If you run the simulation a few times, you should see that for this particular pair of block and disc the block's position has the same numerical value as the disc's angular position, and the block's velocity has the same numerical value as the disc's angular velocity. Which of the following pairs of quantities must also have the same numerical values for the block and disc (assuming we use standard meter-kilogram-second units)? Select all that apply.

a. The work done by the force on the block and the work done by the torque on the disc.
b. The acceleration of the block and the angular acceleration of the disc.
c. The mass of the block and the mass of the disc.
d. The translational kinetic energy of the block and the rotational kinetic energy of the disc.
e. The momentum of the block and the angular momentum of the disc

Explanation:

C) is incorrect.  The mass is equivalent to the moment of inertia in an angular formula.

The others appear all correct.

T * theta = F * r * theta = F * (r theta) = fx = work

This is the only one that needed explanation.  The others should be obvious.

1 the 4 is sig

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.