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marishachu [46]
2 years ago
11

When hydrogen is fused into helium, energy is released from Choose one: A. the increase in pressure. B. the decrease in the grav

itational field. C. conversion of mass to energy. D. gravitational collapse.
Physics
1 answer:
Tatiana [17]2 years ago
8 0

The nuclear fusion of hydrogen atoms releases a huge amount of energy. So the correct choice is C. Conversion of mass to energy.

What is nuclear fusion?

When two small nuclei join to form a new nucleus, then this process is termed nuclear fusion. A huge amount of energy is released when there occurs nuclear fusion between the two nuclei. And a new element is formed.

It has been observed that the amount of energy released in nuclear fusion is equal to the mass difference between the mass of the formed nucleus and the total mass of old nuclei. Hence in the nuclear fusion of hydrogen nuclei to form a helium nucleus, the energy is released due to the conversion of mass into energy.

The pressure is increased to make the hydrogen atoms fuse but this change in pressure does not contribute to the energy released in the fusion of hydrogen.

The magnitude of the gravitational field is too low and it does not contribute to the energy released in the fusion of hydrogen.

The gravitational collapse does not occur between the two hydrogen atoms. This phenomenon occurs in celestial bodies so this also does not contribute to the energy released in the fusion of hydrogen.

Learn more about nuclear fusion here:

brainly.com/question/10165218

#SPJ4

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Answer:

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Explanation:

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Degger [83]

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3 years ago
A light bulb dissipates 100 Watts of power when it is supplied a voltage of 220 volts.
ycow [4]

Given Information:

Power = P = 100 Watts

Voltage = V = 220 Volts

Required Information:

a) Current = I = ?

b) Resistance = R = ?

Answer:

a) Current = I = 0.4545 A

b) Resistance = R = 484 Ω

Explanation:

According to the Ohm’s law, the power dissipated in the light bulb is given by

P = VI

Where V is the voltage across the light bulb, I is the current flowing through the light bulb and P is the power dissipated in the light bulb.

Re-arranging the above equation for current I yields,

I = \frac{P}{V}  \\\\I = \frac{100}{220} \\\\I = 0.4545 \: A \\\\

Therefore, 0.4545 A current is flowing through the light bulb.

According to the Ohm’s law, the voltage across the light bulb is given by

V = IR

Where V is the voltage across the light bulb, I is the current flowing through the light bulb and R is the resistance of the light bulb.

Re-arranging the above equation for resistance R yields,

R = \frac{V}{I} \\\\R = \frac{220}{0.4545} \\\\R = 484 \: \Omega

Therefore, the resistance of the bulb is 484 Ω

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3 years ago
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