The magnitude of force on the elevator cable that would be needed to lower the cat/elevator pair is 198 Newton.
<u>Given the following data:</u>
- Acceleration = 2
To determine the magnitude of force on the elevator cable that would be needed to lower the cat/elevator pair, we would apply Newton's Second Law of Motion:
First of all, we would calculate the total mass of the cat/elevator pair.
Total mass = 99 kilograms
Mathematically, Newton's Second Law of Motion is given by this formula;
Substituting the given parameters into the formula, we have;
Net force = 198 Newton
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Answer:
The average force ≅ 519.44 N.
Explanation:
Impulse = change in momentum of a body
i.e Ft = m(v - u)
where F is the force, t is the time, m is the mass of the body, v is the final velocity and u is the initial velocity.
m = 55.0 g (0.055 Kg), t = 0.00360 s, v = 34.0 m/s, since the ball was initially at rest; u = 0 m/s
So that,
F x 0.00360 = 0.055(34 - 0)
F x 0.00360 = 0.055 x 34
= 1.87
F = 
= 519.4444
The average force exerted on the ball by the club is approximately 519.44 N.
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:
its the top 3 can confirm on plato
Explanation:
V=IR
The more V, the more I
