>T</i> → <i>m</i><sub><i>a</i></sub><i>gh</i> = ½ <i>m</i><sub><i>a</i></sub><i>v</i><sub><i>a</i></sub>² → <i>v</i><sub><i>a</i></sub><i> </i>= (2<i>gh</i>)<sup>½</sup><br />
<br />
after the collision<br />
<i>U'</i><sub><i> </i></sub>= <i>T<span style="font-size: 13.3333px;">' </span></i>→ <i>m</i><sub><i>a</i></sub><i>gh'</i> = ½ <i>m</i><sub><i>a</i></sub><i>v</i><sub><i>a</i></sub><i>'</i>² → <i>v</i><sub><i>a</i></sub><i>' </i>= (2<i>gh'</i>)<sup>½</sup><br />
<br />
Thus, (Eq. 4):<br />
<i>v</i><sub><i>a </i></sub> = − 3<i>v</i><sub><i>a</i></sub><i>' </i><br />
(2<i>gh</i>)<sup>½</sup><sub><i> </i></sub> = − 3(2<i>gh'</i>)<sup>½</sup><i> </i><br />
[(2<i>gh</i>)<sup>½</sup>]² = [− 3(2<i>gh'</i>)<sup>½</sup>]²<br />
<i><br /></i>
<i>h</i><sub><i> </i></sub> = 9<i>h'</i><br />
<i>h' </i>= ¹⁄₉<i>h</i><sub><i> </i></sub><br />
<b><br /></b>
<b>Answer: A</b><br />
<br /><p></p><p></p></div>
Unknownnoreply@blogger.com0tag:blogger.com,1999:blog-3004213209691817978.post-16698646837446684262016-03-05T17:47:00.002-08:002016-03-05T17:47:55.539-08:00Classical Mechanics - Harmonic Oscillator<div dir="ltr" style="text-align: left;" trbidi="on">
<div style="text-align: justify;">
A particle of mass <i>m</i> undergoes harmonic oscillation with period <span style="text-align: left;"><i>T</i></span><sub style="text-align: left;">0</sub>. A force <i>f</i> proportional to the speed <i>v</i> of the particle, <i>f</i> = <span style="text-align: left;">−</span><i>bv</i>, is introduced. If the particle continues to oscillate, the period with <i>f</i> acting is</div>
<br />
A. Larger than <i>T</i><sub>0</sub><br />
B. Smaller than <i>T</i><sub>0</sub><br />
C. Independent of <i>b
</i><br />
D. Dependent linearly on <i>b
</i><br />
E. Constantly changing<br />
<div style="text-align: right;">
<b><span style="color: red;"><b>(GR9677 #08)
</b></span></b></div>
<b>
</b>
<br />
<div class="separator" style="clear: both; text-align: center;">
</div>
<b>Solution:</b>
<br />
<br />
<span style="text-align: justify;"> </span><i style="text-align: justify;">f</i><span style="text-align: justify;"> = </span><span style="text-align: left;">−</span><i style="text-align: justify;">bv </i>→ minus sign means <i style="text-align: justify;">f</i><span style="text-align: justify;"> </span>is restoring force (damped oscillation).<br />
The oscillation is getting slower (larger period) before it finally comes to stop.<br />
<br />
<b>Answer: A</b><br />
<t_0 br="">
<br />
</t_0></div>
Unknownnoreply@blogger.com0tag:blogger.com,1999:blog-3004213209691817978.post-9017945853955951862016-03-05T17:47:00.001-08:002016-03-05T17:47:11.166-08:00Nuclear & Particle Physics - Hydrogen Spectrum<div dir="ltr" style="text-align: left;" trbidi="on">
<div style="text-align: justify;">
In the spectrum of Hydrogen, what is the ratio of the longest wavelength in the Lyman series (<span style="text-align: left;"><i>n</i></span><sub style="text-align: left;"><i>f </i> </sub>= 1) to the longest wavelength in the Balmer series (<span style="text-align: left;"><i>n</i></span><sub style="text-align: left;"><i>f </i></sub> = 2)?</div>
<br />
A. 5/27 <br />
B. 1/3 <br />
C. 4/9 <br />
D. 3/2 <br />
E. 3
<br />
<div style="text-align: right;">
<b><span style="color: red;"><b>(GR9677 #09)
</b></span></b></div>
<b>
</b>
<br />
<div class="separator" style="clear: both; text-align: center;">
</div>
<b>Solution:</b>
<br />
<br />
Rydberg formula:<br />
<br />
<a href="http://www.codecogs.com/eqnedit.php?latex=\frac{1}{\lambda_{vac}}=R_H\bigg(\frac{1}{n^2_1}-\frac{1}{n^2_2}\bigg)" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\frac{1}{\lambda_{vac}}=R_H\bigg(\frac{1}{n^2_1}-\frac{1}{n^2_2}\bigg)" title="\frac{1}{\lambda_{vac}}=R_H\bigg(\frac{1}{n^2_1}-\frac{1}{n^2_2}\bigg)" /></a><br />
<br />
<i>λ</i><sub>vac</sub> = the wavelength of the light emitted in vacuum<br />
<i>R</i><sub>H</sub> = Rydberg constant for Hydrogen<br />
<i>n</i><sub>1</sub> and <i>n</i><sub>2</sub> are integers such that <i>n</i><sub>1 </sub><img src="https://blogger.googleusercontent.com/img/proxy/AVvXsEiFDjnIc6isxO2l4Vk2OY2qurRhUNli3DML-DziNn3t-c6SewQ7mY_bDNCNo8onzNhwMESS0Se04GMRAx_uhCPcCMMzSLOvKvvsyH0_HaDmMcwJG7drogL1JXwVOIrkF9oA9ojnpQ=" /> <i>n</i><sub>2</sub><br />
<br />
<span style="text-align: justify;">For the longest wavelength take </span><i>n</i><sub>2</sub> = ∞<br />
<span style="text-align: justify;"><br /></span>
For Lyman-radiation (<i>n</i><sub>2</sub> = ∞ → <i>n</i><sub>1 </sub>= 1):<br />
<br />
1/<i>λ</i><sub>L</sub> = <i>R</i><sub>H</sub> (1/1 − 0) = <i>R</i><sub>H</sub><br />
<br />
For Balmer-radiation (<i>n</i><sub>2</sub> = ∞ → <i>n</i><sub>1</sub> = 2):<br />
<br />
1/<i>λ</i><sub>B</sub> = <i>R</i><sub>H</sub> (¼ − 0) = ¼<i>R</i><sub>H</sub><br />
<br />
The Ratio:<br />
<br />
<sup><i>λ</i><sub>L</sub></sup>/<sub><i>λ</i><sub>B<i> </i></sub></sub>= (<sup>1</sup>/<sub><i>R</i><sub><i>H</i></sub></sub>)(<sup><i>R</i><sub><i>H</i></sub></sup>/<sub>4</sub>) = ¼ ≈ 5/27<br />
<br />
<b>Answer: A</b></div>
<!-- Blogger automated replacement: "https://images-blogger-opensocial.googleusercontent.com/gadgets/proxy?url=http%3A%2F%2Flatex.codecogs.com%2Fgif.latex%3F%3C&container=blogger&gadget=a&rewriteMime=image%2F*" with "https://blogger.googleusercontent.com/img/proxy/AVvXsEiFDjnIc6isxO2l4Vk2OY2qurRhUNli3DML-DziNn3t-c6SewQ7mY_bDNCNo8onzNhwMESS0Se04GMRAx_uhCPcCMMzSLOvKvvsyH0_HaDmMcwJG7drogL1JXwVOIrkF9oA9ojnpQ=" -->Unknownnoreply@blogger.com0tag:blogger.com,1999:blog-3004213209691817978.post-52405213945669019902016-03-05T17:24:00.000-08:002016-03-07T10:32:11.064-08:00Nuclear & Particle Physics - Internal Conversion<div dir="ltr" style="text-align: left;" trbidi="on">
<div style="text-align: justify;">
Internal conversion is the process whereby an excited nucleus transfers its energy directly to one of the most tightly bound atomic electrons, causing the electron to be ejected from the atom and leaving the atom in an excited state. The most probable process after an internal conversion electron is ejected from an atom with a high atomic number is that the
</div>
<ol type="A">
<li>atom returns to its ground state through inelastic collisions with either atoms </li>
<li>atom emits one or several X-rays </li>
<li>nucleus emits a gamma-ray </li>
<li>nucleus emits an electron </li>
<li>nucleus emits a positron </li>
</ol>
<div style="text-align: right;">
<span style="color: red;"><b>(GR9677 #10)
</b></span></div>
<div class="separator" style="clear: both; text-align: center;">
</div>
<b>Solution:</b>
<br />
<br />
Electron transitions in atom (internal conversion) = X-ray production<br />
