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3 years ago

14

A ball is dropped and falls with an acceleration of 9.8m/s2 downward. it hit the ground with a velocity of 49 m/s downward. how

long did it take the ball to fall to the ground?

1 answer:

Lelechka [254]3 years ago

4 0

V=u+at
49=0+9.8t
t=49/9.8
t = 5 (seconds)

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Answer:

5 Km/h

Explanation:

From the question given above, the following data were obtained:

Distance travelled = 10 Km

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Distance travelled = 10 Km

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The pupil of the human eye can vary in diameter from 2.00 mm in bright light to 8.00 mm in dim light. The eye has a focal length

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The indicated data are of clear understanding for the development of Airy's theory. In optics this phenomenon is described as an optical phenomenon in which The Light, due to its undulatory nature, tends to diffract when it passes through a circular opening.

The formula used for the radius of the Airy disk is given by,

$y_r=1.22\frac{\lambda f}{d}$

Where,

$y_r =$ Range of the radius

$\lambda =$ wavelength

f= focal length

Our values are given by,

State 1:

$d=2.00mm = 2*10^{-3}m$

$f= 25mm = 25*10^{-3}m$

$\lambda = 750nm = 750*10^{-9}m$

State 2:

$d=8.00mm = 8*10^{-3}m$

$f= 25mm = 25*10^{-3}m$

$\lambda = 390nm = 390*10^{-9}$

Replacing in the first equation we have:

$y_{r1} = 1.22\frac{(750*10^{-9})(25*10^{-3})}{2*10^{-3}}$

$y_{r1}= 11.4\mu m$

And also for,

$y_{r2} =1.22\frac{(390*10^{-9})(25*10^{-3})}{8*10^{-3}}$

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Therefor, the airy disk radius ranges from $1.49\mu m$ to $11.4\mu m$

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