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soldi70 [24.7K]
3 years ago
13

Creativity is always discouraged in scientific endeavors.

Physics
2 answers:
IRISSAK [1]3 years ago
4 0
The correct response for Question 9 is:  False.
shepuryov [24]3 years ago
3 0
Absolutely false!! Creativity, curiosity encouraged in Scientific endeavors. Science is all about exploring knowledge, it just promotes intelligency!

In Short, Your Answer would be "False"

Hope this helps!
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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

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f =\dfrac{1}{P}

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f=\dfrac{100}{2.55}

f=39.21\ cm

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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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4 years ago
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Marizza181 [45]

Answer:

A. 4.67 m/s²

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Identify the law, write the equation and calculate the answer to the problem below.
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