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jenyasd209 [6]
3 years ago
7

Kepler modified Copernicus's model of the universe by proposing that the A. Planets follow a circular orbit around the sun. B. P

aths of the planets follow an elliptical orbit around the sun. C. Planets have their own orbits around themselves as they orbit the sun. D. Planets follow an elliptical orbit every leap year.
Physics
2 answers:
lisabon 2012 [21]3 years ago
5 0

Answer:

I believe the answer is B, the paths of the planets follow an elliptical orbit around the sun.

tresset_1 [31]3 years ago
4 0

Answer:

B. Paths of the planets follow an elliptical orbit around the sun.

Explanation:

As per Copernicus model of the universe he explained that all planets revolves around the sun in circular orbit with sun at the center of the of the path.

Now as per his theory Radius of orbit of all planets are different and the centripetal force provided by the sun for the circular path of the planets

Now as per his theory all planets must have to move with uniform speed around the sun but this was not true as we can see that the speed of all planets are different at different positions.

So here in order to correct his theory Kepler gives his law of planetary motion that all planets revolves around the sun in elliptical orbit with position of sun as one of its focus.

This path verify all the experimental results of planetary motion and hence correct answer will be

B. Paths of the planets follow an elliptical orbit around the sun.

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3 years ago
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8 0
3 years ago
Through what vertical height is a 50 N object moved if 250 J of work is done lifting it against the gravitational field of Earth
zimovet [89]

5m

Explanation:

Given parameters:

Weight of object = 50N

Work done in lifting object = 250J

Unknown:

Vertical height = ?

Solution:

The work done on an object is the force applied to lift a body in a specific direction.

   Work done = force x distance

  Weight is a force in the presence of gravity;

  Work done = weight x height of lifting

Height of lifting = \frac{work done }{weight}

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The vertical height through which the object was lifted is 5m

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6 0
3 years ago
14. A rocket is shot up into the air and then comes back down and hits the ground 9.2 second later.
sineoko [7]

Answer:

105.8 m

46 m/s

Explanation:

From the time the rocket is launched to the time it reaches its maximum height:

v = 0 m/s

a = -10 m/s²

t = 9.2 s / 2 = 4.6 s

Find: Δy and v₀

Δy = vt − ½ at²

Δy = (0 m/s) (4.6 s) − ½ (-10 m/s²) (4.6 s)²

Δy = 105.8 m

v = at + v₀

0 m/s = (-10 m/s²) (4.6 s) + v₀

v₀ = 46 m/s

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3 years ago
Initial velocity=3m/s, final velocity=5m/s, displacement=2m. Find the acceleration.
shusha [124]

Answer:

Explanation:

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