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3 years ago

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How does alpha particles transfer energy to their surroundings?

1 answer:

Juli2301 [7.4K]3 years ago

4 0

Answer :

A Alpha Radiation. As charged particles, such as alpha particles, move through material, energy is transferred from the radiation to the atoms or molecules that make up the material. The major energy-loss mechanisms are electronic excitation and ionization.

Hope this helps :)

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A worker drives a 0.554 kg spike into a rail tie with a 2.30 kg sledgehammer. The hammer hits the spike with a speed of 62.7 m/s

Goryan [66]

Answer:

Increase in total energy will be equal to the increase in the internal energy i.e $1130.246$ Joules

Explanation:

Given

Weight of sledge hammer $= 2.30$ kilogram

Speed of sledge hammer $= 62.7$ meter per second

Kinetic energy is equal to half the product of mass and velocity square

$K E = \frac{1}{2} mv^2$

Substituting the value of mass and velocity, we get -

$KE = 0.5 * 2.30 * 62.7^2\\KE = 4520.9835$

It is given that one fourth of the energy is converted into internal energy

One fourth of kinetic energy is equal to

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Increase in total energy will be equal to the increase in the internal energy i.e $1130.246$ Joules

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The thermal efficiency of a power cycle operating in a reversible manner is found to be 50%. Assuming that the same 2 thermal re

inna [77]

Answer:

Explanation:

The thermal efficiency of a Power cycle $\eta = \dfrac{Q_H -Q_c}{Q_H}$

where;

$\eta = 50\% = 0.5$

$Q_H = Heat \ flow \ from \ higher \ temperature$

$Q_c = Heat \ flow \ from \ lower \ temperature$

$0.5 = \dfrac{Q_H -Q_c}{Q_H}$

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$Q_c = 0.5 Q_H$ ---- (2)

The coefficient of performance is:

$COP_R = \dfrac{Q_c}{Q_H -Q_c}$

let replace the value of $Q_c = 0.5 Q_H$ in the above equation then;

$COP_R = \dfrac{0.5Q_H}{Q_H -0.5 Q_H}$

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The

On the other hand, the heat pump

$COP_{HP} = \dfrac{Q_H}{Q_H -Q_c}$

By replacing equation (1) into the above equation; we have:

$COP_{HP} = \dfrac{Q_H}{0.5Q_{H}}$

$COP_{HP} = \dfrac{1}{0.5}$

$COP_{HP} =2$

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