A person sitting on the outer edge of a merry-go-round travels ____________

Light incident on a lake surface is partly reflected and partly refracted.What is the differences between the reflected ray and

iren2701 [21]

Answer: As per the question, a ray of light is incident on a surface and it is partly reflected and refracted. The incident light is an unpolarised light. The reflected light is partially polarised.

If the angle of incidence becomes equal to the Brester angle (polarising angle), then the reflected light is completely plane polarised.

5 0

3 years ago

Read 2 more answers

Letter C on the map labels which feature of the Middle East?

likoan [24]

Honest, the map is so tiny, and so fuzzy when I blow it up, I really can't see anything on it clearly. But I think maybe I do see a letter ' C ' in the eastern Mediterranean, with a curved line over to the southern Gaza strip, where it meets Sinai. So I'll say it's the Gaza Strip.

6 0

3 years ago

The table below shows a sample wavelength of each type of electromagnetic radiation, along with its frequency and energy.

trasher [3.6K]

Answer:

D lower energy waves is most likely the safest if one is exposed to.

5 0

3 years ago

The suspension system of a 1700 kg automobile "sags" 7.7 cm when the chassis is placed on it. Also, the oscillation amplitude de

spin [16.1K]

Answer:

the spring constant k = $5.409*10^4 \ N/m$

the value for the damping constant $\\ \\b = 1.518 *10^3 \ kg/s$

Explanation:

From Hooke's Law

$F = kx\\\\k =\frac{F}{x}\\\\where \ F = mg\\\\k = \frac{mg}{x}\\\\given \ that:\\\\mass \ of \ each \ wheel = 425 \ kg\\\\x = 7.7cm = 0.077 m\\\\g = 9.8 \ m/s^2\\\\Then;\\\\k = \frac{425 \ kg * 9.8 \ m/s^2}{0.077 \ m}\\\\k = 5.409*10^4 \ N/m$

Thus; the spring constant k = $5.409*10^4 \ N/m$

The amplitude is decreasing 37% during one period of the motion

$e^{\frac{-bT}{2m}}= \frac{37}{100}\\\\e^{\frac{-bT}{2m}}= 0.37\\\\\frac{-bT}{2m} = In(0.37)\\\\\frac{-bT}{2m} = -0.9943\\\\b = \frac{2m(0.9943)}{T}\\\\b = \frac{2m(0.9943)}{\frac{2 \pi}{\omega}}\\\\b = \frac{m(0.9943) \ ( \omega) )}{ \pi}$

$b = \frac{m(0.9943)(\sqrt{\frac{k}{m})}}{\pi}\\\\b = \frac{425*(0.9943)(\sqrt{\frac{5.409*10^4}{425}) } }{3.14}\\\\b = 1518.24 \ kg/s\\\\b = 1.518 *10^3 \ kg/s$