Here's a basketball problem: a 87.2 kg basketball player is running in the positive direction at 7.0 m/s. she is met head-on by a 102.0 kg player traveling at 5.2 m/s toward her. if the 102.0 kg player is knocked backwards at 2.9 m/s, what is the resulting velocity of the 87.2 kg player?

2.47 m/s backwards

Explanation:

From the law of conservation of momentum,

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂ Equation 1

Where m₁ and m₂ = mass of the first basketball player and second basket ball player respectively, u₁ and u₂ = initial velocity of the first basket player and the second basketball player respectively, v₁ and v₂ = The final velocity of the first basket ball player and second basket ball player respectively.

Making v₁ the subject of the equation,

v₁ = (m₁u₁ + m₂u₂ - m₂v₂)/m₁ Equation 2.

Given: m₁ = 87.2 kg, m₂ = 102.0 kg, u₁ = 7.0 m/s, u₂ = -5.2 m/s, v₂ = 2.9 m/s

Note: u₂ is negative because it moves towards the first basket ball player.

Substitute into equation 2

v₁ = [87.2(7.0)+102(-5.2) - (102×2.9)]/87.2

v₁ = (610.4-530-295.8)/87.2

v₁ = -215.4/87.2

v₁ = -2.47 m/s.

Thus the velocity of the 87.2 kg player = 2.47 m/s backwards.

answer: decibel

explanation:

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.