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erma4kov [3.2K]
3 years ago
15

A person with a near point of 85 cm, but excellent distant vision, normally wears corrective glasses. But he loses them while tr

aveling. Fortunately, he has his old pair as a spare.
(a) If the lenses of the old pair have a power of +2.25 diopters, what is his near point (measured from his eye) when he is wearing the old glasses if they rest 2.0 cm in front of his eye?
(b) What would his near point be if his old glasses were contact lenses instead?
Physics
1 answer:
svp [43]3 years ago
7 0

Answer:

a)   p = 95.66 cm, b) p = 93.13 cm

Explanation:

For this problem we use the  constructor equation

         \frac{1}{f} = \frac{1}{p} + \frac{1}{q}

where f is the focal length, p and q are the distances to the object and the image, respectively

the power of the lens is

         P = 1 / f

         f = 1 / P

         f = 1 / 2.25

         f = 0.4444 m

the distance to the object is

         \frac{1}{p} = \frac{1}{f} -\frac{1}{q}

the distance to the image is

          q = 85 -2

           q = 83 cm

we must have all the magnitudes in the same units

           f = 0.4444 m = 44.44 cm

we calculate

           \frac{1}{p} = \frac{1}{44.44} - \frac{1}{83}

           1 / p = 0.010454

            p = 95.66 cm

b) if they were contact lenses

            q = 85 cm

            \frac{1}{p} = \frac{1}{44.44} - \frac{1}{85}

             1 / p = 0.107375

             p = 93.13 cm

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Here we will the speed of seagull which is v = 9 m/s

this is the speed of seagull when there is no effect of wind on it

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if effect of wind is in opposite direction then it travels 6 km in 20 min

so the average speed is given by the ratio of total distance and total time

v_{avg} = \frac{6000}{20*60}

v_{avg} = 5m/s

now since effect of wind is in opposite direction then we can say

V_{net} = v_{bird} - v_{wind}

5 = 9 - v_{wind}

v_{wind}= 4 m/s

Part b)

now if bird travels in the same direction of wind then we will have

v_{net}= v_{bird} + v_{wind}

v_{net} = 9 + 4 = 13 m/s

now we can find the time to go back

time = \frac{distance}{speed}

time = \frac{6000}{13}

time = 7.7 minutes

Part c)

Total time of round trip when wind is present

T = t_1 + t_2

T = 20 + 7.7 = 27.7 min

now when there is no wind total time is given by

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T = 22.22 min

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How much thermal energy is needed to melt 1.25 kg of water at its melting point? Use Q = masslaten heat of fusion.
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Answer:

Latent heatnof fusion = 417.5 J

Explanation:

Specific latent heat of fusion of water is 334kJ.kg-1.

The heat required to melt water when it's ice I called latent heat because there is no temperature change, the only change observed is change in physical structure.

The amount of heat required to change 1 kg of solid to its liquid state (at its melting point) at atmospheric pressure is called Latent heat of Fusion.

Latent heat = ML

Latent heat= 1.25 kg * 334kJ.kg-1

Latent heat = 1.25*334 *(J/kg)*kg

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Explanation:

I'm not sure to be honest lol

4 0
3 years ago
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Answer:

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Explanation:

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