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:
Final speed of car = 12 m/s
Explanation:
We have equation of motion v = u + at, where v is final velocity, u is initial velocity, a is acceleration and t is time.
a) A cart starts from rest and accelerates at 4.0 m/s² for 5.0 s
v = ?
u = 0 m/s
a = 4.0 m/s²
t = 5 s
v = u + at = 0 + 4 x 5 = 20 m/s
b) Then maintains that velocity for 10 s
v = ?
u = 20 m/s
a = 0 m/s²
t = 10 s
v = u + at = 20 + 0 x 10 = 20 m/s
c) Then decelerates at the rate of 2.0 m/s² for 4.0 s
v = ?
u = 20 m/s
a = -2.0 m/s²
t = 4 s
v = u + at = 20 + -2 x 4 = 12 m/s
Final speed of car = 12 m/s
Answer:
28√3 m
Explanation:
A = vertex where receiver is placed
S = focus
Bp = r = radius of the outside edge
Bc = 2r = diameter
The full explanation is shown in the picture attached herewith. Thank you and i hope it helps.
Answer: The paper airplane will create a curved path towards the floor as it is pulled toward <u><em>Earth's center.</em></u>
Explanation: The paper airplane will be pulled to the center because <u><em>Earth has a much greater mass than objects on its surface.</em></u> And it will curve because of the amount of <u><em>force</em></u> you are putting onto the plane.
Rw^2 = GmM/r^2
<span> Leads to
</span><span> w^2 r^3 = GM
</span><span> (2pi /T) ^2 r^3 = GM
</span><span> 4pi^2 r^3 = GM T^2
</span><span> r^3 = GM T^2 / 4pi^2
</span><span> Work out r^3 then r.
</span> T = 125 min = 125(60) = 7500 s
<span> R = 6.38E6 m
</span><span> m = 5.97E24 kg
</span><span> G = 6.673E-11
</span> r=<span>
8279791.78</span><span> m
Since r = radius R of Earth + height above urface,h
</span><span> h = r - R = </span><span>
8279791.78 - </span>6.38E6 = <span>
<span>1899791.78 m
h=</span></span><span>
<span>1899.79178 Km</span></span>
