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
Complete Question
The complete question is shown on the first uploaded image
Answer:
a

b

Explanation:
From the question we are told that
The image distance is 
The value of the image is negative because it is on the same side with the corrective glasses
The object distance is 
The reason object distance is because the object father than it being picture by the eye
General focal length is mathematically represented as

substituting values

=> 
Generally the power of the corrective lens is mathematically represented as

substituting values


Since I'm assuming that its perfectly elastic, considering there's not enough information given, so I think that no energy is dissipated in the collision
hmax = h - d + { [ mpvp - mb√(2gd) ] / (mp+mb) }² / (2g)
Answer:
Explanation:
Potential due to a charged metallic sphere having charge Q and radius r on its surface will be
v = k Q / r . On the surface and inside the metallic sphere , potential is the same . Outside the sphere , at a distance R from the centre potential is
v = k Q / R
a ) On the surface of the shell , potential due to positive charge is
V₁ = 
On the surface of the shell , potential due to negative charge is
V₁ = 
Total potential will be zero . they will cancel each other.
b ) On the surface of the sphere potential
= 
= 22.5 x 10⁵ V
On the surface of the sphere potential due to outer shell
= 
= -9 x 10⁵
Total potential
=( 22.5 - 9 ) x 10⁵
= 13.5 x 10⁵ V
c ) In the space between the two , potential will depend upon the distance of the point from the common centre .
d ) Inside the sphere , potential will be same as that on the surface that is
13.5 x 10⁵ V.
e ) Outside the shell , potential due to both positive and negative charge will cancel each other so it will be zero.