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3 years ago

calculate the momentum of 5.07 kg egg that is dropped from a roof and falls 2 seconds before hitting the ground.

1 answer:

8 0

Anything that's dropped on or near the Earth accelerates at the rate of
9.8 m/s² on its way down. The mass of the object makes no difference.
After falling for 2 seconds, it's falling at the rate of 19.6 m/s.

Momentum = (mass) x (speed)

The egg's momentum is (5.07 kg) x (19.6 m/s) = 99.4 kg-m/s .

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Answer:

Explanation:

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$E=13.6 (\frac{1}{n_1^2 } - \frac{1}{n_2^2})eV$

For layman series n₁ = 1 and n₂ = 2 , 3 , 4 , ... etc

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$E_1=13.6 (\frac{1}{1^2 } - \frac{1}{2 ^2})$

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energy of 2 nd line

$E_2=13.6 (\frac{1}{1^2 } - \frac{1}{3 ^2})$

= 12.08 eV

wavelength of photon = 12375 / 12.08 = 1024.4 A

energy of third line

$E_3=13.6 (\frac{1}{1^2 } - \frac{1}{4 ^2})$

12.75 e V

wavelength of photon = 12375 / 12.75 = 970.6 A

energy of fourth line

$E_4=13.6 (\frac{1}{1^2 } - \frac{1}{5 ^2})$

= 13.056 eV

wavelength of photon = 12375 / 13.05 = 948.3 A

energy of fifth line

$E_5=13.6 (\frac{1}{1^2 } - \frac{1}{6 ^2})$

13.22 eV

wavelength of photon = 12375 / 13.22 = 936.1 A

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3 years ago