Answer:
r = 4.44 m
Explanation:
For this exercise we use the Archimedes principle, which states that the buoyant force is equal to the weight of the dislodged fluid
B = ρ g V
Now let's use Newton's equilibrium relationship
B - W = 0
B = W
The weight of the system is the weight of the man and his accessories (W₁) plus the material weight of the ball (W)
σ = W / A
W = σ A
The area of a sphere is
A = 4π r²
W = W₁ + σ 4π r²
The volume of a sphere is
V = 4/3 π r³
Let's replace
ρ g 4/3 π r³ = W₁ + σ 4π r²
If we use the ideal gas equation
P V = n RT
P = ρ RT
ρ = P / RT
P / RT g 4/3 π r³ - σ 4 π r² = W₁
r² 4π (P/3RT r - σ) = W₁
Let's replace the values
r² 4π (1.01 10⁵ / (3 8.314 (70 + 273)) r - 0.060) = 13000
r² (11.81 r -0.060) = 13000 / 4pi
r² (11.81 r - 0.060) = 1034.51
As the independent term is very small we can despise it, to find the solution
r = 4.44 m
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:
1) The magnetic field outside the loop is zero.
In region III the magnetic fields due to the two wire loops point in the opposite direction andhence cancel each other. Therefore the magnetic field is zero in region I, III and V
The diagram is attached
<span>As a sound source gets closer, both the volume and the pitch of the sound increased. Then, as the sound source passed by you, both the volume and the pitch of the sound decreased.
Hope this helps</span>