→ An orbital electron is absorbed and ejected along with an X-ray<br />
<br />
compared to:<br />
<br />
Nuclear transitions = Gamma, <i>γ</i> Ray production<br />
<div>
→<i> </i>The excited nucleus jumps to a lower level and emits a photon <i>γ</i></div>
<div>
<br /></div>
<b>Answer: B</b><br />
<b><br /></b>
<b><span style="color: red;">Note: </span></b><br />
<br />
1. (C), (D), and (E) are products of radioactive decay which are results of unstable nuclei.<br />
<br />
2. In internal conversion:<br />
<ul>
<li>For low atomic number, it will produce the Auger effect and ionize the outside electron.</li>
<li>For high atomic number, it will only emit X-rays.</li>
</ul>
</div>
Unknownnoreply@blogger.com0tag:blogger.com,1999:blog-3004213209691817978.post-25961379129991386452016-03-05T17:23:00.001-08:002016-03-05T17:23:55.644-08:00Nuclear & Particle Physics - Stern-Gerlach Experiment<div dir="ltr" style="text-align: left;" trbidi="on">
<div style="text-align: justify;">
A beam of neutral hydrogen atoms in their ground state is moving into the plane of this page and passes through a region of a strong inhomogeneous magnetic field that is directed upward in the plane of the page. After the beam passes through this field, a detector would find that it has been</div>
<br />
A. deflected upward
<br />
B. deflected to the right
<br />
C. undeviated
<br />
D. split vertically into two beams
<br />
E. split horizontally into three beams
<br />
<div style="text-align: right;">
<span style="color: red;"><b>(GR9677 #11)
</b></span></div>
<div class="separator" style="clear: both; text-align: center;">
</div>
<b>Solution:</b>
<br />
<br />
The Stern–Gerlach experiment → spin discovery<br />
<br />
A beam of neutral atom passes through inhomogeneous magnetic field will split vertically into 2 beams representing spin-up and spin-down electrons.<br />
<br />
<b>Answer: D</b><br />
<br /></div>
Unknownnoreply@blogger.com0tag:blogger.com,1999:blog-3004213209691817978.post-76173346943192283572016-03-05T16:39:00.000-08:002016-03-05T16:39:23.063-08:00Nuclear & Particle Physics - Positronium<div dir="ltr" style="text-align: left;" trbidi="on">
The ground-state energy of positronium is most nearly equal to<br />
<br />
A. − 27.2 eV <br />
B. − 13.6 eV <br />
C. − 6.8 eV
<br />
D. − 3.4 eV
<br />
E. 13.6 eV
<br />
<div style="text-align: right;">
<span style="color: red;"><b>(GR9677 #12)
</b></span></div>
<div class="separator" style="clear: both; text-align: center;">
</div>
<b>Solution:</b>
<br />
<br />
Energy levels of Positronium is half those of Hydrogen (See <a href="http://physics-problems-solutions.blogspot.com/2014/02/nuclear-particle-physics-positronium.html" target="_blank">GR8677 #99</a>)<br />
<br />
<i style="text-align: justify;">E</i><sub><i>n</i>(H) </sub><span style="text-align: justify;">=</span><sub> </sub>− 13.6 / <i>n</i>²<br />
<i style="text-align: justify;">E</i><sub><i>n</i>(Ps) </sub><span style="text-align: justify;">= </span>½ <i style="text-align: justify;">E</i><sub><i>n</i>(H) </sub><br />
<sub><br /></sub>
<span style="text-align: justify;">For the ground-state </span><i>n </i><span style="text-align: justify;">= </span><span style="text-align: justify;">1 </span>→ <i style="text-align: justify;">E</i><sub>Ps </sub><span style="text-align: justify;">= </span>− ½ × 13.6 eV<sup> </sup><span style="text-align: justify;">= </span>− 6.8 eV<br />
<br />
<b>Answer: C</b></div>
Unknownnoreply@blogger.com0tag:blogger.com,1999:blog-3004213209691817978.post-61927792721309457632016-03-05T16:35:00.000-08:002016-03-05T20:04:50.651-08:00Thermal Physics - Power<div dir="ltr" style="text-align: left;" trbidi="on">
<div style="text-align: justify;">
A 100-watt electric heater element is placed in a pan containing one liter of water. Although the heating element is on for a long time, the water, though close to boiling does not boil. When the heating element is removed, approximately how long will it take the water to cool by 1 degree Celsius? (Assume that the specific heat for water is 4.2 kJ/kg<sup style="text-align: left;">o</sup>C)</div>
<br />
A. 20 s
<br />
B. 40 s
<br />
C. 60 s
<br />
D. 130 s
<br />
E. 200 s
<br />
<div style="text-align: right;">
<span style="color: red;"><b>(GR9677 #13)
</b></span></div>
<div class="separator" style="clear: both; text-align: center;">
</div>
<b>Solution:</b>
<br />
<br />
<i>P</i> = 100 W<br />
<i>V</i> = 1 L = 1 m<sup>3</sup> → <i>m </i>= 1 kg (STP)<br />
Δ<i>T </i>= <span style="text-align: justify;">1</span><sup>o</sup><span style="text-align: justify;">C</span><br />
<i>c = </i><span style="text-align: justify;">4.2 kJ/kg</span><sup style="text-align: left;">o</sup><span style="text-align: justify;">C</span><br />
<span style="text-align: justify;"><br /></span>
<i>Q</i> = <i>mc</i>Δ<i>T </i>= <i>Pt </i><br />
1 × 4200 × 1 = 100<i>t</i><br />
<i>t</i> = 42 s<br />
<br />
<b>Answer: B</b><br />
<br /></div>
Unknownnoreply@blogger.com0tag:blogger.com,1999:blog-3004213209691817978.post-81602225076162095962016-03-05T16:15:00.000-08:002016-07-01T10:02:16.809-07:00Thermal Physics - Specific Heat<div dir="ltr" style="text-align: left;" trbidi="on">
<div style="text-align: justify;">
Two identical 1 kg blocks of copper metal one initially at a temperature <span style="text-align: left;"><i>T</i></span><sub style="text-align: left;">1 </sub><span style="text-align: left;">= 0</span><sup style="text-align: left;">o</sup>C and the other initially at a temperature <i style="text-align: left;">T</i><sub style="text-align: left;">2 </sub><span style="text-align: left;">= 100</span><sup style="text-align: left;">o</sup>C are enclosed in a perfectly insulating container. The two blocks are initially separated. When the blocks are placed in contact, they come to equilibrium at a final temperature <i style="text-align: left;">T</i><sub style="text-align: left;"><i>f</i></sub>. The amount of heat exchanged between the two blocks in this process is equal to which of the following? (The specific heat of copper metal is equal to 0.1 kilocalorie/kg<sup style="text-align: left;">o</sup>K)</div>
<br />
A. 50 Kcal <br />
B. 25 Kcal <br />
C. 10 Kcal
<br />
D. 5 Kcal
<br />
E. 1 Kcal<br />
<div style="text-align: right;">
<span style="color: red;"><b>(GR9677 #14)
</b></span></div>
<div class="separator" style="clear: both; text-align: center;">
</div>
<b>Solution:</b>
<br />
<br />
<i>Q </i>= <i>mc</i>Δ<i>T</i><br />
|<i>Q</i>|<sub style="text-align: left;">gain<i> </i></sub>= |<i>Q</i>|<sub style="text-align: left;">lost<i> </i></sub><br />
