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expeople1 [14]
3 years ago
12

Many spacecraft have visited Mars over the years. Mars is smaller than the earth and has correspondingly weaker surface gravity.

On Mars, the free-fall acceleration is only 3.8 m
Physics
1 answer:
sertanlavr [38]3 years ago
7 0

Incomplete question.The complete one is here

Spacecraft have been sent to Mars in recent years. Mars is smaller than Earth and has correspondingly weaker surface gravity. On Mars, the free-fall acceleration is only 3.8m/s2. What is the orbital period of a spacecraft in a low orbit near the surface of Mars?

Answer:

T=5900s=99min

Explanation:

Given

r_{satelite}=r_{mars}=3.37*10^{6}m\\ g_{mars}=3.8m/s^{2}\\

To find

orbital period of a spacecraft T

Solution

An the initial calculating is computing the angular velocity of satellite :

w=\frac{2\pi }{T}\\ w=\frac{2\pi }{110min}(1min/60s)\\ w=9.52*10^{-4}rad/s

Computing T

T=\frac{2\pi }{w}\\ as\\w=\sqrt{\frac{a}{r} }\\ So\\T=\frac{2\pi }{\sqrt{\frac{a}{r} }} \\T=\frac{2\pi }{\sqrt{\frac{3.8m/s^{2} }{3.37*10^{6} m} }}\\T=5900s=99min

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3 years ago
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USPshnik [31]

Given Information:  

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speed of sound = v = 344 m/s

linear density = μ = 0.660 g/m = 0.00066 kg/m

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f = 344/0.391

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The wavelength of the string is

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and finally the length of the vibrating string is

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L = λ/2

L = 0.5596/2

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Therefore, the vibrating section of the violin string is 0.28 m long.

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