Consider a motor that exerts a constant torque of 25.0 nâ‹…m to a horizontal platform whose moment of inertia is 50.0 kgâ‹…m2. assume that the platform is initially at rest and the torque is applied for 12.0 rotations. neglect friction.

W = 1884J

Explanation:

This question is incomplete. The original question was:

Consider a motor that exerts a constant torque of 25.0 N.m to a horizontal platform whose moment of inertia is 50.0kg.m^2 . Assume that the platform is initially at rest and the torque is applied for 12.0rotations . Neglect friction.

How much work W does the motor do on the platform during this process?  Enter your answer in joules to four significant figures.

The amount of work done by the motor is given by:

Where I = 50kg.m^2 and ωo = rad/s. We need to calculate ωf.

By using kinematics:

But we don't have the acceleration yet. So, we have to calculate it by making a sum of torque:

    =>    

Now we can calculate the final velocity:

Finally, we calculate the total work:

Since the question asked to "Enter your answer in joules to four significant figures.":

W = 1884J

d

explanation:

v=uf/(u-f)

= 5.10/(5-10) convex lens has positive focal length

=50/-5

=-10 cm (formed at other side)

can be answered using momentum.

momentum = p

p1= mass1 * velocity1

p1 = p2

therefore

p1 = 6kg * 4m/s

p1 = 24kgm/s

p2=m2 * v2

24 kgm/s = 7kg * v2

v2 = 24 kgm/s / 7kg

v2= 3.42 m/s