<i>m</i><sub style="text-align: left;">1</sub><i>c</i><sub style="text-align: left;">1</sub>Δ<i>T</i><sub style="text-align: left;">1</sub> = <i>m</i><sub style="text-align: left;">2</sub><i>c</i><sub style="text-align: left;">2</sub>Δ<i>T</i><sub style="text-align: left;">2</sub><br />
<br />
<i>m</i><sub>1</sub> = <i>m</i><sub>2 </sub>= 1 kg<br />
<i>c</i><sub>1 </sub>= <i>c</i><sub>2 </sub>= <i>c</i><sub>copper </sub>= 0.1 kcal/kg<sup>o</sup>K<br />
<div style="text-align: justify;">
<span style="text-align: left;"><i>T</i></span><sub style="text-align: left;">1 </sub><span style="text-align: left;">= 0</span><sup style="text-align: left;">o</sup>C = 273<sup style="text-align: left;">o</sup>K</div>
<div style="text-align: justify;">
<i style="text-align: left;">T</i><sub style="text-align: left;">2 </sub><span style="text-align: left;">= 100</span><sup style="text-align: left;">o</sup>C = 373<sup style="text-align: left;">o</sup>K</div>
<sub style="text-align: left;"><i><br /></i></sub>Δ<i>T</i><sub style="text-align: left;">1</sub> = Δ<i>T</i><sub style="text-align: left;">2</sub><br />
<i style="text-align: left;">T</i><sub style="text-align: left;"><i>f </i></sub>− <i style="text-align: left;">T</i><sub style="text-align: left;">1 </sub>= <i style="text-align: justify;">T</i><sub style="text-align: justify;">2 </sub>− <i style="text-align: left;">T</i><sub style="text-align: left;"><i>f </i></sub><br />
<i style="text-align: left;">T</i><sub style="text-align: left;"><i>f </i></sub><sub style="text-align: left;"> </sub>= (<i style="text-align: justify;">T</i><sub style="text-align: justify;">2 </sub>+ <i style="text-align: left;">T</i><sub style="text-align: left;">1</sub>)/2 = (100<sub style="text-align: justify;"> </sub>+ 0)/2 = 50<sup style="text-align: justify;">o</sup><span style="text-align: justify;">C </span><span style="text-align: justify;">= 323</span><sup style="text-align: justify;">o</sup><span style="text-align: justify;">K</span><br />
<span style="text-align: justify;"><br /></span>|<i>Q</i>|<sub style="text-align: left;">gain<i> </i></sub>= |<i>Q</i>|<sub style="text-align: left;">lost<i> </i></sub> = <i>m</i><sub style="text-align: left;">1</sub><i>c</i><sub style="text-align: left;">1</sub>Δ<i>T</i><sub style="text-align: left;">1</sub> = 1 × <span style="text-align: justify;">0.1 </span>× (<span style="text-align: justify;">323 </span>− <span style="text-align: justify;">273</span>) = 5kcal<br />
<br />
<b>Answer: D</b><br />
<br /></div>
Unknownnoreply@blogger.com0tag:blogger.com,1999:blog-3004213209691817978.post-81361686723149009712016-03-05T16:12:00.001-08:002023-12-23T18:08:21.494-08:00Thermal Physics - Isothermal<div dir="ltr" style="text-align: left;" trbidi="on">
<div style="text-align: center;"><br /></div>
<div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjrJwrFwVQr-0BS7zX1h4whGGJBuuPEg0DYMbkW3G0FcCuG4Ak3od0aWddSWAAA1G72tQjwtS7F4vuBIId-Ftsb_1nDVpO2SQYrBgaeSqEBIDlvb10bb0cDwKFICHI15YW0hTvzgRCTvANWpsOSRpom_EvehD8dGAn3Pix0Qls0QVOaEDpBSlmeTjDhrgy8/s863/GR9677-15.JPG" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="716" data-original-width="863" height="265" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjrJwrFwVQr-0BS7zX1h4whGGJBuuPEg0DYMbkW3G0FcCuG4Ak3od0aWddSWAAA1G72tQjwtS7F4vuBIId-Ftsb_1nDVpO2SQYrBgaeSqEBIDlvb10bb0cDwKFICHI15YW0hTvzgRCTvANWpsOSRpom_EvehD8dGAn3Pix0Qls0QVOaEDpBSlmeTjDhrgy8/s320/GR9677-15.JPG" width="320" /></a></div><br /><div style="text-align: justify;">Suppose one mole of an ideal gas undergoes the reversible cycle ABCA shown in the P-V diagram above, where AB is an isotherm. The molar heat capacities are <i style="text-align: left;">C</i><sub style="text-align: left;"><i>p</i></sub> at constant pressure and <i style="text-align: left;">C</i><sub style="text-align: left;"><i>v</i></sub> at constant volume. The net heat added to the gas during the cycle is equal to</div>
<br />
A. <i>RT</i><sub><i>h</i></sub> (<i>V</i><sub>2</sub>/<i>V</i><sub>1</sub>)<br />
B. −<i>C</i><sub><i>p</i></sub>(<i>T</i><sub><i>h</i></sub> − <i>T</i><sub><i>c</i></sub>)
<br />
C. <i>C</i><sub><i>p</i></sub>(<i>T</i><sub><i>h</i></sub> − <i>T</i><sub><i>c</i></sub>)
<br />
D. <i>RT</i><sub><i>h</i></sub> ln (<i>V</i><sub>2</sub>/<i>V</i><sub>1</sub>) − <i>C</i><sub><i>p</i></sub>(<i>T</i><sub><i>h</i></sub> − <i>T</i><sub><i>c</i></sub>)<br />
E. <i>RT</i><sub><i>h</i></sub> ln (<i>V</i><sub>2</sub>/<i>V</i><sub>1</sub>) − <i>R</i>(<i>T</i><sub><i>h</i></sub> − <i>T</i><sub><i>c</i></sub>)<br />
<div style="text-align: right;">
<span style="color: red;"><b>(GR9677 #15)
</b></span></div>
<div class="separator" style="clear: both; text-align: center;">
</div>
<b>Solution:</b>
<br />
<br />
<div style="text-align: justify;">
AB Isotherm <span style="text-align: left;">→ </span><i style="text-align: left;">T</i><span style="text-align: left;"> </span><span style="text-align: left;">Constant = </span><i style="text-align: left;">T</i><sub style="text-align: left;"><i>h</i></sub></div>
<br />
Ideal Gas: <i>PV</i> = <i>nRT </i><br />
<i>P </i>= <i>nRT / V</i><br />
<i><br /></i>
<i>n </i>= 1 mole,<br />
<br />
<i>W</i><sub>AB</sub> = <sub><sub><i>V</i><sub>1</sub></sub></sub><span style="font-size: small;"><b>∫</b></span><sup><sup><i>V</i><sub>2 </sub></sup></sup><i>P dV</i> = <i>RT</i><sub><i>h</i></sub><i> </i><sub><sub><i>V</i><sub>1</sub></sub></sub><span style="font-size: small;"><b>∫</b></span><sup><sup><i>V</i><sub>2 </sub></sup></sup>(1<i>/V</i>)<i> dV </i> =<i> </i><i>RT</i><sub><i>h</i></sub> ln (<i>V</i><sub>2</sub>/<i>V</i><sub>1</sub>)<br />
<div style="text-align: justify;">
<br /></div>
<div style="text-align: justify;">
BC Isobaric <span style="text-align: left;">→ </span><i style="text-align: left;">P</i><span style="text-align: left;"> c</span><span style="text-align: left;">onstant </span><span style="text-align: left;">=</span><span style="text-align: left;"> </span><i style="text-align: left;">P</i><sub style="text-align: left;">2</sub></div>
<div style="text-align: justify;">
<span style="text-align: left;"><br /></span></div>
<div style="text-align: justify;">
