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3 years ago

Please answer correctlyWill give the brainliest!

Physics

1 answer:

Oksi-84 [34.3K]3 years ago

3 0

In the first case:

when we heat any gas, the Kinetic Energy of the molecules increases, making it collide more frequently with the surface, increasing the pressure

more collisions with the surface means more force applied on it, which would push the piston harder than before, moving it outwards.

In the second case:

since the molecules inside the beaker have no way to escape, they would keep compressing the more you push the beaker downwards.

since there is the same number of molecules and lesser volume to cover, the molecules will start colliding with the surfaces more frequently, which would resist the downward force.

<em>another way to think about it is to imagine yourself where the trapped air is. you would be happy when the room is spacious but if the wall starts moving towards you, you would resist the change by your body because you need space to exist. making it harder for the wall to move.</em>

<em>pushing the beaker downwards will keep getting harder and harder the more you push until you reach a point where the molecules will be completely compact. applying even more force forces the molecules to enter water, removing the air that was resisting it all and making you able to get the beaker in water.</em>

Third case:

just like in the first case, the heated air will apply force on the surface, including the cork. which would pop off when enough force is applied.

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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

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lorasvet [3.4K]

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Answer:

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Explanation:

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3 years ago

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Answer:

So waves are everywhere. But what makes a wave a wave? What characteristics, properties, or behaviors are shared by the phenomena that we typically characterize as being a wave? How can waves be described in a manner that allows us to understand their basic nature and qualities?

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Hope That Helps!!

Explanation:

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3 years ago

Please answer correctlyWill give the brainliest!​

Please answer correctlyWill give the brainliest!