Answer:
correct option is b. 31.3 m/s
Explanation:
given data
artificial gravity a1 = 1 g
artificial gravity a2 = 2 g
diameter = 100 m
radius r= 50 m
speed v1 = 22.1 m/s
solution
As acceleration is ∝ v²
so we can say
.....................1
put here value
solve it
v2 =
× 22.1
v2 = 31.25 m/s
so correct option is b. 31.3 m/s
Complete Question
The complete question is shown on the first uploaded image
Answer:
The workdone is 
Explanation:
From the question we are told that
The initial Volume is 
The final volume is 
The external pressure is
Generally the change in volume is

Substituting values we have


Generally workdone is mathematically represented as

W is negative because the working is done on the environment by the system which is indicated by volume increase
Substituting values


Now 
Therefore 

Answer:
E = 0 r <R₁
Explanation:
If we use Gauss's law
Ф = ∫ E. dA =
/ ε₀
in this case the charge is distributed throughout the spherical shell and as we are asked for the field for a radius smaller than the radius of the spherical shell, therefore, THERE ARE NO CHARGES INSIDE this surface.
Consequently by Gauss's law the electric field is ZERO
E = 0 r <R₁
Answer:

Explanation:
From the concept of fluids mechanics we know that if a tank has a hole at the bottom, the equation that we need to use is:

Since we know gravity and its hight
