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AveGali [126]
2 years ago
5

If the moon's acceleration due to gravity caused by its gravitational field is one-sixth that of the earth, what is its accelera

tion at a point in space 3R (three moon radii) from its center? (Use the earth's value of "g" as your significant figure reference; i.e., use 2 SF.)
_______ m/s^2
Physics
1 answer:
EastWind [94]2 years ago
7 0

To make it easy, let's call the moon's surface gravity 'Q'.  We know that Q is 1/6 of 'g', but let's just hold onto that for a minute.   Let's first work out what it is at 3 moon radii from the moon's center, and once we have that, relate it back to the Earth.

We know that the strength of gravity is inversely proportional to the square of the distance between the centers of the two objects.  On the moon's surface, you're 1 Moon radius from the center.  At 3 Moon radii from the center, you're 3 times as far from the center, so the gravity out there is (1/3²) = 1/9 of the gravity on the surface.

So at 3 Moon radii from the surface, the Moon's gravity is ( Q/9 ) .  

So far, so good. Now, we know that Q is (1/6) x (Earth 'g').

Moon's gravity at 3 Moon radii = Q/9

Substitute g/6 for Q.

Moon's gravity at 3 Moon radii = (g/6) / 9 .

Moon's gravity at 3 Moon radii = g/54

(9.8 m/s) / 54 = <em>0.18 m/s²</em>

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Star 1 - 4 hours right ascension
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 Subtracting hours right ascension
 4 hours right ascension - 3 hours right ascension = 1 hours right ascension.
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 star 1 will rise 1 hour before star 2
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How does Electromagnetic Energy Travel?
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Perpendicular to the electric field and magnetic field

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they does not require any media to travel. It has two perpendicular electric field and the magnetic field which are perpendicular to each other

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3 years ago
Water at the top of a slope has potential energy. true or false
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You place a light bulb 8 cm in front of a concave mirror. You then move a sheet of paper back and forth in front of the mirror.
Alika [10]

sorry - late reply...just stumbled across tis...hope u can still use it :)


By the mirror equation: 1/di + 1/do = 1/f

<span>
</span>

<span>where di = distance to image = +12cm (+ for real image)</span>


and do = distance to object = +8cm


Substitute and solve for f, the focal length

<span><span>
1/12 + 1/8 = 1/f
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1/f = (8 + 12) / 12 * 8 = 20/96
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5 0
3 years ago
A flutist assembles her flute in a room where the speed of sound is 342 m/s. When she plays the note A, it is in perfect tune wi
sertanlavr [38]

Answer:

5.15348 Beats/s

4.55 mm

Explanation:

v_1 = Velocity of sound = 342 m/s

v_2 = Velocity of sound = 346 m/s

f_1 = First frequency = 440 Hz

Frequency is given by

f_2=\frac{v_2}{2L_1}\\\Rightarrow f_2=\frac{346}{2\times 0.38863}\\\Rightarrow f_2=445.15348\ Hz

Beat frequency is given by

|f_1-f_2|=|440-445.15348|=5.15348\ Beats/s

Beat frequency is 5.15348 Hz

Wavelength is given by

\lambda_1=\frac{v_1}{f}\\\Rightarrow \lambda_1=\frac{342}{440}\\\Rightarrow \lambda_1=0.77727\ m

Relation between length of the flute and wavelength is

\lambda_1=2L_1\\\Rightarrow L_1=\frac{\lambda_1}{2}\\\Rightarrow L_1=\frac{0.77727}{2}\\\Rightarrow L_1=0.38863\ m

At v = 346 m/s

\lambda_2=\frac{v_2}{f}\\\Rightarrow \lambda_2=\frac{346}{440}\\\Rightarrow \lambda_1=0.78636\ m

L_2=\frac{\lambda_2}{2}\\\Rightarrow L_2=\frac{0.78636}{2}\\\Rightarrow L_2=0.39318\ m

Difference in length is

\Delta L=L_2-L_1\\\Rightarrow \Delta L=0.39318-0.38863\\\Rightarrow \Delta L=0.00455\ m=4.55\ mm

It extends to 4.55 mm

7 0
3 years ago
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