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Setler79 [48]
3 years ago
12

A farsighted woman breaks her current eyeglasses and is using an old pair whose refractive power is 1.570 diopters. Since these

eyeglasses do not completely correct her vision, she must hold a newspaper 42.00 cm from her eyes in order to read it. She wears the eyeglasses 2.00 cm from her eyes. How far is her near point from her eyes
Physics
1 answer:
vaieri [72.5K]3 years ago
3 0

Answer:

Explanation:

The refractive index of the lens is

d = 1.570 dioptres

Therefore, her focal length is

f = 1 / d

f = 1 / 1.570

f = 0.6369 m

f = 63.69cm

The distance between the newspapers and her eyes is 42cm

Distance from her eye to the glasses is 2cm

We want to calculate her near point?

So, the distance between the newspaper and the glass

do = 42 - 2 = 40cm

The distance between the glass and the virtual image formed is given as

di = f•do / (do —f)

di = 63.69 × 40 / (40 — 63.69)

di = 2547.6cm / -23.69

di = —107.54cm

The relation use is obtain from the lens equation and the negative shows that virtual image formed.

The near point from her eye is

107.54 + 2 = 109.54cm

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Answer:

A) v = 28.3 m/s

B) t =  4.64 s

Explanation:

A)

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        v_{f}^{2} - v_{o}^{2} = 2* g* \Delta h  (1)

  • Taking into account that at this point, the speed of the rock is just zero, this means vf=0 in (1), so replacing by the givens and solving for Δh, we get:

       \Delta h = \frac{-v_{o} ^{2}}{2*g} = \frac{-(17.0m/s)^{2} }{2*(-9.8m/s2)} = 14.8 m (2)

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B)

  • In order to find the total elapsed from when the rock is thrown until it hits the street, we can divide this time in two parts:
  • 1) Time elapsed from the the rock is thrown, until it reaches to its maximum height, when vf =0
  • 2) Time elapsed from this point until it hits the street, with vo=0.
  • For the first part, we can simply use the definition of acceleration (g in this case), making vf =0, as follows:

       v_{f} = v_{o} + a*\Delta t = v_{o} - g*\Delta t = 0 (5)

  • Replacing by the givens in (5) and solving for Δt, we get:

       \Delta t = \frac{v_{o}}{g} = \frac{17.0m/s}{9.8m/s2} = 1.74 s (6)

  • For the second part, since we know the total vertical displacement from (3), and that vo = 0 since it starts to fall, we can use the kinematic equation for displacement, as follows:

       \Delta h = \frac{1}{2} * g * t^{2}  (7)

  • Replacing by the givens and solving for t in (7), we get:

       t_{fall} =\sqrt{\frac{2*\Delta h}{g}} = \sqrt{\frac{2*40.8m}{9.8m/s2} } = 2.9 s (8)

  • So, total time is just the sum of (6) and (8):
  • t = 2.9 s + 1.74 s = 4.64 s
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