Where might water be found on the moon

Physics

2 answers:

Cerrena [4.2K]3 years ago

8 0

Water Could be Found in the frozen Ice on the moon. it would most likely be underground however

NNADVOKAT [17]3 years ago

4 0

I think water would be found in the middle of the moon aka the core of the moon. If there was water on the moon it could either be found in rocks or in the core. The reason i said core instead of water is because the moon doesn't really have rocks like Mars does it's sandy or ashy on the moon.

Hope this helped :)

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raketka [301]

Answer:

its A

Explanation:

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2 years ago

Show solution for # 4

velikii [3]

M1 = 750Kg, v1 = 10m/s
m2 = 2500Kg , v2= 0 (because in problem say cuz that object don t move).

The momentum before colision is equal with the momentum after colision:

m1v1 + m2v2 = (m1+m2)v3 => v3 is the velocity after colison and that s u want to caluclate for your problem

=> m1v1 = (m1+m2)v3 => v3 = m1v1/(m1+m2) now u should do the math i think v3 prox 2,4 but not sure u should caculate

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3 years ago

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Delicious77 [7]

Answer: It should be the 3rd option down!

Explanation:

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2 years ago

Read 2 more answers

The frequency of a wave is 200 Hz. The wavelength is 0.1 m. What is the period of the wave?

Aleks04 [339]

The formula for the period of wave is: wave period is equals to 1 over the frequency. $waveperiod=\frac{1}{frequency}$
To get the value of period of wave you need to divide 1 by 200 Hz. However, beforehand, you have to convert 200 Hz to cycles per second. So that would be, 200 cyles per second or 200/s.
By then, you can start the computation by dividing 1 by 200/s. Since 200/s is in fractional form, you have to find its reciprocal form and multiply it to one which would give you 1 (one) second over 200. This would then lead us to the value 0.005 seconds as the wave period.

wave period= 1/200 Hz
Convert Hz to cycles per second first
200 Hz x 1/s= 200/second
Make 200/second as your divisor, so:

wave period= 1/ 200/s

get the reciprocal form of 200/s which is s/200

then you can start the actual computation:

wave period= 1 x s divided by 200

this would give us an answer of 0.005 s.

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3 years ago

On a horizontal surface is located

Ierofanga [76]

By Newton's second law, the net vertical force acting on the object is 0, so that

n - w = 0

where n = magnitude of the normal force of the surface pushing up on the object, and w = weight of the object. Hence n = w = mg = 196 N, where m = 20 kg and g = 9.80 m/s².

The force of static friction exerts up to 80 N on the object, since that's the minimum required force needed to get it moving, which means the coefficient of static friction µ is such that

80 N = µ (196 N) → µ = (80 N)/(196 N) ≈ 0.408

Moving at constant speed, there is a kinetic friction force of 40 N opposing the object's motion, so that the coefficient of kinetic friction ν is

40 N = ν (196 N) → ν = (40 N)/(196 N) ≈ 0.204

And so the closest answer is C.

(Note: µ and ν are the Greek letters mu and nu)

3 0

3 years ago