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pychu [463]
3 years ago
10

10 PTS! HELP!

Physics
1 answer:
tino4ka555 [31]3 years ago
6 0
Shortsightedness is corrected<span> using a concave (curved inwards) lens which is placed in front of a myopic eye, moving the image back to the retina and making it clearer. Longsightedness is </span>corrected <span>using a </span>convex <span>Hyperopia is known to you probably as farsightedness(</span>outward facing) lens.
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A physicist's right eye is presbyopic (i.e., farsighted). This eye can see clearly only beyond a distance of 97 cm, which makes
Ivan

Answer:

f = 19,877 cm   and  P = 5D

Explanation:

This is a lens focal length exercise, which must be solved with the optical constructor equation

        1 / f = 1 / p + 1 / q

where f is the focal length, p is the distance to the object and q is the distance to the image.

In this case the object is placed p = 25 cm from the eye, to be able to see it clearly the image must be at q = 97 cm from the eye

let's calculate

        1 / f = 1/97 + 1/25

        1 / f = 0.05

        f = 19,877 cm

the power of a lens is defined by the inverse of the focal length in meters

         P = 1 / f

         P = 1 / 19,877 10-2

         P = 5D

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Your roommate leaves a 120W fan running in your apartment.Over the course of an hour,how much thermal energy does the fan add to
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Answer:

4.32\cdot 10^5 J

Explanation:

Power is related to energy by the following relationship:

P=\frac{E}{t}

where

P is the power used

E is the energy used

t is the time elapsed

In this problem, we know that

- the power of the fan is P = 120 W

- the fan has been running for one hour, which corresponds to a time of

t = 1 h \cdot (60 min/h)(60 s/min)=3600 s

So we can re-arrange the previous equation to find E, the energy (in the form of thermal energy) released by the fan:

E=Pt=(120 W)(3600 s)=4.32\cdot 10^5 J

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