<i style="text-align: left;">W</i><sub style="text-align: left;">BC</sub><span style="text-align: left;"> = </span><sub style="text-align: left;"><sub><i>V</i><sub>2</sub></sub></sub><span style="font-size: small; text-align: left;"><b>∫</b></span><sup style="text-align: left;"><sup><i>V</i><sub>1 </sub></sup></sup><i style="text-align: left;">P dV </i><span style="text-align: left;">=</span><span style="text-align: left;"> </span><i style="text-align: left;">P</i><sub style="text-align: left;">2</sub><span style="text-align: left;"> </span><span style="text-align: left;">(</span><i style="text-align: left;">V</i><sub style="text-align: left;">1</sub><span style="text-align: left;">− </span><i style="text-align: left;">V</i><sub style="text-align: left;">2</sub><span style="text-align: left;">) </span><span style="text-align: left;">= </span><sub style="text-align: left;"> </sub><i style="text-align: left;">P</i><sub style="text-align: left;">2</sub><i style="text-align: left;">V</i><sub style="text-align: left;">1</sub><span style="text-align: left;"> </span><span style="text-align: left;">−</span><sub style="text-align: left;"> </sub><span style="text-align: left;"> </span><i style="text-align: left;">P</i><sub style="text-align: left;">2</sub><i style="text-align: left;">V</i><sub style="text-align: left;">2</sub></div>
<div style="text-align: justify;">
<span style="text-align: left;"><br /></span></div>
<div style="text-align: justify;">
<span style="text-align: left;">From the diagram:</span></div>
<div style="text-align: justify;">
<i style="text-align: left;">P</i><sub style="text-align: left;">2</sub><i style="text-align: left;">V</i><sub style="text-align: left;">1</sub><span style="text-align: left;"> = </span><i style="text-align: left;">nR</i><i style="text-align: left;">T</i><sub style="text-align: left;"><i>c</i></sub><i style="text-align: left;"> </i></div>
<div style="text-align: justify;">
<i style="text-align: left;">P</i><sub style="text-align: left;">2</sub><i style="text-align: left;">V</i><sub style="text-align: left;">2</sub><span style="text-align: left;"> = </span><i style="text-align: left;">nR</i><i style="text-align: left;">T</i><sub style="text-align: left;"><i>h</i></sub></div>
<div style="text-align: justify;">
<i style="text-align: left;"><br /></i></div>
<div style="text-align: justify;">
<i style="text-align: left;">W</i><sub style="text-align: left;">BC</sub><span style="text-align: left;"> =</span><span style="text-align: left;"> </span><i style="text-align: left;">nR</i><span style="text-align: left;">(</span><i style="text-align: left;">T</i><sub style="text-align: left;"><i>c</i></sub><span style="text-align: left;"> − </span><i style="text-align: left;">T</i><sub style="text-align: left;"><i>h</i></sub><span style="text-align: left;">) </span><span style="text-align: left;"> = </span><i style="text-align: left;">R</i><span style="text-align: left;">(</span><i style="text-align: left;">T</i><sub style="text-align: left;"><i>c</i></sub><span style="text-align: left;"> − </span><i style="text-align: left;">T</i><sub style="text-align: left;"><i>h</i></sub><span style="text-align: left;">)</span></div>
<br />
<i>W</i><sub>CA</sub> = 0 since <i>V</i> constant<br />
<br />
Total <i>W </i>= <i>W</i><sub>AB </sub>+ <i>W</i><sub>BC</sub> = <i>RT</i><sub><i>h</i></sub> ln (<i>V</i><sub>2</sub>/<i>V</i><sub>1</sub>) + <i>R</i>(<i>T</i><sub><i>c</i></sub> − <i>T</i><sub><i>h</i></sub>)<br />
<br />
or<br />
<br />
<i>W </i>= <i>RT</i><sub><i>h</i></sub> ln (<i>V</i><sub>2</sub>/<i>V</i><sub>1</sub>) − <i>R</i>(<i>T</i><sub><i>h</i></sub> − <i>T</i><sub><i>c</i></sub>)<br />
<br />
<b>Answer: E</b><br />
<br /></div>
Unknownnoreply@blogger.com0tag:blogger.com,1999:blog-3004213209691817978.post-71755480026788503252016-03-05T16:08:00.001-08:002016-03-05T16:14:32.812-08:00Thermal Physics - Mean Free Path<div dir="ltr" style="text-align: left;" trbidi="on">
<div style="text-align: justify;">
The mean free path for the molecules of a gas is approximately given by 1/<i style="text-align: left;">η</i><i style="text-align: left;">σ</i>, where <i style="text-align: left;">η</i> is the number density and <i style="text-align: left;">σ </i>is the collision cross section. The mean free path for air molecules at room conditions is approximately</div>
<br />
A. 10<sup>−4</sup> m<br />
B. 10<sup>−7</sup> m
<br />
C. 10<sup>−10</sup> m
<br />
D. 10<sup>−13</sup> m
<br />
E. 10<sup>−16</sup> m
<br />
<div style="text-align: right;">
<span style="color: red;"><b>(GR9677 #16)
</b></span></div>
<div class="separator" style="clear: both; text-align: center;">
</div>
<b>Solution:</b>
<br />
<br />
<span style="text-align: justify;">Mean free path = </span><span style="text-align: justify;">1/</span><i style="text-align: left;">η</i><i style="text-align: left;">σ</i><br />
<br />
<span style="text-align: left;"><span style="text-align: justify;">Number density,<i> </i></span><i>η</i></span><span style="text-align: justify;"> </span><span style="text-align: justify;">= <i>N/V</i> </span><span style="text-align: justify;"> </span><br />
<span style="text-align: left;"><span style="text-align: justify;">Cross section area,<i> </i></span><i>σ </i></span><span style="text-align: justify;">= </span><i>π</i><span style="text-align: justify;"><i>r</i></span><sup>2</sup><i> </i><br />
<br />
For Ideal Gas: <i>PV</i> = <i>NkT</i><br />
<br />
<span style="text-align: justify;">1/</span><i style="text-align: left;">η</i><i style="text-align: left;">σ </i><span style="text-align: justify;">= <i>V </i></span><span style="text-align: justify;">/ </span><i style="text-align: left;">N</i><i>π</i><span style="text-align: justify;"><i>r</i></span><sup>2 </sup><i style="text-align: left;"> </i><span style="text-align: justify;">= </span><i>NkT </i><span style="text-align: justify;">/ <i>P</i></span><i style="text-align: left;">N</i><i>π</i><span style="text-align: justify;"><i>r</i></span><sup>2 </sup><span style="text-align: justify;">= </span><i>kT </i><span style="text-align: justify;">/ <i>P</i></span><i>π</i><span style="text-align: justify;"><i>r</i></span><sup>2 </sup><span style="text-align: justify;"><i> </i></span><i style="text-align: left;"> </i><br />
<br />
<i>k </i>= 1.38 ×<sup> </sup>10<sup>−2</sup><sup>3 </sup><span style="text-align: justify;">Joule/K</span><br />
Radius of atom is in order of Angstrom: 10<sup>−10 </sup>m<br />
<i>P </i>(STP) = 1 atm = 10<sup>5</sup><sup> </sup><span style="text-align: justify;">Newton/meter</span><sup>2</sup><br />
<i>T</i> (STP) = 0 °C = 32 °F = 273.15 K ≈ 2 × 10<sup>2 </sup>K<br />
<i><br /></i>
<i>kT </i><span style="text-align: justify;">/ <i>P</i></span><i>π</i><span style="text-align: justify;"><i>r</i></span><sup>2 </sup><span style="text-align: justify;">= (</span>1.38 ×<sup> </sup>10<sup>−2</sup><sup>3 </sup>× <span style="text-align: justify;"><span style="text-align: left;">2 ×10</span><sup style="text-align: left;">2</sup>) / </span><span style="text-align: justify;">(</span><i>π </i>×10<sup>5 </sup>× 1<span style="text-align: justify;"><span style="text-align: left;">0</span><sup style="text-align: left;">−20</sup>)</span><i style="text-align: left;"> </i><br />
