Fddgfjdgkdghsreawoejfkurbgkb,xv biusghorsnj lsjrfgWIL4RGBOHGVkwdc
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.
Seems to me that it flies 400 m/s there and 600 m/s back the same distance.
therefore the average of 400 and 600 is 500 m/s. The distance is the same so the normal formula of (d2-d1)/(t2-t1) is applicable.
Answer:
Velocity of the vehicle (v) = 6.5 m/s
Explanation:
Mass of automobile (m) = 1200 kg
Momentum of vehicle (p) = 7800 kg m/s
By using formula of momentum:

"1 watt" means 1 joule of energy per second.
75 W means 75 joules/sec .
Energy = (75 Joule/sec) x (12 min) x (60 sec/min)
Energy = (75 x 12 x 60) (Joule-<em>min-sec</em> / <em>sec-min</em>)
<em>Energy = 54,000 Joules</em>