Questions 41-42
A. 80 Nm
B. 40 Nm
C. 32 Nm
D. 16 Nm
E. 8 Nm
}{t}=4\frac{\left&space;(&space;40&space;-&space;80&space;\right&space;)}{10}=-16&space;\text{&space;Nm} "\tau = I\alpha = I \frac{\left ( \omega_f - \omega_i \right )}{t}=4\frac{\left ( 40 - 80 \right )}{10}=-16 \text{ Nm}")
Answer: D
If the cylinder takes 10 seconds to reach 40 radian per second, the magnitude of the applied torque is
A. 80 Nm
B. 40 Nm
C. 32 Nm
D. 16 Nm
E. 8 Nm
(GR9277 #42)
Solution:
Answer: D
No comments :
Post a Comment