By v = u - at
<span>=>8 = 12 - a x 0.25 </span>
<span>=>a = 4/0.25 km/hr/sec </span>
<span>=>a = 16km/hr/sec
I hope this helped!</span>
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:
The answer is 100J.
Explanation:
In classical mechanics, kinetic energy (KE) is equal to half of an object's mass (1/2*m) multiplied by the velocity squared. In this question, the mass is equals to 50kg and the velocity is 2m/s
Now,
25kg×4m/s^2 = 100kgm/s^2 or 100J
The magnetic field or force seems to be associated with the lineup of electrons withim the magnet
Answer:
a) load in Newton is 96,138 b) 129.314mm
Explanation:
Stress = force/ area (cross sectional area of the bronze)
Force(load) = 294*10^6*327*10^-6 = 96138N
b) modulus e = stress/ strain
Strain = stress/ e = (294*10^6)/ (121*10^ 9) = 2.34* 10^ -3
Strain = change in length/ original length = DL/ 129
Change in length DL = 129 * 2.34*10^ -3 = 0.31347
Maximum length = change in length + original length = 129.314mm