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3 years ago

Which most likely could cause acceleration? Check all that apply.

Physics

1 answer:

Lerok [7]3 years ago

5 0

<u>Answer </u>

Wind

Water movement

A hill

<u>Explanation. </u>

The question does not specify it is the acceleration of what is to be accelerated. So my answer will be general. Acceleration is the change of velocity with time.

Wind can cause acceleration of sound. If the sound is travelling in the direction of sound, its speed wound increase. The reverse is also true.

Water movement can also cause acceleration. When a bout is moving in the direction of water movement, then it will be expected that the resultant speed of the bout to be higher. If the bout was moving against the water movement, its speed would be decreased.

A hill would reduce the speed of a moving vehicle. This is also an acceleration that is negative (deceleration).

A stop light and a flat road cannot cause acceleration.

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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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Centripetal acceleration=(6 x 6) ➗ 0.85

Centripetal acceleration=36 ➗ 0.85

Centripetal acceleration=42.4

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3 years ago