Answer:
I belive it would be "C"
Explanation:
If it was any of the other answers "B" it would instantly stop. "A" it would roll forever.
Answer:
Magnitude of force on wheel B is 4 N
Explanation:
Given that

For wheel A
m= 1 kg
d= 1 m,r= 0.5 m
F=1 N
We know that
T= F x r
T=1 x 0.5 N.m
T= 0.5 N.m
T= I α
Where I is the moment of inertia and α is the angular acceleration


T= I α
0.5= 0.25 α

For Wheel B
m= 1 kg
d= 2 m,r=1 m


Given that angular acceleration is same for both the wheel

T= I α
T= 1 x 2
T= 2 N.m
Lets force on wheel is F then
T = F x r
2 = F x 1
So F= 2 N
Magnitude of force on wheel B is 2 N
Your diagram should include four forces:
• the box's weight, pointing down (magnitude <em>w</em> = 43.2 N)
• the normal force, pointing up (mag. <em>n</em>)
• the applied force, pointing the direction in which the box is sliding (mag. <em>p</em> = 6.30 N, with <em>p</em> for "pull")
• the frictional force, pointing oppoiste the applied force (mag. <em>f</em> )
The box is moving at a constant speed, so it is inequilibrium and the net forces in both the vertical and horizontal directions sum to 0. By Newton's second law, we have
<em>n</em> + (-<em>w</em>) = 0
and
<em>p</em> + (-<em>f</em> ) = 0
So then the forces have magnitudes
<em>w</em> = 43.2 N
<em>n</em> = <em>w</em> = 43.2 N
<em>p</em> = 6.30 N
<em>f</em> = <em>p</em> = 6.30 N
Answer:
The the maximum force acting on the crate is 533.12 newtons.
Explanation:
It is given that,
Mass of the wooden crate, m = 136 kg
The coefficient of static friction, 
The coefficient of kinetic friction, 
We need to find the maximum force exerted horizontally on the crate without moving it. As the crate is not moving than the coefficient of static friction will act and the force is given by :


F = 533.12 N
So, the maximum force acting on the crate is 533.12 newtons. Hence, this is the required solution.
Asteroids are too small to have atmospheres