Answer:
Ф_cube /Ф_sphere = 3 /π
Explanation:
The electrical flow is
Ф = E A
where E is the electric field and A is the surface area
Let's shut down the electric field with Gauss's law
Фi = ∫ E .dA =
/ ε₀
the Gaussian surface is a sphere so its area is
A = 4 π r²
the charge inside is
q_{int} = Q
we substitute
E 4π r² = Q /ε₀
E = 1 / 4πε₀ Q / r²
To calculate the flow on the two surfaces
* Sphere
Ф = E A
Ф = 1 / 4πε₀ Q / r² (4π r²)
Ф_sphere = Q /ε₀
* Cube
Let's find the side value of the cube inscribed inside the sphere.
In this case the radius of the sphere is half the diagonal of the cube
r = d / 2
We look for the diagonal with the Pythagorean theorem
d² = L² + L² = 2 L²
d = √2 L
we substitute
r = √2 / 2 L
r = L / √2
L = √2 r
now we can calculate the area of the cube that has 6 faces
A = 6 L²
A = 6 (√2 r)²
A = 12 r²
the flow is
Ф = E A
Ф = 1 / 4πε₀ Q/r² (12r²)
Ф_cubo = 3 /πε₀ Q
the relationship of these two flows is
Ф_cube /Ф_sphere = 3 /π
Ans) A) Centripetal force will be doubled.
See centripetal force F = mv^2/r
That means centripetal force is directly proportional to the mass of the particle
So, if we double the mass, centripetal force will be increased by twofolds.
So, option A) is correct.
Now, looking at the other options,
B) says centripetal force is unaltered which is incorrect as centripetal force has been altered and increased twofold.
Option C) and D) reduces centripetal force which are also not possible here.
So, only Option A) is correct
Explanation:
It is given that,
Length of the helicopter, l = 3.1 m
The helicopter rotates, the length of helicopter will become the radius of circular path, r = 3.1 m
Angular speed of the helicopter, 
(a) The centripetal acceleration in terms of angular velocity is given by :



(b) Let v is the linear speed of the tip. The relation between the linear and angular speed is given by :


v = 90.89 m/s

Hence, this is the required solution.
Answer:
Light consists of photons, which are produced when an object's atoms heat up. Light travels in waves and is the only form of energy visible to the human eye.
Explanation: