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3 years ago

Determine the momentum of a helium atom (m = 6.696 x 10-27 kg) if the atom is traveling at 1.03% of the speed of light.

Physics

1 answer:

rjkz [21]3 years ago

7 0

Are there any options you can choose from?

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A gas always spreads out to fill all available space.<br> True or False <br> Science Hw

GREYUIT [131]

Answer:

true

Explanation:

i know this im in 6th grade

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3 years ago

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Which of the following is true concerning nuclear fusion?

Zielflug [23.3K]

Choice-'a' is a slippery, misleading, ambiguous statement,
but it's less wrong than any of the other choices on this list.

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3 years ago

How does a sound wave transfer energy to your ears ?

horrorfan [7]

Answer:

A. Particles in air move in circles as the wave moves forward.

B. Particles in air move forward but not backward as the wave moves

forward.

C. Particles in air move up and down as the wave moves forward.

✔ D. Particles in air move forward and backward as the wave moves

forward.

Explanation:

The waves transfer energy from the source of the sound, e.g. a drum, to its surroundings. Your ear detects sound waves when vibrating air particles cause your ear drum to vibrate. The bigger the vibrations the louder the sound.

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3 years ago

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A car is moving in uniform circular motion. If the car's speed were to double to keep the car moving with the same radius, the a

Anestetic [448]

Answer:

Increase by a factor of 4.

Explanation:

The acceleration of a car moving with speed v in a circle of radius R is given by:

$a=\frac{v^2}{R}.$

Now if we double the speed $v$ in the equation above, it becomes $2v$ . Thus:

$a=\frac{v^2}{R}\:\: {\rightarrow}\:\: a_n=\frac{(2v)^2}{R}=4\frac{v^2}{R}=4a.$

Therefore the acceleration is increased by a factor of 4.

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3 years ago

A 0.290 cm diameter plastic sphere, used in a static electricity demonstration, has a uniformly distributed 30.0 pC charge on it

mojhsa [17]

Answer:

Therefore,

The potential (in V) near its surface is 186.13 Volt.

Explanation:

Given:

Diameter of sphere,

d= 0.29 cm

$radius=\dfrac{d}{2}=\dfrac{0.29}{2}=0.145\ cm$

$r = 0.145\ cm = 0.145\times 10^{-2}\ m$

Charge ,

$Q = 30.0\ pC=30\times 10^{-12}$

To Find:

Electric potential , V = ?

Solution:

Electric Potential at point surface is given as,

$V=\dfrac{1}{4\pi\epsilon_{0}}\times \dfrac{Q}{r}$

Where,

V= Electric potential,

ε0 = permeability free space = 8.85 × 10–12 F/m

Q = Charge

r = Radius

Substituting the values we get

$V=\dfrac{1}{4\times 3.14\times 8.85\times 10^{-12}}\times \dfrac{30\times 10^{-12}}{0.145\times 10^{-2}}$

$V=\dfrac{30}{16.117\times 10^{-2}}=186.13\ Volt$

Therefore,

The potential (in V) near its surface is 186.13 Volt.

3 0

3 years ago