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1 year ago

A particle is moving at a speed less than c / 2 . If the speed of the particle is doubled, what happens to its momentum?

Physics

1 answer:

Semenov [28]1 year ago

3 0

If the speed of the particle is doubled. Then the value of the momentum also gets doubled.

<h3>What is momentum?</h3>

The sum of an entity's speed and mass is known as its linear momentum, translational momentum, or simply momentum in Newton's laws of motion. It has both a magnitude and a direction, making it a vector quantity. The SI unit of momentum is kg-m / s.

Let m be the mass and v be the velocity of an object. Then the momentum is given as,

P = mv kg-m / s

A particle is moving at a speed less than c / 2. If the speed of the particle is doubled.

P = m(2v)

P = 2mv

If the particle's speed is twice. The momentum's value is then increased by two.

More about the momentum link is given below.

brainly.com/question/24030570

#SPJ1

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Answer:

A.The positive z-direction

Explanation:

We are given that

Linear charge density of long line which is located on the x-axis= $\lambda_1$

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We have to find the direction of electric field at z=a on the positive z-axis if $\lambda_1$ and $\lambda_2$ are positive.

The direction of electric field at z=a on the positive z-axis is positive z-direction .

Because $\lambda_1$ and $\lambda_2$ are positive and the electric field is applied away from the positive charge.

Hence, option A is true.

A.The positive z-direction

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3 years ago

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Answer:

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3 years ago

A heavy box weighing 1000 newtons sits on the floor. You lift upward on the box with a force of 450 newtons, but the box does no

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Answer:

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Explanation:

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3 years ago

Read 2 more answers

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Answer:

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Explanation:

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2 years ago

One gram of Uranium averages release 1.01 KJ (10^7) of energy. How much mass could be converted to energy to release this much e

frutty [35]

Answer:

The amount of mass that needs to be converted to release that amount of energy is $1.122 X 10^{-7} kg$

Explanation:

From Albert Einstein's Energy equation, we can understand that mass can get converted to energy, using the formula

$E= \Delta mc^{2}$

where $\Delta m$ = change in mass

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Making m the subject of the formula, we can find the change in mass to be

$\Delta m = \frac{E}{c^{2}}= \frac{1.01 \times 10^{3} \times 10^{7}}{(3 \times 10^{8})^{2}}= 1.122 \times 10 ^{-7}kg$

There fore, the amount of mass that needs to be converted to release that amount of energy is 1.122 X 10 ^-7 kg

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3 years ago