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ludmilkaskok [199]
3 years ago
13

The sidereal period of the moon around the Earth is 27.3 days. Suppose a satellite were placed in Earth orbit, halfway between E

arth's center and the moon's orbit. Use Kepler's third law to find the period of this satellite. (Just use T2/r3 = constant. No need for Earth's mass or the value of G.) days.
Physics
1 answer:
icang [17]3 years ago
7 0

Answer: 9.7 days

Explanation:

Applying Kepler's 3rd law, we can write the following proportion:

(Tm)² / (dem)³ = (Tsat)² / (dem/2)³

(As the satellite is placed in an orbit halfway between Erth's center and the moon's orbit).

Simplifyng common terms, and solving for Tsat, we have:

Tsat = √((27.3)²/8) = 9.7 days

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Problem 4: a long wire carries current towards east. a positive charge moves westward and just north from the wire. what is the
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The direction of the force experienced by the positive charge is upward.

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(if it was a negative charge, we should have taken the opposite direction)
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3 years ago
Which conditions are low air pressure systems usually associated with?
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cloudy, wet weather                                          

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3 years ago
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What was the long-term effect of the Northwest Ordinance of 1787?
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Answer:

Established a government for the Northwest Territory, outlined the process for admitting a new state to the Union, and guaranteed that newly created states would be equal to the original thirteen states

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Goooogle, so I hope this helps somewhat

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3 years ago
Which choice correctly ranks these items from smallest to largest?
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6 0
3 years ago
An open organ pipe of length 0.47328 m and another pipe closed at one end of length 0.702821 m are sounded together. What beat f
sineoko [7]

Answer:

fb = 240.35 Hz

Explanation:

In order to calculate the beat frequency generated by the first modes of each, organ and tube, you use the following formulas for the fundamental frequencies.

Open tube:

f=\frac{v_s}{2L}         (1)

vs: speed of sound = 343m/s

L: length of the open tube = 0.47328m

You replace in the equation (1):

f=\frac{343m/s}{2(0.47228m)}=362.36Hz      

Closed tube:

f'=\frac{v_s}{4L'}

L': length of the closed tube = 0.702821m

f'=\frac{343m/s}{4(0.702821m)}=122.00Hz

Next, you use the following formula for the beat frequency:

f_b=|f-f'|=|362.36Hz-122.00Hz|=240.35Hz

The beat frequency generated by the first overtone pf the closed pipe and the fundamental of the open pipe is 240.35Hz

7 0
3 years ago
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