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3 years ago

A rock hits the ground with a speed of 13 m s−1

Physics

1 answer:

polet [3.4K]3 years ago

4 0

Answer:

3.55kg

Explanation:

K. E = ½mv²

Where m = mass, v = velocity

K. E = 300J

v = 13m/s

Therefore,

K. E = ½mv²

300 = ½ × m × 13²

Multiply both sides by 2

600 = m × 169

Divide both sides by 169

m = 600 ÷ 169

mass = 3.55kg

I hope this was helpful, please mark as brainliest

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Consider a hydrogen atom in the n = 1 state. The atom is placed in a uniform B field of magnitude 2.5 T. Calculate the energy di

dlinn [17]

Answer:

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Explanation:

For n-=1 state hydrogen energy level is split into three componets in the presence of external magnetic field. The energies are,

$E^{+}=E+\mu B$ ,

$E^{-}=E-\mu B$ ,

$E^{0}=E$

Here, E is the energy in the absence of electric field.

And

$E^{+} and E^{-}$ are the highest and the lowest energies.

The difference of these energies

$\Delta E=2\mu B$

$\mu=9.3\times 10^{-24}J/T$ is known as Bohr's magneton.

B=2.5 T,

Therefore,

$\Delta E=2(9.3\times 10^{-24}J/T)\times 2.5 T\\\Delta E=46.5\times 10^{-24}J$

Now,

$Delta E=46.5\times 10^{-24}J(\frac{1eV}{1.6\times 10^{-9}J } )\\Delta E=29.05\times 10^{-5}eV\\Delta E\simeq29\times 10^{-5}eV$

Therefore, the energy difference between highest and lowest energy levels in presence of magnetic field is $E=29\times 10^{-5}eV$

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3 years ago

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miss Akunina [59]

Answer:

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Answer:

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