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Vinil7 [7]
3 years ago
13

What is the magnitude of the gravitational force acting on the sun due to the earth?The earth does not exert any gravitational f

orce on the sun._______________The earth exerts some force on the sun, but less than 3.53×1022N because the earth, which is exerting the force, is so much less massive than the sun._______________The earth exerts 3.53×1022N of force on the sun, exactly the same amount of force the sun exerts on the earth found in Part A._______________The earth exerts more than 3.53×1022N of force on the sun because the sun, which is experiencing the force, is so much more massive than the earth._______________
Physics
1 answer:
Sholpan [36]3 years ago
4 0

I haven't worked on Part-A, and I don't happen to know the magnitude of the gravitational force that the Sun exerts on the Earth.  

But whatever it is, it's exactly, precisely, identical, the same, and equal to the magnitude of the gravitational force that the Earth exerts on the Sun.

I think that's the THIRD choice here, but I'm not sure of that either.

You might be interested in
Describe the water cycle process starting from an afternoon thunderstorm. There are many different variations that could happen.
spin [16.1K]

Explanation:

The water cycle basically involves five steps:

  • evaporation and transpiration ⇄
  • condensation, ⇄
  • precipitation, ⇄
  • runoff, ⇄
  • infiltration ⇄

So when a <u>thunderstorm </u>occurs it <em>helps in completing the precipitation process </em>by enabling the release of water vapor stored up in the atmosphere to fall on the ground as rain.

After this, the water <em>runoffs </em><em>to the surface of the ground, on plants, into rocks, rivers, and lakes.</em>

Next, the <em>Infiltration process</em> enables the water on the ground surface to enter the soil some of which becomes groundwater.

The cycle begins again as the<em> </em><em>evaporation and transpiration</em> <em>process </em>begins, where the groundwater as a result of heat from the sun is taken back into the atmosphere, while water in plants by means of transpiration goes back <em>into the atmosphere</em>.

It then <em>condenses </em>and falls back as precipitation again.

3 0
3 years ago
A bag of marbles is hanging motionless on a spring scale. What prevents the bag from falling?​
Tanya [424]

Answer:

The vector sum of all forces acting on it is zero, its at equilibrium.

Explanation:

The bag of marbles hanging on a spring scale applies its weight downwards, which was counterbalanced by the reaction from the spring scale (obeying the Newton's third law of motion). And since no external forces are applied to the system, thus the equilibrium of the system.

If the weight of the bag is greater than the reaction from the spring scale, the scale breaks and the system would not be balanced.

7 0
3 years ago
A plane traveled for 3 hours at a velocity of 1200 km/hr North. What is the distance traveled?
Daniel [21]

Answer:

the answer is 3600 kilometers

4 0
3 years ago
The longer the time between the arrival of the P-wave and S-wave, the _______ is the epicenter.
Brrunno [24]

The longer the time between the arrival of the P-wave and S-wave, the <u>farther away</u> is the epicenter.

<h3>What is epicenter and the relation between P-wave and S-wave?</h3>
  • The point on the earth's surface vertically above the hypocenter (or focus), point in the crust where a seismic rupture begins is said to be epicenter.
  • There are two types of waves during earthquakes, they are:
  1. P - wave
  2. S - wave
  • Each seismograph records the times when the first (P waves) and second (S waves) seismic waves arrive.
  • From the graph, through the information, scientists can determine how fast the waves are traveling.
  • The longer the time between the arrival of the P-wave and S-wave, the farther away is the epicenter.

Hence, Option B is the correct answer.

Learn more about epicenter,

brainly.com/question/28136716

#SPJ1

7 0
2 years ago
Please show steps as to how to solve this problem <br> Thank you!
bezimeni [28]

Explanation:

Let x = distance of F_1 from the fulcrum and let's assume that the counterclockwise direction is positive. In order to attain equilibrium, the net torque \tau_{net} about the fulcrum is zero:

\tau_{net} = -F_1x + F_2d_2 = 0

-m_1gx + m_2gd_2 = 0

m_1x = m_2d_2

Solving for <em>x</em>,

x = \dfrac{m_2}{m_1}d_2

\:\:\:\:=\left(\dfrac{105.7\:\text{g}}{65.7\:\text{g}} \right)(13.8\:\text{cm}) = 22.2\:\text{cm}

4 0
3 years ago
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