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aniked [119]
2 years ago
8

A person drops a pebble of mass m1 from a height h, and it hits the floor with kinetic energy KE. The person drops another pebbl

e of mass m2 from a height of 4h, and it hits the floor with the same kinetic energy KE. How do the masses of the pebbles compare
Physics
1 answer:
Leona [35]2 years ago
8 0
<h2>Answer:</h2>

<em>Hello, </em>

<h3><u>QUESTION)</u></h3>

<em>✔ We have: KE = PE (potential energy) </em>

<em>PE = m x g x h </em>

The potential energy that the pebble of mass 1 has is called PE1 and the potential energy that the pebble of mass 2 has is called PE2  

PE1 = PE2 ⇔ PE1/PE2 = 1

\frac{m_1\times g\times h}{m_2\times g\times 4h} = 1 \\ \\  \frac{m_1}{m_2\times 4} = 1 \\ \\  \frac{m_1}{m_2} = 4

The mass m1 is therefore 4 times greater than that of the stone of mass m2.

 

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Afina-wow [57]

Answer:

For A

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For B

Displacement= 1/2*4*4= 8m

5 0
3 years ago
A less massive moving object has an elastic collision with a more massive object that is not moving. Compare the initial velocit
stepladder [879]

Assume that the small-massed particle is m and the heavier mass particle is M.

Now, by momentum conservation and energy conservation:

   mv = mv_{m} + Mv_{M}

   mv^{2} = mv^{2}_{m} + Mv^{2}_{M}

Now, there are 2 solutions but, one of them is useless to this question's main point so I excluded that point. Ask me in the comments if you want the excluded solution too.

   v_{m} = v\frac{m - M}{m + M}\\\\\\v_{M} = v\frac{m + M}{m + M}\\

So now, we see that v_{m} < 0 and v_{M} > 0. So therefore, the smaller mass recoils out.

Hope this helps you!

Bye!

7 0
3 years ago
Which quantity is measured in newton seconds (Ns)?
pshichka [43]

Answer:

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Explanation:

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7 0
2 years ago
A real heat engine operates between temperatures tc and th. during a certain time, an amount qc of heat is released to the cold
tino4ka555 [31]

q_{c} = Heat released to cold reservoir

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t_{c} = temperature of cold reservoir

t_{h} = temperature of hot reservoir

we know that

\frac{q_{c}}{q_{h}}=\frac{t_{c}}{t_{h}}

q_{h} = (\frac{t_{h}}{t_{c}})q_{c}                                eq-1

maximum work is given as

W_{max} = q_{h} - q_{c}

using eq-1

W_{max} =  (\frac{t_{h}}{t_{c}})q_{c} - q_{c}



6 0
3 years ago
Plz help 20 points!!
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Answer:

1-d

2- a

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This is the correct sequence

5 0
2 years ago
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