A force of 1.50 N acts on a 0.20 kg trolley so as to accelerate it along an air track. The track and force are horizontal and in line. How fast is the trolley going after acceleration from rest through 30 cm, if friction is negligible?

The trolley is moving at 2.12m/s

Explanation:

Given from the question that the track and force are horizontal and inline, we have that

F= ma

where F= force= 1.50 N

m= mass = 0.2kg

therefore Acceleration , a = F/ m= 1.50/0.2 =7.5 m/s^2

To find how fast the trolley  is going ie the Velocity, v

Having that

initial velocity at rest , u = 0,

acceleration a = 7.5 and

and distance, s = 30 cm =  30/100 = 0.30 m

we use motion equation that

v² = u² + 2 a s

v² = 0² + 2 x 7.5  x 0.30

v² = 4.5

v =

v = 2.12 m/s

work: w = fd (given)

work done by chet   =50*1 = 50 j

work done by mika =40*2 = 80 j

work done by sara =30*1.5 = 45 j

work done by bill =60*0.5 = 30 j

students in order from the greatest amount of work done to the least is mika> chet> sara> bill

hence option (b) is correct.

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.