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Bess [88]
3 years ago
11

Clouds absorb outgoing radiation emitted by earth and reradiate a portion of it back to the surface during _____.

Physics
1 answer:
GarryVolchara [31]3 years ago
4 0
I would say the correct answer is nighttime. The clouds in the atmosphere would absorb the radiation that is emitted by the Earth and at at night 30% of these would be radiated back to the Earth's surface. Hope this answers the question. <span />
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3 years ago
How long is the image of a metrestick formed by a plane mirror?
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The formulas used to analyze the horizontal and vertical motion of projectiles launched at an angle involve the use of which fun
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7 0
3 years ago
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What is the equations to prove the answer is A?
Pepsi [2]

Explanation:

Let's say your mass is M, the ball's mass is m, and the ball's speed is v.  Let's say your final speed is V.

Momentum is conserved.  If you catch the ball, your speed is:

mv = (m + M)V

V = mv / (m + M)

If you deflect the ball, your speed is:

mv = MV + m(-v)

V = 2mv / M

To compare these, we need the common denominator.

If you catch the ball:

V = mMv / (M (m + M))

If you throw the ball:

V = 2m(m + M)v / (M (m + M))

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7 0
3 years ago
50POINTS! What is the escape velocity for lunar module? Lunar module mass 15,200 kg radius of moon 1.74x106m, mass of moon 7.34x
Airida [17]

Answer:

2.73 km/s

Explanation:

The escape velocity of an object in the gravitational field of the moon is (on the surface of the planet)

v=\sqrt{\frac{2GM}{r}}

where

G=6.67\cdot 10^{-11} m^3 kg^{-1} s^{-2} is the gravitational constant

M=7.34\cdot 10^{22} kg is the mass of the Moon

r=1.74\cdot 10^6 m is the radius of the Moon

As we can see, the escape velocity does not depend on the mass of the lunar module.

Substituting the numbers into the formula, we find

v=\sqrt{\frac{2(6.67\cdot 10^{-11})(7.34\cdot 10^{22}kg)}{(1.74\cdot 10^6 m)}}=2732 m/s=2.73 km/s

7 0
3 years ago
Read 2 more answers
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