Two tiny metal spheres A and B, mass mA $ 5.00 g and mB $ 10.0 g, have equal positive charge q $ 5.00 mC. The spheres are connected by a massless nonconducting string of length d $ 1.00 m, which is much greater than the radii of the spheres. (a) What is the electric potential energy of the system? (b) Suppose you cut the string. At that instant, what is the acceleration of each sphere? (c) A long time after you cut the string, what is the speed of each sphere?

Given Information:  

Mass of sphere A = m1 = 5 g

Mass of sphere B = m2 = 10 g

Charge = q1 = q2 = q = 5 mC = 5x10⁻³ C

Length = d = 1 m

Required Information:  

Potential energy = ?  

Acceleration = ?

Speed = ?

U = 2.247x10⁵ j

a1 = 4.49x10⁷ m/s²

a2 = 2.247x10⁷ m/s²

v1 = -7740.8 m/s

v2 = 3870.4 m/s

Explanation:  

(a) What is the electric potential energy of the system?

The electric potential energy can be found using

U = q²/4πe₀d

Where 1/4πe₀ is equal to 8.99x10⁹

U = ( 5x10⁻³)²*8.99x10⁹/1

U = 2.247x10⁵ j

(b) Suppose you cut the string.At that instant, what is the acceleration of each sphere?

Sphere A:

As we know from Newton's second law of motion

F1 = m1a1

We need the force, that can be found using Coulomb's law

F1 = q²/4πe₀d²

F1 = (5x10⁻³)²*8.99x10⁹/1²

F1 = 2.247x10⁵ N

a1 = F1/m1

a1 = 2.247x10⁵/0.005

a1 = 4.49x10⁷ m/s²

Sphere B:

F2 = q²/4πe₀d²

F2 = (5x10⁻³)²*8.99x10⁹/1²

F2 = 2.247x10⁵ N

a2 = F2/m2

a2 = 2.247x10⁵/0.010

a2 = 2.247x10⁷ m/s²

(c) A long time after you cut the string, what is the speed of each sphere?

The potential energy would be converted into kinetic energy

0.5m1v1² + 0.5m2v2² = 2.247x10⁵

According to the equation of conservation of linear momentum

m1v1 + m2v2 = 0

v1 = -m2/m1(v2)

v1² = m2²/m1²(v2²)

0.5m1(m2²/m1²(v2²))

so

0.5v2²((m2²/m1)+m2) = 0.5v2²((0.01²/0.005)+0.01) = 0.5v2²(0.03)

0.5v2²(0.03) = 2.247x10⁵

v2² = 2.247x10⁵/0.015

v2 = √1.49x10⁷

v2 = 3870.4 m/s

v1 = -m2/m1(v2)

v1 = -0.01/0.005(3870.4)

v1 = -7740.8 m/s

The negative sign means that both spheres move in opposite directions.

scattering of light by tiny particles in air to produce the colors we see on sky, is dependent on the following factors:

(i) wavelength of light

(ii) size of the scattering particle  

(iii) distance traveled by light through the atmosphere.

according to rayleigh's law of scattering, the amount of scattering is inversely proportional to the fourth power of its wavelength. shorter the wavelength, greater is its scattering. hence violet/blue light is scattered more than red.

sun's light is a composite light ( visible light) with wavelengths in the range 400 nm (violet) to 700 nm (red). as sunlight passes through our atmosphere, the molecules of air scatter light according to their wavelengths. during sunrise and sunset, the sun's position is below the horizon. hence light from the sun travels a greater distance in the atmosphere when compared to its position a noon.

as sunlight enters the earth's atmosphere, blue light is scattered away and at the end of its journey in the sky, only red and orange wavelengths remain in the beam.

hence at sunrise and sunset, the sky appears reddish orange.

note: the given question is incomplete.

complete question is given below.

which is an example of kinetic energy?

a. the energy stored in ethanol  

b. a hockey puck sliding across the ice  

c. a compressed spring  

d. a ball sitting at the top of a ramp

solution: the kinetic energy is the energy posses by the virtue of the motion of the object.

in the given options, a, c and d are the example of potential energy because it is the energy posses by the virtue of the rest of the object. but option b has the kinetic energy as the object is in motion.

hence, a hockey puck sliding across the ice   is the apex example of kinetic energy.

so, the option (b) a hockey puck sliding across the ice   is the correct option.