The electric flux through the hole is
.
- Electric flux is the number of electric field lines cutting through the surface and is measured as surface intregal of electric field over that surface
- Mathematically it is given by
where E is the electric field and A is the area. - Gauss's law states that electric flux through closed surface is equal to the 1 / ε₀ times the charge enclosed by that surface which is given by Ф = q / ε₀ where q is the central charge and ε₀ is the permittivity of the medium.
It is given , hollow sphere of radius 10.0cm surrounds a 10.0-μC charge.
The whole surface of hollow sphere 

Area of the hole ( both side ) 

According to Gauss's theorem, the flow from a particular charge in the center is given by

This flux flows through the surface of the sphere, so the flux per unit area which is given by

Flux through area of hole is given by :

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The answer is : We’ll see the bell move, but we won’t hear it ring. This is because light can travel through vacuum but sound cannot. Sound waves are vibrations of particles in any media, so sound requires a medium to travel, and it cannot travel in a vacuum as there is no particles to vibrate.
Most of the energy will be absorbed by the materials that make up the cars, causing them to deform. The energy will also be converted into sound energy, causing a loud bang upon collision. Also, some energy will be converted to thermal energy, which will cause the cars to heat up slightly.
Work needed: 720 J
Explanation:
The work needed to stretch a spring is equal to the elastic potential energy stored in the spring when it is stretched, which is given by

where
k is the spring constant
x is the stretching of the spring from the equilibrium position
In this problem, we have
E = 90 J (work done to stretch the spring)
x = 0.2 m (stretching)
Therefore, the spring constant is

Now we can find what is the work done to stretch the spring by an additional 0.4 m, that means to a total displacement of
x = 0.2 + 0.4 = 0.6 m
Substituting,

Therefore, the additional work needed is

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Answer:
The puck moves a vertical height of 2.6 cm before stopping
Explanation:
As the puck is accelerated by the spring, the kinetic energy of the puck equals the elastic potential energy of the spring.
So, 1/2mv² = 1/2kx² where m = mass of puck = 39.2 g = 0.0392 g, v = velocity of puck, k = spring constant = 59 N/m and x = compression of spring = 1.3 cm = 0.013 cm.
Now, since the puck has an initial velocity, v before it slides up the inclined surface, its loss in kinetic energy equals its gain in potential energy before it stops. So
1/2mv² = mgh where h = vertical height puck moves and g = acceleration due to gravity = 9.8 m/s².
Substituting the kinetic energy of the puck for the potential energy of the spring, we have
1/2kx² = mgh
h = kx²/2mg
= 59 N/m × (0.013 m)²/(0.0392 kg × 9.8 m/s²)
= 0.009971 Nm/0.38416 N
= 0.0259 m
= 2.59 cm
≅ 2.6 cm
So the puck moves a vertical height of 2.6 cm before stopping