Answer:
4611.58 ft/s²
Explanation:
t = Time taken
u = Initial velocity
v = Final velocity
s = Displacement
a = Acceleration due to gravity = 32.174 ft/s²
Equation of motion


Magnitude of acceleration while stopping is 4611.58 ft/s²
Answer:
B. 0 degrees
Explanation:
The magnetic flux through a certain area enclosed by a loop of wire immersed in a region with magnetic field is given by:

where
B is the strength of the magnetic field
A is the area enclosed by the coil
is the angle between the direction of the magnetic field and the direction of the normal to the coil
From the equation above, we see that the magnetic flux is maximum when

So when the field is parallel to the normal to the coil (so, when it is perpendicular to the coil), therefore when:

So, the correct option is
B. 0 degrees
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