The force of the tripped catch exerted on the 2.5 Kg ball moving at 8.5 m/s to the Left is 160 N
<h3>Data obtained from the question </h3>
- Initial velocity (u) = 8.5 m/s
- Final velocity (v) = 7.5 m/s
- Time (t) = 5 ms = 0.25 s
- Mass (m) = 2.5 Kg
- Force (F) = ?
<h3>How to determine the force</h3>
The force exerted on the ball can be obtained as follow:
F = m(v + u) / t
F = [2.5(7.5 + 8.5)]/ 0.25
F = 40 / 0.25
F = 160 N
Thus, the force exerted on the ball is 160 N
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To solve this problem it is necessary to apply the concepts related to Newton's second Law and the force of friction. According to Newton, the Force is defined as
F = ma
Where,
m= Mass
a = Acceleration
At the same time the frictional force can be defined as,

Where,
Frictional coefficient
N = Normal force (mass*gravity)
Our values are given as,

By condition of Balance the friction force must be equal to the total net force, that is to say



Re-arrange to find acceleration,



Therefore the acceleration the horse can give is 
It is actually B. Promoter 2.
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Answer:
The magnitude is "3.8 m/s²", in the upward direction.
Explanation:
The given values are:
Mass,
m = 88 kg
Scale reads,
T = 900 N
As we know,
⇒ 
On substituting the given values, we get
⇒ 
⇒ 
Now,
⇒ 
On substituting the given values in the above equation, we get
⇒ 
On subtracting "862.4" from both sides, we get
⇒ 
⇒ 
⇒ 
⇒
(upward direction)
Answer:
2.000.000.000
Explanation:
Apply the formula:
Work = Force . Displacement
W = 500.10 . 400.000 (the 10 come from gravity)
W = 5000 . 400.000
W = 2.000.000.000 Joules
I think it is that, can be wrong.