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kykrilka [37]
2 years ago
5

Would the phases of the moon be affected if the moon did not make one rotation for each revolution of earth. Explain .

Physics
1 answer:
elixir [45]2 years ago
3 0

Answer:

No. If the moon did not rotate at all we would still see the same visible "fraction" of the moon's surface.

At the full moon  phase we would still the entire surface of  1/2 of the moon regardless of which side of the moon we were observing.

Actually the moon does not make one rotation for each revolution of the earth, but rather, one rotation for each complete revolution of the moon around the the earth. That's why we always see the same side of the moon from earth. The back side of the moon is never visible from earth.

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An infinite conducting cylindrical shell of outer radius r1 = 0.10 m and inner radius r2 = 0.08 m initially carries a surface ch
gayaneshka [121]

Answer:

Explanation:

Solution is in the picture attached

8 0
3 years ago
Which dog has the most kinetic energy? A. A dog of mass 12 kg running with speed 6 m/s B. A dog of mass 10 kg running with speed
Nonamiya [84]

Answer:

A

Explanation:

KE = 1/2 m v^2

     the greatest is  A

5 0
1 year ago
Juan is receiving bills from the hospital even though his insurance company said his procedure would be covered in full. What st
BaLLatris [955]

==> Jot down notes before and after making each call, and avoid calling during business hours when people are busy.

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5 0
2 years ago
Except for the nodes on a standing wave, what is the frequency f of the points executing simple harmonic motion?
Katen [24]

Take into account that in a standing wave, the frequency f of the points executing simple harmonic motion, is simply a multiple of the fundamental harmonic fo, that is:

f = n·fo

where n is an integer and fo is the first harmonic or fundamental.

fo is given by the length L of a string, in the following way:

fo = v/λ = v/(L/2) = 2v/L

becasue in the fundamental harmonic, the length of th string coincides with one hal of the wavelength of the wave.

6 0
1 year ago
A skateboarder, starting from rest, rolls down a 12.8-m ramp. When she arrives at the bottom of the ramp her speed is 8.89 m/s.
melamori03 [73]

Answer:

a) a = 3.09 m/s²

b) aₓ = 2.60 m/s²

Explanation:

a) The magnitude of her acceleration can be calculated using the following equation:

V_{f}^{2} = V_{0}^{2} + 2ad

<u>Where</u>:

V_{f}: is the final speed = 8.89 m/s

V_{0}: is the initial speed = 0 (since she starts from rest)

a: is the acceleration

d: is the distance = 12.8 m    

a = \frac{V_{f}^{2}}{2d} = \frac{(8.89 m/s)^{2}}{2*12.8 m} = 3.09 m/s^{2}

Therefore, the magnitude of her acceleration is 3.09 m/s².              

b) The component of her acceleration that is parallel to the ground is given by:

a_{x} = a*cos(\theta)

<u>Where</u>:

θ: is the angle respect to the ground = 32.6 °

a_{x} = 3.09 m/s^{2}*cos(32.6) = 2.60 m/s^{2}

Hence, the component of her acceleration that is parallel to the ground is 2.60 m/s².

I hope it helps you!

7 0
3 years ago
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