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soldi70 [24.7K]

3 years ago

8

An object of 6 cm height is 16 cm in front of a converging lens with focal length of magnitude 4 cm. What is the dimension of th

e image(include the unit and if it is the case the sign)?

1 answer:

Pani-rosa [81]3 years ago

3 0

Answer:

$- 2$ cm

Explanation:

$h_{o}$ = height of the object = 6 cm

$d_{o}$ = distance of the object from the lens = 16 cm

$f$ = focal length of the lens = 4 cm

$d_{i}$ = image distance

using the lens equation

$\frac{1}{d_{o}} + \frac{1}{d_{i}} = \frac{1}{f}$

$\frac{1}{16} + \frac{1}{d_{i}} = \frac{1}{4}$

$d_{i} = \frac{16}{3}$ cm

$h_{i}$ = height of the image

Using the equation

$\frac{h_{i}}{h_{o}} = \frac{- d_{i}}{d_{o}}$

$\frac{h_{i}}{6} = \frac{- \frac{16}{3}}{16}$

$h_{i} =-2$ cm

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a)

The acceleration of gravity for an object near a planet is given by

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$R=6.37\cdot 10^6 m$ (Earth's radius)

$h=2.60\cdot 10^6 m$ (altitude of the satellite)

Substituting,

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The critical velocity for a satellite orbiting around a planet is given by

$v=\sqrt{\frac{GM}{R+h}}$

where we have again:

$M=5.98\cdot 10^{24} kg$ (mass of the Earth)

$R=6.37\cdot 10^6 m$ (Earth's radius)

$h=2.60\cdot 10^6 m$ (altitude of the satellite)

Substituting,

$v=\sqrt{\frac{(6.67\cdot 10^{-11})(5.98\cdot 10^{24}}{(6.37\cdot 10^6 + 2.60\cdot 10^6)}}=6668 m/s$

Converting into km/h,

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c)

The period of the orbit is given by the circumference of the orbit divided by the velocity:

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where

$R=6.37\cdot 10^6 m$

$h=2.60\cdot 10^6 m$

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Substituting,

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One day consists of:

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While the period of the orbit is

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The escape velocity for an object in the gravitational field of a planet is given by

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where here we have:

$M=5.98\cdot 10^{24} kg$

$R=6.37\cdot 10^6 m$

$h=2.60\cdot 10^6 m$

Substituting, we find

$v=\sqrt{\frac{2(6.67\cdot 10^{-11})(5.98\cdot 10^{24}}{(6.37\cdot 10^6 + 2.60\cdot 10^6)}}=9430 m/s$

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We can apply again the formula to find the escape velocity for the rocket:

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