<i style="text-align: left;">= </i><span style="text-align: justify;">(</span>1.38 × 2 / <i>π</i><span style="text-align: justify;">) </span>10<sup>−2</sup><sup>3+2</sup><sup>−</sup><sup>5+20 </sup><i style="text-align: left;"><span style="font-style: normal;">≈ </span></i>10<sup>−7</sup><br />
<b><br /></b>
<b>Answer: B</b><br />
<br /></div>
Unknownnoreply@blogger.com0tag:blogger.com,1999:blog-3004213209691817978.post-81893097650089773472016-03-05T16:07:00.000-08:002023-12-23T18:34:01.379-08:00Quantum Mechanics - Probability<div dir="ltr" style="text-align: left;" trbidi="on">
<div style="text-align: justify;">
<div style="text-align: center;"><br /></div><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgh6y0T5GDUKHQ3jVXrmNCRo3B5guGJaxTr9OwrEFK9PQhZOG9naWvsAFTsxL4uIyktgcE1ARzRBFvZS5IDmYPj9CYr097Y9cA8nmkDg_OIwCwkPieDLgHaKy4zdU3jI4WpvYiun9h8gxGiQC4J1gN5e1FEe8rnNoy3SGRaFvr42s1612ru8b2EhPUUbgkf/s641/GR9677-17.JPG" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="406" data-original-width="641" height="203" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgh6y0T5GDUKHQ3jVXrmNCRo3B5guGJaxTr9OwrEFK9PQhZOG9naWvsAFTsxL4uIyktgcE1ARzRBFvZS5IDmYPj9CYr097Y9cA8nmkDg_OIwCwkPieDLgHaKy4zdU3jI4WpvYiun9h8gxGiQC4J1gN5e1FEe8rnNoy3SGRaFvr42s1612ru8b2EhPUUbgkf/s320/GR9677-17.JPG" width="320" /></a></div><br /><div style="text-align: justify;">The wave function for a particle constrained to move in one dimension is shown in the graph (<span style="text-align: left;">Ψ = 0 </span>for <i>x</i> <span style="text-align: left;">≤ </span> 0 and <i>x</i> <span style="text-align: left;">≥</span> 5). What is the probability that the particle would be found between <i>x</i> = 2 and <i>x</i> = 4?</div></div>
<div style="text-align: justify;">
<br /></div>
A. 17/64
<br />
B. 25/64
<br />
C. 5/8 <br />
D. v(5/8)
<br />
E. 13/16
<br />
<div style="text-align: right;">
<span style="color: red;"><b>(GR9677 #17)
</b></span></div>
<div class="separator" style="clear: both; text-align: center;">
</div>
<b>Solution:</b>
<br />
<br />
Probability, <i>P </i>∼ <i>ψ</i><sup>2</sup><br />
<br />
Probability to find the particle between <i>x </i>= 2 and <i>x</i> = 4 (unnormalized probability):<br />
<i>ψ</i><sup>2 </sup>= 2<sup>2 </sup> + 3<sup>2 </sup>= 13<br />
<br />
Total probability (normalized probability):<br />
<i>ψ</i><sup>2 </sup>= 1<sup>2 </sup> + 1<sup>2 </sup> + 2<sup>2 </sup> + 3<sup>2 </sup> + 1<sup>2 </sup>= 16<br />
<i><br /></i>
<i>P = </i>unnormalized probability / normalized probability = 13/16<br />
<br />
<b>Answer: E</b><br />
<br /></div>
Unknownnoreply@blogger.com0tag:blogger.com,1999:blog-3004213209691817978.post-63118065222982784682016-03-05T16:06:00.001-08:002024-01-22T07:25:08.075-08:00Quantum Mechanics - Potential Wall<div dir="ltr" style="text-align: left;" trbidi="on">
<div style="text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhYAqWp6ji5L1Qdo-s2czy7ttqfjA4a0P9PzoiLR1qsmX4nMFbuzJl4AHF7DvBeRN4Cvnh47b9RkfvifNvGUrvmShSYe3K-ArGKleL_m2YW5sZLwx6QN1igkLoC9rs8tRGRJ456z9fJF7tpkYG17eql13kp-VHoau3Bl_lrBXVl62IrVRcqWy0XmBWXH9iT/s353/GR9677-18a.JPG" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="246" data-original-width="353" height="223" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhYAqWp6ji5L1Qdo-s2czy7ttqfjA4a0P9PzoiLR1qsmX4nMFbuzJl4AHF7DvBeRN4Cvnh47b9RkfvifNvGUrvmShSYe3K-ArGKleL_m2YW5sZLwx6QN1igkLoC9rs8tRGRJ456z9fJF7tpkYG17eql13kp-VHoau3Bl_lrBXVl62IrVRcqWy0XmBWXH9iT/s320/GR9677-18a.JPG" width="320" /></a></div><br /><div dir="ltr" style="text-align: left;" trbidi="on"><br /></div>
Consider a potential of the form <br />
<br />
<i>V</i>(<i>x</i>) = 0, <i>x</i> ≤ <i>a</i><br />
<i>V</i>(<i>x</i>) = <i>V</i><sub>0, </sub><i>a < </i><i>x <</i><x b="" br=""> <i>b</i></x><br />
<x b="" br=""><i>V</i>(<i>x</i>) = 0, </x><i>x</i> <x b="" br="">≥ <i>b</i></x><br />
<x b="" br=""><br /></x>
<x b="" br="">
As shown in the figure above. Which of the following wave functions is possible for a particle </x>incident from the left with energy <i>E </i>< <i>V</i><sub>0</sub>.<br />
<br />
<br /><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEg_YYheL-bo83cxyWkePc1BKdc2VNYJRiUQp6UuS4tmJh5RUAskaTpw0MuC7jgevKCu2qor0qnaGqUCzCzOFzO1ETDXjiatTmP4Q5uFyGJc9WuQdK6UbBByLsE_IpuUg07ufYstaTt7F1b9Nkj6ik5COrej017GGN2PwPDHrULOpjZ4-nzsMe4r4cu67Cwi/s740/GR9677-18b.JPG" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="740" data-original-width="308" height="735" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEg_YYheL-bo83cxyWkePc1BKdc2VNYJRiUQp6UuS4tmJh5RUAskaTpw0MuC7jgevKCu2qor0qnaGqUCzCzOFzO1ETDXjiatTmP4Q5uFyGJc9WuQdK6UbBByLsE_IpuUg07ufYstaTt7F1b9Nkj6ik5COrej017GGN2PwPDHrULOpjZ4-nzsMe4r4cu67Cwi/w305-h735/GR9677-18b.JPG" width="305" /></a></div><br /><div style="text-align: center;"><br /></div>
<div style="text-align: right;">
<span style="color: red;"><b>(GR9677 #18)
</b></span></div>
<div class="separator" style="clear: both; text-align: center;">
</div>
<b>Solution:</b>
<br />
<br />
A. Classic not QM potential → FALSE<br />
B. No decrease in amplitude → FALSE<br />
C. Decay exponentially inside the wall, decrease amplitude (fits <i>E </i>< <i>V</i><sub>0</sub>) → TRUE</div><div dir="ltr" style="text-align: left;" trbidi="on">
D. QM Oscillator harmonics → FALSE<br />
E. Cosine wave function not QM potential → FALSE<br />
<v_0 amplitude.="" amplitude="" an="" barrier="" be="" decrease="" e="" emerges="" even="" for="" function="" good="" graph="" have="" his="" however="" is="" it="" of="" part="" should="" smaller="" that="" the="" true="" tunneled="" wave="" when="" would=""><x amplitude="" and="" b="" br="" exit.="" tunneled-decrease="" when=""><br />
<b>Answer: C</b></x></v_0></div>
Unknownnoreply@blogger.com0tag:blogger.com,1999:blog-3004213209691817978.post-60864246699498395322016-03-05T16:06:00.000-08:002023-12-23T18:40:48.404-08:00Nuclear & Particle Physics - Alpha/Rutherford Scattering<div dir="ltr" style="text-align: left;" trbidi="on">
<div style="text-align: justify;">
When alpha particles are directed onto atoms in a thin metal foil, some make very close collisions with the nuclei of the atoms and are scattered at large angles. If an alpha particles with an initial kinetic energy of 5 MeV happens to be scattered through an angle of <span style="text-align: left;">180</span><sup style="text-align: left;">o</sup>, which of the following must have been its distance of the closest approach to the scattering nucleus? (Assume that the metal foil is made of silver, with <i>Z</i> = 50.)
