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disa [49]
3 years ago
12

The earth remains in orbit around the sun due to the force of gravity. How does the force of gravity exerted by the sun on the e

arth compare to the force of gravity exerted by the earth on the sun?
Physics
2 answers:
Katen [24]3 years ago
4 0

The sun is bigger than the earth, therefore it has more gravitational force than the earth.

netineya [11]3 years ago
4 0

Answer:

The force remains same.

Explanation:

Let the mass of sun is M and the mass of earth is m. The distance between them is d.

According to the Newton's law of gravitation, the force between the two bodies is directly proportional to the product of their masses and inversely proportional to the square of distance between them.

Thus, the force exerted by the sun on earth is equal to the force exerted by the earth on sun are same which is given by

F = G\frac{M\times m}{d^{2}}

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A railroad track and a road cross at right angles. An observer stands on the road and watches an eastbound train traveling at 60
mamaluj [8]

Answer:

After 4 s of passing through the intersection, the train travels with 57.6 m/s

Solution:

As per the question:

Suppose the distance to the south of the crossing watching the east bound train be x = 70 m

Also, the east bound travels as a function of time and can be given as:

y(t) = 60t

Now,

To calculate the speed, z(t) of the train as it passes through the intersection:

Since, the road cross at right angles, thus by Pythagoras theorem:

z(t) = \sqrt{x^{2} + y(t)^{2}}

z(t) = \sqrt{70^{2} + 60t^{2}}

Now, differentiate the above eqn w.r.t 't':

\frac{dz(t)}{dt} = \frac{1}{2}.\frac{1}{sqrt{3600t^{2} + 4900}}\times 2t\times 3600

\frac{dz(t)}{dt} = \frac{1}{sqrt{3600t^{2} + 4900}}\times 3600t

For t = 4 s:

\frac{dz(4)}{dt} = \frac{1}{sqrt{3600\times 4^{2} + 4900}}\times 3600\times 4 = 57.6\ m/s

4 0
3 years ago
How is velocity and instantaneous speed alike
allsm [11]
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5 0
3 years ago
Can you please answer this question?
Sidana [21]
I think its b too but i may be wrong
5 0
3 years ago
(15 Points)
oksian1 [2.3K]

The vertical weight carried by the builder at the rear end is F = 308.1 N

<h3>Calculations and Parameters</h3>

Given that:

The weight is carried up along the plane in rotational equilibrium condition

The torque equilibrium condition can be used to solve

We can note that the torque due to the force of the rear person about the position of the front person = Torque due to the weight of the block about the position of the front person

This would lead to:

F(W*cosθ) = mgsinθ(L/2) + mgcosθ(W/2)

F(1cos20)= 197/2(3.10sin20 + 2 cos 20)

Fcos20= 289.55

F= 308.1N

Read more about vertical weight here:

brainly.com/question/15244771

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5 0
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