I think it's C, longer wave length.
Answer:
![\mu_k=0.18](https://tex.z-dn.net/?f=%5Cmu_k%3D0.18)
Explanation:
First, we write the equations of motion for each axis. Since the crate is sliding with constant speed, its acceleration is zero. Then, we have:
![x: T+F-f_k=0\\\\y:N-mg=0](https://tex.z-dn.net/?f=x%3A%20T%2BF-f_k%3D0%5C%5C%5C%5Cy%3AN-mg%3D0)
Where T is the tension in the rope, F is the force exerted by the first worker, f_k is the frictional force, N is the normal force and mg is the weight of the crate.
Since
and
, we can rewrite the first equation as:
![T+F-\mu_k mg=0](https://tex.z-dn.net/?f=T%2BF-%5Cmu_k%20mg%3D0)
Now, we solve for
and calculate it:
![\mu_k=\frac{T+F}{mg}\\ \\\mu_k =\frac{220N+390N}{(350kg)(9.8m/s^{2})} =0.18](https://tex.z-dn.net/?f=%5Cmu_k%3D%5Cfrac%7BT%2BF%7D%7Bmg%7D%5C%5C%20%5C%5C%5Cmu_k%20%3D%5Cfrac%7B220N%2B390N%7D%7B%28350kg%29%289.8m%2Fs%5E%7B2%7D%29%7D%20%3D0.18)
This means that the crate's coefficient of kinetic friction on the floor is 0.18.
Dependent on what you are measuring and what took you are using. Please be more specific.
Explanation:
Acceleration is change in velocity over change in time:
a = Δv / Δt
a = (10 m/s - 25 m/s) / (240 s - 0 s)
a = -0.0625 m/s²
So the car decelerates at 0.0625 m/s².