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kenny6666 [7]
3 years ago
15

What scientist was the first to propose the heliocentric model of the universe

Physics
1 answer:
Schach [20]3 years ago
3 0
This theory was first proposed by Nicolaus Copernicus. Copernicus was a Polish astronomer. He first published the heliocentric system in his book: De revolutionibus <span>orbium </span>coelestium<span> , "On the revolutions of the heavenly bodies," which appeared in 1543.</span>
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Suppose a free-fall ride at an amusement park starts at rest and is in free fall. What is the velocity of the ride after 2.3 s?
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Answer:

V = a * t = 9.8 m/s^2 * 2.3 s = 22.5 m/s   velocity after 2.3 s

S = 1/2 g t^2      since initial speed is zero

S = 1/2 * 9.8 m/s^2 * 5.29 s^2 = 25.9 m

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What happens to an iron bar when it is placed within the coils of a solenoid
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22.3 kg•m/s

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What is the rising of rock layers called
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The answer of this question is called uplift i.e the rising of rock layers in geology is called uplift.

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magine an astronaut on an extrasolar planet, standing on a sheer cliff 50.0 m high. She is so happy to be on a different planet,
Mama L [17]

Answer:

\Delta t=(\frac{20}{g'}+\sqrt{\frac{400}{g'^2}+\frac{100}{g'}  }  )-(\frac{20}{g}+\sqrt{\frac{400}{g^2}+\frac{100}{g}  }  )

Explanation:

Given:

height above which the rock is thrown up, \Delta h=50\ m

initial velocity of projection, u=20\ m.s^{-1}

let the gravity on the other planet be g'

The time taken by the rock to reach the top height on the exoplanet:

v=u+g'.t'

where:

v= final velocity at the top height = 0 m.s^{-1}

0=20-g'.t' (-ve sign to indicate that acceleration acts opposite to the velocity)

t'=\frac{20}{g'}\ s

The time taken by the rock to reach the top height on the earth:

v=u+g.t

0=20-g.t

t=\frac{20}{g} \ s

Height reached by the rock above the point of throwing on the exoplanet:

v^2=u^2+2g'.h'

where:

v= final velocity at the top height = 0 m.s^{-1}

0^2=20^2-2\times g'.h'

h'=\frac{200}{g'}\ m

Height reached by the rock above the point of throwing on the earth:

v^2=u^2+2g.h

0^2=20^2-2g.h

h=\frac{200}{g}\ m

The time taken by the rock to fall from the highest point to the ground on the exoplanet:

(50+h')=u.t_f'+\frac{1}{2} g'.t_f'^2 (during falling it falls below the cliff)

here:

u= initial velocity= 0 m.s^{-1}

\frac{200}{g'}+50 =0+\frac{1}{2} g'.t_f'^2

t_f'^2=\frac{400}{g'^2}+\frac{100}{g'}

t_f'=\sqrt{\frac{400}{g'^2}+\frac{100}{g'}  }

Similarly on earth:

t_f=\sqrt{\frac{400}{g^2}+\frac{100}{g}  }

Now the required time difference:

\Delta t=(t'+t_f')-(t+t_f)

\Delta t=(\frac{20}{g'}+\sqrt{\frac{400}{g'^2}+\frac{100}{g'}  }  )-(\frac{20}{g}+\sqrt{\frac{400}{g^2}+\frac{100}{g}  }  )

3 0
3 years ago
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