Aflywheel in the form of a heavy circular disk of diameter 0.612 m and mass 301 kg is mounted on a frictionless bearing. a motor connected to the flywheel accelerates it from rest to 1060 rev/min. a) what is the moment of inertia of the flywheel? answer in units of kg · m2 .b)how much work is done on it during this acceleration? answer in units of j. c)after 1060 rev/min is achieved, the motor is disengaged. a friction brake is used to slow the rotational rate to 502 rev/min. what is the magnitude of the energy dissipated as heat from the friction brake? answer in units of j.

a)   I= 56.37 kg m², b)  W = 3.47 10⁵ J

c) ΔK = - 2.69 10⁵ J

Explanation:

a) The moment of inertia of the steering wheel is the moment of inertia of a disc with a shaft that passes through its center

      I = ½ M R²

      I = ½ 301  0.612²

      I= 56.37 kg m²

b) Work is equal to the change in kinetic energy

     W = ΔK = ½ I w² - ½ I w₀²

As part of the rest the initial angular velocity is zero

     W = ½ I w²

Let's reduce the angular velocity to SI units

     w = 1060 rev / min (2π rad / 1rev) (1min / 60s) = 111 rad / s

     W = ½ 56.37 111²

     W = 3.47 10⁵ J

c) Dissipated energy is equal to work

    W = ΔK

    W = ½ I ² - ½ I w₀²

For this case

    w₀ = 111 rad / s

      = 502 rav / min (2pirad / 1rev) (1min / 60s) = 52,569 rad / s

     ΔK = ½ 56.37 (52,569² - 111²)

     ΔK = - 2.69 10⁵ J

nah.   i'm just sleepy.

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.