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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.

You might be interested in
The astronomical unit (AU) is defined as the mean center-to-center distance from Earth to the Sun, namely 1.496x10^(11) m. The p
Rudiy27

Answer:

a) How many parsecs are there in one astronomical unit?

4.85x10^{-6}pc

(b) How many meters are in a parsec?

3.081x10^{16}m

(c) How many meters in a light-year?

9.46x10^{15}m

(d) How many astronomical units in a light-year?

63325AU

(e) How many light-years in a parsec?

3.26ly

Explanation:

The parallax angle can be used to find out the distance using triangulation. Making a triangle between the nearby star, the Sun and the Earth, knowing that the distance between the Earth and the Sun (1.496x10^{11} m) is defined as 1 astronomical unit:

\tan{p} = \frac{1AU}{d}

Where d is the distance to the star.

Since p is small it can be represent as:

p(rad) = \frac{1AU}{d}  (1)

Where p(rad) is the value of in radians

However, it is better to express small angles in arcseconds

p('') = p(rad)\frac{180^\circ}{\pi rad}.\frac{60'}{1^\circ}.\frac{60''}{1'}

p('') = 2.06x10^5 p(rad)

p(rad) = \frac{p('')}{2.06x10^5} (2)

Then, equation 2 can be replace in equation 1:

\frac{p('')}{2.06x10^5} = \frac{1AU}{d}  

\frac{d}{1AU} = \frac{2.06x10^5}{p('')}  (3)

From equation 3 it can be see that 1pc = 2.06x10^5 AU

<em>a) How many parsecs are there in one astronomical unit? </em>

1AU . \frac{1pc}{2.06x10^5AU} ⇒ 4.85x10^{-6}pc

<em>(b) How many meters are in a parsec? </em>

2.06x10^{5}AU . \frac{1.496x10^{11}m}{1AU} ⇒ 3.081x10^{16}m

<em>(c) How many meters in a light-year? </em>

To determine the number of meters in a light-year it is necessary to use the next equation:

x = c.t

Where c is the speed of light (c = 3x10^{8}m/s) and x is the distance that light travels in 1 year.

In 1 year they are 31536000 seconds

x = (3x10^{8}m/s)(31536000s)

x = 9.46x10^{15}m

<em>(d) How many astronomical units in a light-year?</em>

9.46x10^{15}m . \frac{1AU}{1.496x10^{11}m} ⇒ 63325AU

<em>(e) How many light-years in a parsec?</em>

2.06x10^{5}AU . \frac{1ly}{63235AU} ⇒ 3.26ly

5 0
3 years ago
Bromium has two naturally occurring isotopes: 79br, with an atomic weight of 78.918 amu, and 81br, with an atomic weight of 80.9
saul85 [17]

The two different isotopes have weights :

w1 = 78.918 amu

w2 = 80.916 amu

average weight w3 = 79.903 amu

The mixing of two components can be modeled as

let the fraction of w1 be 'x'

hence w1. x + w2.(1-x)  = w3

now this is a linear equation in 'x'. Substituting the values we get

x = 0.507

hence the percentage of Br79 = 50.7% and the percentage of BR81 = 49.3%

8 0
3 years ago
How long will it take an object to hit the ground if dropped from 100 meters
mash [69]

Answer:

The answer depends on what object you are dropping. Are you dropping a balloon or a car? (I'm joking 'bout that one.) If the mass of the object is very little, then it might drop slower. If the mass is bigger, then it might drop faster.

Good luck!

Explanation:

8 0
3 years ago
Read 2 more answers
Five physical properties of metals
Alekssandra [29.7K]
Metals are malleable and ductile.
Metals are good conductors of heat and electricity.
Metals are lustrous (shiny) and can be polished.
Metals are solids at room temperature (except mercury, which is liquid).
Metals are tough and strong.
hope this helps!
3 0
2 years ago
A 91.0-kg hockey player is skating on ice at 5.50 m/s. another hockey player of equal mass, moving at 8.1 m/s in the
never [62]

The momentum before the collision velocity after the collision will be 1237.6 kg m/s² and 6.8 m/sec.

<h3>What is the law of conservation of momentum?</h3>

According to the law of conservation of momentum, the momentum of the body before the collision is always equal to the momentum of the body after the collision.

The given data in the problem is;

(m₁) is the mass of hockey player 1= 91.0-kg

(m₂) is the mass of hockey player 2=  91.0-kg

(u₁) is the velocity before collision of hockey player 1 = 5.50 m/s.

(u₂) is the velocity before the collision of hockey player 2=?

a)

Momentum before the collision;

\rm  m_1u_1 + m_2u_2 \\\\ 91.0 \times 5.50 + 91.0 \times 8.1 \\\\ 1237.6 kg m/s^2

Momentum before the collision = 1237.6 kg m/s².

b)

The velocity of the two hockey players after the collision from the law of conservation of the momentum as:

Momentum before collision = Momentum after the collision

1237.6 kg m/s² = (m₁+m₂)V

1237.6 kg m/s² =(2 ×91.0-kg )V

V=6.8 m/sec.

Hence, momentum before the collision velocity after the collision will be 1237.6 kg m/s² and 6.8 m/sec.

To learn more about the law of conservation of momentum refer;

brainly.com/question/1113396

#SPJ1

8 0
2 years ago
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