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arlik [135]
3 years ago
14

A mirror with a flat surface is a

Physics
2 answers:
USPshnik [31]3 years ago
7 0
A- plane mirror because its surface is plane
meriva3 years ago
4 0

Answer:

plane mirror

Explanation:

gradpoint

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is there anything that is a hint? like a picture or a story?

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3 years ago
To a stationary observer, a man jogs east at 2.5 m/s and a woman jogs west at 1.5 m/s. from the woman's frame of reference, what
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T o a stationary observer, a man jogs east at 2.5 m/s and a woman jogs west at 1.5 m/s. from the woman's frame of reference, what is the man's velocity? it is 4m/s east
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4 years ago
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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.

5 0
4 years ago
What is the ionisation energy of an atom
V125BC [204]

Explanation:

its the minimum amount of energy required to remove the most loosely bound electron

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3 years ago
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