</div>
<br />
A. 1.22 × 50<sup>1/3</sup> fm<br />
B. 2.9 × 10<sup>−14</sup> m
<br />
C. 1.0 × 10<sup>−12</sup> m<br />
D. 3.0 × 10<sup>−8</sup> m
<br />
E. 1.7 × 10<sup>−7</sup> m<br />
<div style="text-align: right;">
<span style="color: red;"><b>(GR9677 #19)
</b></span></div>
<div class="separator" style="clear: both; text-align: center;">
</div>
<b>Solution:</b><br />
<br />
"<span style="text-align: justify;">When alpha particles are directed onto atoms in a thin metal foil, some make very close collisions with the nuclei of the atoms and are scattered at large angles."</span>→ Rutherford Scattering: the discovery of nucleus.<br />
<br />
Rutherford estimated the radius of a silver nucleus to be 2 × 10<sup>−14</sup> m, by observing the angular dependence of alpha-particle scattering (<a href="http://web.pdx.edu/~egertonr/ph311-12/nuclear.htm" target="_blank">source</a>).<br />
<br />
<div style="text-align: left;">
<span style="font-weight: bold;">Answer: B</span></div>
<br />
<br />
<b><span style="color: red;">Calculation:</span></b><br />
<br />
Conservation of Energy: <i>U </i>= <i>T</i><br />
<br />
<i>U </i>= <i>kq</i><sub><i>α </i></sub><i>q</i><sub><i>s </i></sub>/ <i>r </i>= <i>T</i><br />
<i>r </i>= <i>kq</i><sub><i>α </i></sub><i>q</i><sub><i>s </i></sub>/ <i>T</i><br />
<br />
Given:<br />
<i>T </i>= <span style="text-align: justify;">5 M<i>e</i>V = 5 </span>×<sup> </sup>10<sup>6 </sup><span style="text-align: justify;"><i>e</i>V</span><br />
<i>Z</i><sub><i>s </i></sub><span style="text-align: justify;">= 50 </span>→ <i>q</i><sub><i>s </i></sub><span style="text-align: justify;">= </span><i>Z</i><sub><i>α</i></sub><span style="text-align: justify;"><i>e </i></span><span style="text-align: justify;">= </span><span style="text-align: justify;">50</span><span style="text-align: justify;"><i>e</i></span><br />
<i>Z</i><sub><i>α </i></sub><span style="text-align: justify;">= 2 </span>→ <i>q</i><sub><i>α </i></sub><span style="text-align: justify;">= </span><i>Z</i><sub><i>α</i></sub><span style="text-align: justify;"><i>e </i></span><span style="text-align: justify;">= </span><span style="text-align: justify;">2</span><span style="text-align: justify;"><i>e </i></span><br />
<span style="text-align: justify;"><i><br /></i></span>
<i>k </i><span style="text-align: justify;">= 1/</span><span style="text-align: center;">4</span><i style="text-align: center;">π</i><i style="text-align: center;">ɛ</i><sub style="text-align: center;">0</sub><span style="text-align: justify;"> </span><span style="text-align: justify;">=</span><span style="text-align: justify;"> 1/(4 </span>× 3.14<sup> </sup>× 8.85 ×<sup> </sup>10<sup>−12</sup>)<sup> </sup>≈ 10<sup>10 </sup><span style="text-align: justify;">Nm</span><sup>2</sup><span style="text-align: justify;">/</span><span style="text-align: justify;">C</span><sup>2</sup><br />
<b><br /></b><i>r </i>= (10<sup>10 </sup><span style="text-align: justify;">Nm</span><sup>2</sup><span style="text-align: justify;">/</span><span style="text-align: justify;">C</span><sup>2</sup>×<span style="text-align: justify;"> </span><span style="text-align: justify;">50</span><i style="text-align: justify;">e</i><span style="text-align: justify;"> </span>×<span style="text-align: justify;"> </span><span style="text-align: justify;">2</span><i style="text-align: justify;">e</i>) / (<span style="text-align: justify;">5 </span>×<sup> </sup>10<sup>6 </sup><span style="text-align: justify;"><i>e</i>V</span>)<br />
= 2 × 10<sup>5 </sup><i style="text-align: justify;">e </i><span style="text-align: justify;">Nm</span><sup>2</sup><span style="text-align: justify;">/V</span><span style="text-align: justify;">C</span><sup>2 </sup><br />
<div style="text-align: justify;"><span style="text-align: left;"><br /></span></div><div style="text-align: justify;"><span style="text-align: left;">With <i>e</i> = 1.60 × 10</span><sup style="text-align: left;">−19 </sup><span style="text-align: left;">C</span></div><div style="text-align: justify;"><span style="text-align: left;">1 Volt = 1 Nm/C</span></div><p>
<i>r </i>= 2 × 10<sup>5 </sup>× 1.60 ×<sup> </sup>10<sup>−</sup><sup>19</sup><i style="text-align: justify;"> </i><span style="text-align: justify;">m</span><span style="text-align: justify;"> </span>≈ 3 × 10<sup>−14 </sup><span style="text-align: justify;">m</span><br />
<br />
<br /></p><p></p><p></p></div>
Unknownnoreply@blogger.com0tag:blogger.com,1999:blog-3004213209691817978.post-61810791448377467862016-03-05T16:04:00.000-08:002016-03-05T16:04:33.009-08:00Classical Mechanics - Linear Momentum<div dir="ltr" style="text-align: left;" trbidi="on">
<div style="text-align: justify;">
A helium atom, mass 4u travels with non relativistic speed <i>v</i> normal to the surface of a certain material, makes an elastic collision with an (essentially free) surface atom, and leaves in the opposite direction with speed 0.6<i>v</i>. The atom on the surface must be an atom of</div>
<br />
A. Hydrogen, mass 1u
<br />
B. Helium, mass 4u
<br />
C. Carbon, mass 12u
<br />
D. Oxygen, mass 16u
<br />
E. Silicon, mass 28u<br />
<div style="text-align: right;">
<span style="color: red;"><b>(GR9677 #20)
</b></span></div>
<div class="separator" style="clear: both; text-align: center;">
</div>
<b>Solution:</b>
<br />
<br />
<i>m</i><sub><i>a</i></sub> = 4u<br />
<i>v</i><sub><i>a </i></sub> = <i>v</i><br />
<i>v</i><sub><i>b </i></sub><sub><i> </i></sub>= 0<br />
<i>v</i><sub><i>a</i></sub><i>' </i>= − <span style="text-align: justify;">0.6</span><i style="text-align: justify;">v</i><br />
<br />
Conservation of momentum of the system:<br />
<br />
<i>m</i><sub><i>a</i></sub><i>v</i><sub><i>a </i></sub>+ <i>m</i><sub><i>b</i></sub><i>v</i><sub><i>b</i></sub> = <i>m</i><sub><i>a</i></sub><i>v</i><sub><i>a</i></sub><i>' </i>+ <i>m</i><sub><i>b</i></sub><i>v</i><sub><i>b</i></sub><i>' </i><br />
4u<i>v</i><i><span style="font-size: 13.3333px;"> </span></i>= 4u(− <span style="text-align: justify;">0.6</span><i style="text-align: justify;">v</i>)<i> </i>+ <i>m</i><sub><i>b</i></sub><i>v</i><sub><i>b</i></sub><i>' </i><br />
4u<i>v</i><i><span style="font-size: 13.3333px;"> </span></i>= − 2<span style="text-align: justify;">.4</span>u<i>v</i><i><span style="font-size: 13.3333px;"> </span></i><i> </i>+ <i>m</i><sub><i>b</i></sub><i>v</i><sub><i>b</i></sub><i>' </i><br />
<i>m</i><sub><i>b</i></sub><i>v</i><sub><i>b</i></sub><i>' </i>= 6.4u<i>v </i>(Eq.1)<br />
<br />
Conservation of kinetic energy of the system:<br />
<br />
½ <i>m</i><sub><i>a</i></sub><i>v</i><sub><i>a</i></sub>² + ½ <i>m</i><sub><i>b</i></sub><i>v</i><sub><i>b</i></sub>² = ½ <i>m</i><sub><i>a</i></sub><i>v</i><sub><i>a</i></sub><i>'</i>² + ½ <i>m</i><sub><i>b</i></sub><i>v</i><sub><i>b</i></sub><i>'</i>²<br />
4u<i>v</i>²<i><span style="font-size: 13.3333px;"> </span></i>= <span style="text-align: justify;">4</span>u(− <span style="text-align: justify;">0.6</span><i>v</i>)² + <i>m</i><sub><i>b</i></sub><i>v</i><sub><i>b</i></sub><i>'</i>²<br />
4u<i>v</i>²<i><span style="font-size: 13.3333px;"> </span></i>= <span style="text-align: justify;">4</span>u(<span style="text-align: justify;">0.36</span><i>v</i>²) + <i>m</i><sub><i>b</i></sub><i>v</i><sub><i>b</i></sub><i>'</i>²<br />
<i>m</i><sub><i>b</i></sub><i>v</i><sub><i>b</i></sub><i>'</i>²<i> </i>= 4u<i>v</i>²<i><span style="font-size: 13.3333px;"> </span></i><i><span style="font-size: 13.3333px;"> </span></i>− 1.44u<i>v</i>²<br />
