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ki77a [65]
3 years ago
9

The moon's orbit around the Earth will advance in one day: 1° 13° 27° 29°

Physics
2 answers:
s2008m [1.1K]3 years ago
4 0
13 is the correct answer
Sever21 [200]3 years ago
3 0
<span>The moon's orbit around the Earth will advance in one day:

1°
13° correct answer 
27°
29°</span><span />
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The answer is "1.144 \times 10^{-4} \ m".

Explanation:

w=\frac{2.44 \lambda L}{D}\\\\D=\frac{2.44 \lambda L}{w}\\\\

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

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

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

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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

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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

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