The magnitude of the change in momentum of the stone is about 18.4 kg.m/s

<h3>Further explanation</h3>
Let's recall Impulse formula as follows:

<em>where:</em>
<em>I = impulse on the object ( kg m/s )</em>
<em>∑F = net force acting on object ( kg m /s² = Newton )</em>
<em>t = elapsed time ( s )</em>
Let us now tackle the problem!

<u>Given:</u>
mass of ball = m = 0.500 kg
initial speed of ball = vo = 20.0 m/s
final kinetic energy = Ek = 70% Eko
<u>Asked:</u>
magnitude of the change of momentum of the stone = Δp = ?
<u>Solution:</u>
<em>Firstly, we will calculate the final speed of the ball as follows:</em>



→ <em>negative sign due to ball rebounds</em>


<em>Next, we could find the magnitude of the change of momentum of the stone as follows:</em>

![\Delta p_{stone} = - [ mv - mv_o ]](https://tex.z-dn.net/?f=%5CDelta%20p_%7Bstone%7D%20%3D%20-%20%5B%20mv%20-%20mv_o%20%5D)
![\Delta p_{stone} = m[ v_o - v ]](https://tex.z-dn.net/?f=%5CDelta%20p_%7Bstone%7D%20%3D%20m%5B%20v_o%20-%20v%20%5D)
![\Delta p_{stone} = m[ v_o + v_o\sqrt{0.7} ]](https://tex.z-dn.net/?f=%5CDelta%20p_%7Bstone%7D%20%3D%20m%5B%20v_o%20%2B%20v_o%5Csqrt%7B0.7%7D%20%5D)
![\Delta p_{stone} = mv_o [ 1 + \sqrt{0.7} ]](https://tex.z-dn.net/?f=%5CDelta%20p_%7Bstone%7D%20%3D%20mv_o%20%5B%201%20%2B%20%5Csqrt%7B0.7%7D%20%5D)
![\Delta p_{stone} = 0.500 ( 20.0 ) [ 1 + \sqrt{0.7} ]](https://tex.z-dn.net/?f=%5CDelta%20p_%7Bstone%7D%20%3D%200.500%20%28%2020.0%20%29%20%5B%201%20%2B%20%5Csqrt%7B0.7%7D%20%5D)


<h3>Learn more</h3>

<h3>Answer details</h3>
Grade: High School
Subject: Physics
Chapter: Dynamics
Answer:
Explanation:
The specific heat of gold is 129 J/kgC
It's melting point is 1336 K
It's Heat of fusion is 63000 J/kg
Assuming that the mixture will be solid, the thermal energy to solidify the gold has to be less than that needed to raise the solid gold to the melting point. So,
The first is E1 = 63000 J/kg x 1.5 = 94500 J
the second is E2 = 129 J/kgC x 2 kg x (1336–1000)K = 86688 J
Therefore, all solid is not correct. You will have a mixture of solid and liquid.
For more detail, the difference between E1 and E2 is 7812 J, and that will melt
7812/63000 = 0.124 kg of the solid gold
Answer:
Explanation:
Using the efficiency formula;
Efficiency = Work done by the machine (output)/work done on the machine (input) ×100%
Efficiency =w/50 ×100
90 = 100w/50
Cross multiply
90×50 = 100W
4500 = 100W
W = 4500/100
W = 45Joules
Hence the lever does 45Joules of work on its load
2) Mechanical Advantage= Load/Effort
Given
MA = 4
Load = 500N
4 = 500/Effort
Effort = 500/4
Effort =125N
Hence the effort required to lift the load is 125N
Answer:
fcosθ + Fbcosθ =Wtanθ
Explanation:
Consider the diagram shown in attachment
fx= fcosθ (fx: component of friction force in x-direction ; f: frictional force)
Fbx= Fbcosθ ( Fbx: component of braking force in x-direction ; Fb: braking force)
Wx= Wtanθ (Wx: component of weight in x-direction ; W: Weight of semi)
sum of x-direction forces = 0
fx+ Fbx=Wx
fcosθ + Fbcosθ =Wtanθ