<i>m</i><sub><i>b</i></sub><i>v</i><sub><i>b</i></sub><i>'</i>²<i> </i>= 2.56u<i>v</i>² (Eq.2)<br />
<br />
(Eq.1) →<span style="text-align: justify;"><i style="text-align: left;"> </i></span>(Eq.2)<br />
6.4<i> </i>(<i>v</i><sub><i>b</i></sub><i>'</i>) = 2.56<i>v</i><br />
<i>v</i><sub><i>b</i></sub><i>'</i><i> </i>= (2.56/6.4)<i>v </i>= 0.4<i>v</i> (Eq.3)<br />
<b><br /></b>
(Eq.3) →<span style="text-align: justify;"><i style="text-align: left;"> </i></span>(Eq.1)<br />
<i>m</i><sub><i>b </i></sub>= 6.4u<i>v </i>/ 0.4<i>v </i>= 16u<br />
<i><br /></i>
<b>Answer: D</b><br />
<br /></div>
Unknownnoreply@blogger.com0tag:blogger.com,1999:blog-3004213209691817978.post-68511953753064553582016-03-05T16:03:00.000-08:002016-03-05T20:19:09.959-08:00Classical Mechanics - Moment Inertia<div dir="ltr" style="text-align: left;" trbidi="on">
<div style="text-align: justify;">
The period of physical pendulum is 2<i style="text-align: left;">π</i><span style="text-align: left;">√</span>(<i>I</i>/<i>mgd</i>), where <i>I</i> is the moment of inertia about the pivot point and <i>d</i> is the distance from the pivot to the center of mass. A circular hoop hangs from a nail on a barn wall. The mass of the hoop is 3 kilograms and its radius is 20 centimeters. If it is displaced slightly by a passing breeze, what is the period of the resulting oscillations?</div>
<br />
A. 0.63 s
<br />
B. 1.0 s
<br />
C. 1.3 s
<br />
D. 1.8 s
<br />
E. 2.1 s
<br />
<div style="text-align: right;">
<span style="color: red;"><b>(GR9677 #21)
</b></span></div>
<div class="separator" style="clear: both; text-align: center;">
</div>
<b>Solution:</b>
<br />
<br />
<i>T</i> = <span style="text-align: justify;">2</span><i style="text-align: left;">π</i><span style="text-align: left;">√</span><span style="text-align: justify;">(</span><i style="text-align: justify;">I</i><span style="text-align: justify;">/</span><i style="text-align: justify;">mgd</i><span style="text-align: justify;">)</span><br />
<i>m </i>= 3 kg<br />
<i>r </i>= 20 cm = 0.2 m<br />
In this case <i>d </i>= <i>r</i><br />
<br />
To find total <i>I</i>:<br />
Parallel axis theorem: <i>I</i> = <i>ml</i><sup>2</sup> + <i>I</i><sub>CM</sub><br />
<i>I</i><sub>CM </sub><i> </i>= <i>I</i><sub>loop </sub>= <i>mr</i><sup>2 </sup><br />
<br />
In this case <i>l</i><i> </i>= <i>r</i><br />
→ <i>I</i> = <i>mr</i><sup>2</sup> + <i>mr</i><sup>2 </sup>= 2<i>mr</i><sup>2</sup><br />
<br />
<div style="text-align: justify;">
<i>T</i> <span style="text-align: justify;">= </span><span style="text-align: justify;">2</span><i style="text-align: left;">π</i><span style="text-align: left;">√</span><span style="text-align: justify;">(</span>2<i>mr</i><sup>2</sup> <span style="text-align: justify;">/ </span><i style="text-align: justify;">mgr</i><span style="text-align: justify;">) </span><br />
<span style="text-align: justify;">= </span><span style="text-align: justify;">2</span><i style="text-align: left;">π</i><span style="text-align: left;">√</span><span style="text-align: justify;">(</span>2<i>r</i><span style="text-align: justify;">/</span><i style="text-align: justify;">g</i><span style="text-align: justify;">) </span><br />
<span style="text-align: justify;">= </span><span style="text-align: justify;">2<i style="text-align: left;">π</i></span><span style="text-align: left;">√[</span>2(0.2)<span style="text-align: justify;">/(10)]</span><span style="text-align: justify;"> </span><br />
<span style="text-align: justify;">= </span>2<i style="text-align: left;">π</i>(0.2)<br />
= 1.25 s</div>
<span style="text-align: justify;"><br /></span>
<b>Answer: C</b></div>
Unknownnoreply@blogger.com0tag:blogger.com,1999:blog-3004213209691817978.post-7407412200715219912016-03-05T16:02:00.000-08:002023-12-24T10:25:38.092-08:00Classical Mechanics - Curved Trajectory<div dir="ltr" style="text-align: left;" trbidi="on">
<div style="text-align: justify;">
The curvature of Mars is such that its surface drops a vertical distance of 2.0 meters for every 3600 meters tangent to the surface. In addition, the gravitational acceleration near its surface is 0.4 times that near the surface of Earth. What is the speed a golf ball would need to orbit Mars near the surface, ignoring the effects of air resistance?</div>
<br />
A. 0.9 km/s
<br />
B. 1.8 km/s
<br />
C. 3.6 km/s
<br />
D. 4.5 km/s
<br />
E. 5.4 km/s
<br />
<div style="text-align: right;">
<span style="color: red;"><b>(GR9677 #22)
</b></span></div>
<div class="separator" style="clear: both; text-align: center;">
</div>
<b>Solution:</b>
<br />
<br /><br />
<div style="text-align: justify;"><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiCLBrRr4TzY62sVMGD_qHPER1tc9K0QQkO9OevlSeS59gy8mYvpdURQ7Nm2Zbe3gZ9IXaqywrzoNxVWl4q8Sf9TmKPmpC9rJ8gQGWohQnDhZqFtKVdUCMkBxIiTPLo2UGb56vDz3O-EWbJkkpiGMAULJRB_HH4S-Ra8hSWl4_tPLBTyRk9L1zFxJPdaVXZ/s558/GR9677-22.JPG" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="178" data-original-width="558" height="102" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiCLBrRr4TzY62sVMGD_qHPER1tc9K0QQkO9OevlSeS59gy8mYvpdURQ7Nm2Zbe3gZ9IXaqywrzoNxVWl4q8Sf9TmKPmpC9rJ8gQGWohQnDhZqFtKVdUCMkBxIiTPLo2UGb56vDz3O-EWbJkkpiGMAULJRB_HH4S-Ra8hSWl4_tPLBTyRk9L1zFxJPdaVXZ/s320/GR9677-22.JPG" width="320" /></a></div>
<i>y</i> = <span style="text-align: left;">½</span> <i>g</i><sub style="text-align: left;">mars</sub><i>t</i><sup>2</sup></div>
<span style="text-align: justify;"><span style="text-align: left;"><span style="text-align: justify;"><span style="text-align: justify;"><i style="text-align: left;"><i>v</i><sub style="font-style: normal;"><i>x </i></sub></i></span><span style="text-align: justify;">=<i> x/t </i></span></span></span></span><br />
<span style="text-align: justify;"><span style="text-align: left;"><span style="text-align: justify;"><span style="text-align: justify;"><i><br /></i></span></span></span></span>
<span style="text-align: justify;"><span style="text-align: left;"><span style="text-align: justify;">Given:</span></span></span><br />
<span style="text-align: justify;"><span style="text-align: left;"><span style="text-align: justify;"><span style="text-align: justify;"><span style="text-align: left;"><span style="text-align: justify;"><i>y</i> </span></span></span>= 2</span></span></span><br />
<span style="text-align: justify;"><span style="text-align: left;"><span style="text-align: justify;"><i>g</i><sub style="text-align: left;">mars </sub></span></span></span><span style="text-align: justify;">= 0.4 g </span><br />
<i style="text-align: justify;">x </i><span style="text-align: justify;">= </span><span style="text-align: justify;">3600</span><br />
<span style="text-align: justify;"><br /></span>
<span style="text-align: justify;"><span style="text-align: left;"><span style="text-align: justify;">2</span><span style="text-align: justify;"> = </span>½<span style="text-align: justify;">(0.4)(10)</span><span style="text-align: justify;">(</span><i style="text-align: justify;">t</i><span style="text-align: justify;">)</span><sup style="text-align: justify;">2</sup></span></span><br />
<span style="text-align: justify;"><i style="text-align: left;"><i style="text-align: justify;">t</i></i></span><span style="text-align: justify;"> </span><span style="text-align: justify;">= 1 s</span><br />
<span style="text-align: justify;"><i style="text-align: left;"><i>v</i><sub style="font-style: normal;"><i>x </i></sub></i></span><span style="text-align: justify;">= 3600/1 = 3600 m/s = </span>3.6 km/s<br />
