Answer:
W = 1884J
Explanation:
This question is incomplete. The original question was:
<em>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.
</em>
<em>
How much work W does the motor do on the platform during this process? Enter your answer in joules to four significant figures.</em>
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 "<em>Enter your answer in joules to four significant figures.</em>":
W = 1884J
C is the best answer
hope it helped
Answer:
56.86153 N
Explanation:
t =Time taken
F = Force
Power

Work done

The magnitude of the force that is exerted on the handle is 56.86153 N
Because the Earth<span> is a sphere, the surface gets much more intense </span>sunlight<span>, hence heat, at </span>the equator<span>than at the poles.</span>
Answer
Given,
y(x, t) = (3.5 cm) cos(2.7 x − 92 t)
comparing the given equation with general equation
y(x,t) = A cos(k x - ω t)
A = 3.5 cm , k = 2.7 rad/m , ω = 92 rad/s
we know,
a) ω =2πf
f = 92/ 2π
f = 14.64 Hz
b) Wavelength of the wave
we now, k = 2π/λ
2π/λ = 2.7
λ = 2 π/2.7
λ = 2.33 m
c) Speed of wave
v = ν λ
v = 14.64 x 2.33
v = 34.11 m/s