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balu736 [363]
3 years ago
11

How would the absence of gravity affect the formation of planets?

Physics
2 answers:
Mumz [18]3 years ago
3 0
In the absence of gravity, t<span>he rocks and debris
would never accrete into a planet. (B)

Also by the way, it wouldn't matter much, because
there wouldn't be a star to orbit around, AND orbits
wouldn't exist either.</span>
11111nata11111 [884]3 years ago
3 0

Answer:

(B)The rocks and debris would never accrete into a planet.

Explanation:

During the formation of planets, rocks and debris aggregate together because of the gravity. Gravity, in fact, is the force exerted between every object that has mass. The magnitude of the force of gravity is given by

F=G\frac{m_1 m_2}{r^2}

where

G is the gravitational constant

m1 and m2 are the masses of the two particles

r is the distance between the two particles

Without the force of gravity, there will be no force that would accrete the rocks and the debris into a planet.


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What was john philip career​
goldfiish [28.3K]

Answer:

John Philip Sousa gained great proficiency on the violin,  age of 13 he was almost persuaded to join a circus band,his father intervened and enlisted him as an apprentice musician in the Marine Band, Louis, he received a telegram offering him the leadership of the Marine Band in Washington

Explanation:

3 0
3 years ago
As shown in (Figure 1), a layer of water covers a slab of material X in a beaker. A ray of light traveling upwards follows the p
Vsevolod [243]

Answer: the index of refraction of material X will be 1.09 and the angle the light makes with the normal in the air is 81.25°.

Explanation: To find the answer, we need to know the Snell's law.

<h3>What is Snell's law of refraction? Using this, how to solve the problem?</h3>
  • The Snell's law for refraction can be written as,

                      \frac{sin (i)}{sin(r)} =\frac{n_r}{n_i}

where, i is the incident angle, r is the refracted angle, n is the refractive index.

  • As we know that the refractive index of water is 1.33
  • For the first case, incident angle from the picture is 65°, and the refracted angle is 48°. Thus, the refractive index of the medium X will be,

                           \frac{sin 65}{sin48} =\frac{n_w}{n_X} \\n_X=\frac{1.33*sin48}{sin65} =1.09\\

  • In the second case, incident angle is 48° and we have to find the refracted angle r for the air.
  • As we know that the refractive index of air is 1.
  • Thus, the refracted angle will be,

                         \frac{sin 48}{sin r}=\frac{1}{1.33} \\\\ sin(r)=\frac{1.33*sin 48}{1}=0.988\\\\r=sin^{-1}(0.988)=81.25 degrees.

Thus, we can conclude that, the index of refraction of material X will be 1.09 and the angle the light makes with the normal in the air is 81.25°.

Learn more about the Snell's law here:

brainly.com/question/28108530

#SPJ4

3 0
1 year ago
Read 2 more answers
A student is blowing over a straw 25 cm long, which is closed at one end. If the sound produced has a frequency of 350 Hz, what
NARA [144]

Answer:

If the end of the tube is uncovered such that the air at the end of the tube can freely vibrate when the sound wave reaches it, then the end is referred to as an open end. If both ends of the tube are uncovered or open, the musical instrument is said to contain an open-end air column.

Explanation: hope this helps:)

7 0
3 years ago
you (60 kg) are standing in a (500 kg) elevator that is moving upwards from a ground floor on a building what is the power ratin
PtichkaEL [24]

Explanation:

Power = work / time

Power = force × distance / time

P = (650 kg) (10 m/s²) (20 m) / (15 s)

P = 8667 W

5 0
3 years ago
A bucket that has a mass of 20 kg when filled with sand needs to be lifted to the top of a 15 meter tall building. You have a ro
prohojiy [21]

Answer:

work done lifting the bucket (sand and rope) to the top of the building,

W=67.46 Nm

Explanation:

in this question we have given

mass of bucket=20kg

mass of rope=.2\frac{kg}{m}

height of building= 15 meter

We have to find the work done lifting the bucket (sand and rope) to the building =work done in lifting the rope + work done in lifting the sand

work done in lifting the rope is given as,W_{1}=Force \times displacement

=\int\limits^{15}_0 {.2x} \, dx ..............(1)

=.1\times 15^2

=22.5 Nm

work done in lifting the sand is given as,W_{2}=Force \times displacement

W_{2}=\int\limits^{15}_0 F \, dx.................(2)

Here,

F=mx+c

here,

c=20-18

c=2

m=\frac{20-18}{15-0}

m=.133

Therefore,

F=.133x+2

Put value of F in equation 2

W_{2}=\int\limits^{15}_0 (.133x+2) \, dx

W_{2}=.133 \times 112.5+2\times15\\W_{2}=14.96+30\\W_{2}=44.96 Nm

Therefore,

work done lifting the bucket (sand and rope) to the top of the building,W=W_{1}+W_{2}

W=22.5 Nm+44.96 Nm

W=67.46 Nm

4 0
3 years ago
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