The given question is incomplete. The complete question is as follows.
A box of oranges which weighs 83 N is being pushed across a horizontal floor. As it moves, it is slowing at a constant rate of 0.90 m/s each second. The push force has a horizontal component of 20 N and a vertical component of 25 N downward. Calculate the coefficient of kinetic friction between the box and the floor.
Explanation:
The given data is as follows.
= 20 N,
= 25 N, a = -0.9
W = 83 N
m = 
= 8.46
Now, we will balance the forces along the y-component as follows.
N = W +
= 83 + 25 = 108 N
Now, balancing the forces along the x component as follows.
= ma
= 7.614 N
Also, we know that relation between force and coefficient of friction is as follows.

= 
= 0.0705
Thus, we can conclude that the coefficient of kinetic friction between the box and the floor is 0.0705.
Answer:
F = 4000 N
Explanation:
given,
mass of rocket (M)= 5000 Kg
10 Kg gas burns at speed (m)= 4000 m/s
time = 10 s
average force = ?
at the end the rocket is at rest
by conservation of momentum
M v + m v' = 0
5000 x v - 10 x 4000 = 0
5000 v = 40000
v = 8 m/s
speed of rocket = 8 m/s
now,
we know
change in momentum = F x Δ t


F = 4000 N
Hence, the average force applied to the rocket is equal to F = 4000 N
It makes no sense how you typed this problem out.
Answer:
9
Explanation:
2.13 rad/s * 26.9 sec
2.13 * 26.9
57.297
3282.88 deg / 360 deg = 9.12
It makes 9 complete revolutoins
The correct answer is letter b.
To find the answer follow the following steps.
1. 6524.96 x .25 = X
2. 1631.24 = X
This works for all of the given answers to find the correct answer.