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Nikitich [7]
3 years ago
9

How and why planets orbit the sun and how and why a planets speed changes orbit

Physics
2 answers:
anygoal [31]3 years ago
7 0

the gravitonal pull

Explanation:

Brut [27]3 years ago
5 0

The details are very long and complicated. The simplest way I can answer those 4 questions is: That's the way gravity works.

If you take Newton's law of gravity and his three laws of motion, and massage them properly using some geometry and some calculus, you can show that planets MUST move like Kepler's laws say they do, and like what we see them actually doing in the sky.

(I used to be able to do this, but it's been a while.)

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A comet is in an elliptical orbit around the Sun. Its closest approach to the Sun is a distance of 4.8 1010 m (inside the orbit
creativ13 [48]

Answer:

Explanation:

From the given information:

Distance d_i = 4.8 \times 10^{10} \ m

Speed of the comet V_i = 9.1 \times 10^{4} \ m/s

At distance d_2 = 6 \times 10^{12} \ m

where;

mass of the sun = 1.98 \times 10^{30}

G = 6.67 \times 10^{-11}

To find the speed V_f:

Using the formula:

E_f = E_i + W \\ \\  where; \  \  W = 0  \ \  \text{since work done by surrounding is zero (0)}

E_f = E_i + 0 \\ \\  K_f + U_f = K_i + U_i  \\ \\ = \dfrac{1}{2}mV_f^2 +  \dfrac{-GMm}{d^2} =  \dfrac{1}{2}mV_i^2+ \dfrac{-GMm}{d_i} \\ \\ V_f = \sqrt{V_i^2 + 2 GM \Big [  \dfrac{1}{d_2}- \dfrac{1}{d_i}\Big ]}

V_f = \sqrt{(9.1 \times 10^{4})^2 + 2 (6.67\times 10^{-11}) *(1.98 * 10^{30} ) \Big [  \dfrac{1}{6*10^{12}}- \dfrac{1}{4.8*10^{10}}\Big ]}

\mathbf{V_f =53.125 \times 10^4 \ m/s}

3 0
3 years ago
A car is initially moving at 10.5 m/s and accelerates uniformly to reach a speed of 21.7 m/s in 4.34 s. How far did the car move
Zarrin [17]

Answer:

A. 69.9m

Explanation:

Given parameters:

Initial velocity = 10.5m/s

Final velocity  = 21.7m/s

Time  = 4.34s

Unknown:

Distance traveled = ?

Solution:

Let us first find the acceleration of the car;

  Acceleration  = \frac{v - u}{t}

  v is final velocity

   u is initial velocity

   t is the time

     Acceleration  = \frac{21.7 - 10.5}{4.34}   = 2.58m/s²

Distance traveled;

     V² = U² + 2aS

    21.7² = 10.5² + 2 x 2.58 x S

   360.64 = 2 x 2.58 x S

     S = 69.9m

3 0
4 years ago
Which forces are shown on a free body diagram?
notsponge [240]

Answer:

What is a Free Body Diagram?

The free body diagram helps you understand and solve static and dynamic problem involving forces. It is a diagram including all forces acting on a given object without the other object in the system. You need to first understand all the forces acting on the object and then represent these force by arrows in the direction of the force to be drawn.

Explanation:

5 0
3 years ago
1. Which cell organelle is popularly known as power house of the cell. Why?
MrRissso [65]
The mitochondria is the powerhouse of the cell because it takes nutrients and breaks it down to provide energy for the cell.
5 0
3 years ago
You are at the edge of a diving board that is 9 meters above the water. If you weigh 500 Newtons, what is your potential energy?
Semenov [28]

Answer:

4500 J

Explanation:

First, let's define some equations and derivations.

Our potential energy formula is:

  • \displaystyle U = mgh

Where <em>m </em>is mass (in kg), <em>g</em> is the gravitational constant (in m/s²), and <em>h</em> is height (in m).

We also know that <em>mg</em> is equal to the weight of an object (in N), from Newton's 2nd Law of Motion: F = ma (Force is equal to [constant] mass times acceleration).

Therefore, we can simply substitute force into the equation:

  • \displaystyle U = Fh

Where <em>F</em> is the force (in N) and <em>h</em> is still height (in m).

Now we can calculate the amount of potential energy in our system, measured in joules.

Substitute in the given variables, F = 500 N and h = 9 m:

  • \displaystyle U = (500 \ N)(9 \ m)

Using simple Pre-Algebra rules, we find that:

  • \displaystyle U = 4500 \ J

This tells us that the we have 4500 joules of potential energy when I am 9 meters above the water on the edge of the diving board.

6 0
3 years ago
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