<br />
<b>Answer: C</b></div><br /><div class="separator" style="clear: both; text-align: center;"><br /></div><br />
Unknownnoreply@blogger.com0tag:blogger.com,1999:blog-3004213209691817978.post-9831933725535391052016-03-05T16:01:00.003-08:002016-03-05T16:01:31.597-08:00Classical Mechanics - Conservative Force<div dir="ltr" style="text-align: left;" trbidi="on">
<div dir="ltr" style="text-align: left;" trbidi="on">
<div style="text-align: justify;">
Suppose that the gravitational force law between two massive objects were<br />
<br /></div>
<div style="text-align: justify;">
<b>F</b><sub>12</sub> = <b>r̂</b><sub>12</sub> <i>Gm</i><sub>1</sub><i>m</i><sub>2</sub>/<i>r</i><sub>12</sub><sup>(</sup><sup>2+ɛ</sup><sup>)</sup> </div>
<div style="text-align: justify;">
<br />
where <i>ɛ</i> is a small positive number. Which of the following statements would be FALSE?</div>
<ol type="A">
<li>The total mechanical energy of the planet-Sun system would be conserved. </li>
<li>The angular momentum of a single planet moving about the Sun would be conserved. </li>
<li>The periods of planets in circular orbits would be proportional to the (3+<i style="text-align: justify;">ɛ</i>)/2 power of their respective orbital radii. </li>
<li>A single planet could move in a stationary non circular elliptical orbit about the Sun. </li>
<li>A single planet could move in a stationary circular orbit about the Sun.</li>
</ol>
<div style="text-align: right;">
<span style="color: red;"><b>(GR9677 #23)
</b></span></div>
<div class="separator" style="clear: both; text-align: center;">
</div>
<b>Solution:</b>
<br />
<br /></div>
(A) TRUE.<br />
Gravitational force is a conservative force.<br />
In conservative field, the total mechanical energy is conserved.<br />
<br />
(B) TRUE<br />
In conservative field, angular momentum, <i>L</i> is conserved.<br />
<br />
(C) TRUE<br />
<i>F</i><sub>g </sub>= <i>F</i><sub>c</sub><br />
<i>GMm</i>/<i style="text-align: justify;">r</i><sup style="text-align: justify;">(</sup><sup style="text-align: justify;">2+ɛ</sup><sup style="text-align: justify;">)</sup><sup> </sup>= <i>mr</i><i>ω</i><sup>2</sup><br />
<i>GMm</i>/<i style="text-align: justify;">r</i><sup style="text-align: justify;">(</sup><sup style="text-align: justify;">2+ɛ</sup><sup style="text-align: justify;">)</sup><sup> </sup>= <i>mr</i>(2<i>π</i>/<i>T</i>)<sup>2</sup><br />
<i>GM</i>/<i style="text-align: justify;">r</i><sup style="text-align: justify;">(</sup><sup style="text-align: justify;">3+ɛ</sup><sup style="text-align: justify;">)</sup><sup> </sup>=<sup> </sup>4<i>π</i><sup>2</sup>/<i>T</i><sup>2</sup><br />
<i>T</i><sup>2 </sup>= 4<i>π</i><sup>2</sup><i style="text-align: justify;">r</i><sup style="text-align: justify;">(</sup><sup style="text-align: justify;">3+ɛ</sup><sup style="text-align: justify;">)</sup>/<i>GM</i><br />
<i>T</i><sup> </sup>∝ <i style="text-align: justify;">r</i><sup style="text-align: justify;">(</sup><sup style="text-align: justify;">3+ɛ</sup><sup style="text-align: justify;">)/2</sup><sup> </sup><br />
<br />
(D) FALSE<br />
Central force = centripetal force (<i>F</i><sub>g </sub>= <i>F</i><sub>c</sub>) produces circular orbit.<br />
Non central forces do not produce circular orbit.<br />
<br />
(E) TRUE<br />
See (D)<br />
<br />
<b>Answer: D</b><br />
<br />
<span style="color: red;"><b>Notes:</b></span><br />
<br />
<div dir="ltr" style="text-align: left;" trbidi="on">
Central force:<br />
<ol type="bullet">
<li>It is a force whose magnitude depends only on the distance between the object and the origin.</li>
<li>It is a conservative field, can be expressed as <i>F</i> = − ∇<i>V </i>(the negative gradient of a potential energy).</li>
<li>Gravitational force, Coulomb force, and Elastic Force (Harmonic Oscillator) are examples of central (conservative) forces.</li>
<li>In conservative field, the <i><span style="font-style: normal;">net work done by the force is zero, </span>W</i> = ∮<sub>c</sub> <b>F</b> ∙ d<b>r</b> = 0 → the total mechanical energy is conserved.</li>
<li>Conservative force is irrotional (torque = 0), since curl ∇<i>V </i>or ∇ <i><span style="font-style: normal;">× ∇</span>V </i>= 0.</li>
<li>Torque, <i>τ = dL/dT </i>= 0 → angular momentum, <i>L</i> is conserved</li>
<li>Central force = centripetal force (<i>F</i><sub>g </sub>= <i>F</i><sub>c</sub>) produces circular orbit.</li>
</ol>
</div>
<br /></div>
Unknownnoreply@blogger.com0tag:blogger.com,1999:blog-3004213209691817978.post-13182470172016857412016-03-04T17:06:00.005-08:002016-03-25T04:12:56.556-07:00Electromagnetism - Electric Force<div dir="ltr" style="text-align: left;" trbidi="on">
<div style="text-align: justify;">
Two identical conducting spheres, A and B, carry equal charge. They are initially separated by a distance much larger than their diameters, and the force between them is <i>F</i>. A third identical conducting sphere, C, is uncharged. Sphere C is first touched to A, then to B, and then removed. As a result, the force between A and B is equal to.
</div>
<br />
A. 0
<br />
B. <i>F</i>/16
<br />
C. <i>F</i>/4
<br />
D. 3<i>F</i>/8<br />
E. <i>F</i>/2<br />
<div style="text-align: right;">
<span style="color: red;"><b>(GR9677 #24)
</b></span></div>
<div class="separator" style="clear: both; text-align: center;">
</div>
<b>Solution:</b>
<br />
<br />
Coulomb's Law, <i>F</i><sub><i>i </i></sub>= <i>kq</i><sub>A</sub><i>q</i><sub>B</sub>/<i>r</i><sup>2 </sup><br />
<i>q</i><sub>A</sub><sub><i> </i></sub>= <i>q</i><sub>B</sub><sub><i> </i></sub>= <i>Q</i><br />
<i>F</i><sub><i>i </i></sub>= <i style="text-align: justify;">F </i>= <i>k</i><i>Q</i><sup>2 </sup>/<i>r</i><sup>2 </sup><br />
<br />
When uncharged s<span style="text-align: justify;">phere C </span><span style="text-align: justify;">touches </span><span style="text-align: justify;">A, charge A distributes itself evenly:</span><br />
<i>q</i><sub>C</sub><sub><i> </i></sub>= <i>q</i><sub>A</sub><i>'</i><sub><i> </i></sub>= <sup>1</sup>/<sub>2</sub> <i>q</i><sub>A</sub><i> </i>= <sup>1</sup>/<sub>2</sub><i>Q</i><br />
<span style="text-align: justify;"><br /></span>
When s<span style="text-align: justify;">phere C </span><span style="text-align: justify;">touches </span><span style="text-align: justify;">B:</span><br />
<i>q</i><sub>C</sub><i>'</i><sub><i> </i></sub><sub><i> </i></sub>= <i>q</i><sub>B</sub><i>'</i><sub><i> </i></sub><sub><i> </i></sub>= <sup>1</sup>/<sub>2</sub>(<i>q</i><sub>C</sub><i> </i>+ <i>q</i><sub>B</sub>) = <sup>1</sup>/<sub>2</sub>(<sup>1</sup>/<sub>2</sub><i>Q </i>+ <i>Q</i>) = <sup>3</sup>/<sub>4</sub><i>Q</i><br />
<span style="text-align: justify;"><br /></span>
Therefore,<br />
<i>F</i><sub><i>f </i></sub>= <i>k</i><i>q</i><sub>A</sub><i>'</i><sub><i> </i></sub><i>q</i><sub>B</sub><i>'</i>/<span style="text-align: center;"><i style="text-align: left;">r</i><sup style="text-align: left;">2</sup> </span><br />
= <i>k </i>(<sup>1</sup>/<sub>2</sub><i>Q</i>) (<sup>3</sup>/<sub>4</sub><i>Q</i>)/<i>r</i><sup>2</sup><br />
= <sup>3</sup>/<sub>8</sub> <i>k</i><i>Q</i><sup>2</sup>/<span style="text-align: center;"><i style="text-align: left;">r</i><sup style="text-align: left;">2</sup> </span><br />
= <sup>3</sup>/<sub>8</sub> <i style="text-align: justify;">F</i><br />
<br />
<b>Answer: D</b><br />
<br /></div>
Unknownnoreply@blogger.com0