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2 years ago

If the volume of a container of gas is reduced, what will happen to the pressure inside the container?.

Physics

1 answer:

AfilCa [17]2 years ago

7 0

Answer:

The pressure of the gas will increase

Explanation:

When gas is put into a container, for example, a balloon, the gas expands to fill the space it can occupy. Since gas is not a solid or a liquid, its particles are all over the place - they are constantly moving and vibrating. As such, when too much gas is blown into a balloon, it will pop. So, when the volume of the container decreases, the pressure of the gas will increase the smaller it gets. Vice versa, the greater the space, the less pressure that will be present in the container.

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A ball is kicked from the origin with an initial speed of 10 = 25.0 /, at an angle with

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Answer:

2.5 2.566335 sec

Explanation:

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2 years ago

A carmaker has designed a car that can reach a maximum acceleration of 12 meters per second the cars mass is 1515 assuming the s

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The answer is attached. Also, you should know that the unit for acceleration is m/s2 and for velocity it is m/s.

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3 years ago

Read 2 more answers

Where is the near point of an eye for which a contact lens with a power of +2.55 diopters is prescribed?Where is the far point o

tankabanditka [31]

Answer:

(a). The eye's near point is 68.98 cm from the eye.

(b). The eye's far point is 33.33 cm from the eye.

Explanation:

Given that,

Power = 2.55 D

Object distance = 25 cm for near point

Object distance = ∞ for far point

Suppose where is the far point of an eye for which a contact lens with a power of -3.00 D is prescribed for distant vision?

(a) We need to calculate the focal length

Using formula of power

$f =\dfrac{1}{P}$

Put the value into the formula

$f=\dfrac{100}{2.55}$

$f=39.21\ cm$

We need to calculate the image distance

Using formula of lens

$\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}$

Put the value into the formula

$\dfrac{1}{39.21}=\dfrac{1}{v}-\dfrac{1}{-25}$

$\dfrac{1}{39.21}-\dfrac{1}{25}=\dfrac{1}{v}$

$\dfrac{1}{v}=-\dfrac{1421}{98025}$

$v=-68.98\ cm$

The eye's near point is 68.98 cm from the eye.

(b). We need to calculate the focal length

Using formula of power

$f =\dfrac{1}{P}$

Put the value into the formula

$f=-\dfrac{100}{3.00}$

$f=-33.33\ cm$

We need to calculate the image distance

Using formula of lens

$\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}$

Put the value into the formula

$\dfrac{1}{-33.33}=\dfrac{1}{v}-\dfrac{1}{\infty}$

$-\dfrac{1}{33.33}+\dfrac{1}{\infty}=\dfrac{1}{v}$

$\dfrac{1}{v}=-\dfrac{1}{33.33}$

$v=-33.33\ cm$

The eye's far point is 33.33 cm from the eye.

Hence, (a). The eye's near point is 68.98 cm from the eye.

(b). The eye's far point is 33.33 cm from the eye.

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3 years ago

Suppose you're on a hot air balloon ride, carrying a buzzer that emits a sound of frequency f. If you accidentally drop the buzz

amid [387]

Answer:

Explanation:

The Doppler effect describes the change in frequency or wavelength of a wave in relation to an observer who is moving relative to the wave source, or the wave source is moving relative to the observer, or both.

if the observer and the source move away from each other as is the case for this problem, the wavelength heard by the observer is bigger.

The frequency is the inverse from the wavelength, so the frequency heard will increase.

The sound intensity depends inversely on the area in which the sound propagates. When the buzzer is close, the area is from a small sphere, but as the buzzer moves further away, the wave area will be from a larger sphere and therefore the intensity will decrease.

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3 years ago

Need help on this please

Ratling [72]

question 2 answer is ALL OF THE ABOVE

question 3 answer is WARM AND MOIST.

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3 years ago