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.
The standard unit is KW/hr, = 1,000W/hr.
(85 + 60) = 145W.
You need to find its fraction of 1,000W., so (145/1000) = 0.145 KWH.
(0.145 x 10p) = 1.45p. per hr.
Answer:
29.4m/s
Explanation:
Given parameters:
Time = 3s
Unknown:
Average velocity = ?
Solution:
To solve this problem, we use the expression below:
v = u + gt
v is the average velocity
u is the initial velocity = 0m/s
g is the acceleration due to gravity = 9.8m/s²
t is the time
So;
v = 0 + (9.8 x 3) = 29.4m/s
The horizontal force applied is 160 N while the velocity is 2.03 m/s.
<h3>What is the speed of the car?</h3>
The work done by the car is obtained as the product of the force and the distance;
W = F x
F = ?
x = 30.0 m
W = 4,800 J
F = 4,800 J/30.0 m
F = 160 N
But F = ma
a = F/m
a = 160 N/2.30 ✕ 10^3-kg
a= 0.069 m/s
Now;
v^2 = u^2 + 2as
u = 0/ms because the car started from rest
v = √2as
v = √2 * 0.069 * 30
v = 2.03 m/s
Learn more about force and work:brainly.com/question/758238
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