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Lilit [14]
3 years ago
12

If a nearsighted person has a far point df that is 3.50m from the eye, what is the focal length f1 of the contact lenses that th

e person would need to see an object at infinity clearly?
If a farsighted person has a near point that is 0.600m from the eye, what is the focal length f2 of the contact lenses that the person would need to be able to read a book held at 0.350m from the person's eyes?

Physics
1 answer:
olga55 [171]3 years ago
3 0

Answer:

f1= -350cm or -3.5m

f2= 22.1cm or 0.221m

Explanation:

A person is nearsighted when the person's far point is less than infinity. A diverging lens is normally used to correct this eye defect. A diverging lens has a negative focal length as seen in the solution attached.

Farsightedness is when a person's near point is farther than 25cm. This eye defect is corrected using a converging lens. The focal length of a converging lens is positive. This is evident in the solution attached. The near point is also referred to as the least distance of distinct vision.

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3 years ago
Calculate the tangential speed of a yo-yo twirled at the end of a 4 meter long string at 2 revolutions per second.
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7. Which law describes when a person lands on a
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Answer:

Newton's Third Law

Explanation:

Newton's third law

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3 years ago
Read 2 more answers
what happens to the current in a circuit if the resitance of the components in the circuit is increased​
Anit [1.1K]

Answer:

The current decreases.

Explanation:

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Rearranging this equation, you get:

I = \frac{V}{R}

and

R = \frac{V}{I}

If the value of voltage in both equations remains constant, and the value of R decreases, the value of I will increase. Conversely, if in the second equation R = \frac{V}{I} , the value of V remains constant the value of I decreases, then the value of R, resistance will increase.

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7 0
3 years ago
Which temperature is the hottest? 98 F or 39 C or 303K?<br> F= 1.8C + 32<br> C= (F-32)/1.8
sergejj [24]

Answer:

The hottest temperature is  T_2 = 39^o C

Explanation:

From the question we are given

    T_1 =  98 F

  T_2 =  39^oC

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Generally converting T_3 to  Fahrenheit

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=> T_3' =  (303 -273 ) * \frac{9}{5}  + 32

=> T_3' = 86 F

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=> T_2' =  39 * \frac{9}{5}  + 32

=> T_2' =102.2 F  

Now comparing  the temperature  in Fahrenheit we see that T_2  is the hottest

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