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photoshop1234 [79]
3 years ago
9

Kepler's third law is founded on a mathematical formula that is based on the inverse relationship between a planet's orbital vel

ocity and its distance from _____. the sun the earth the moon some other entity
Physics
2 answers:
zepelin [54]3 years ago
6 0

Answer:

Kepler's third law is founded on a mathematical formula that is based on the inverse relationship between a planet's orbital velocity and its distance from <u>the Sun</u>.        

Explanation:

According to Kepler's third law of planetary motion, the square of period of orbit of a planet (P) is directly proportional to the cube of distance from the Sun (a).

P^2=\frac{4\pi^2 a^3}{GM}

It was derived from equating gravitation force with centripetal force:

\frac{GMm}{a^2}=\frac{mv^2}{a}

There is inverse relation between orbital velocity (v) and its distance from the sun (r)

Cloud [144]3 years ago
4 0
Kepler's third law is founded on a mathematical formula that is based on
the inverse relationship between a planet's orbital velocity and its distance
from the sun.
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What is the relationship between acceleration and time?
Harlamova29_29 [7]

Answer:

The relationship between acceleration and time relates to the velocity and how it changes throughout the movement of an object.

8 0
3 years ago
Will give correct answer brainliest<br><br>5 kg m/s<br>8kg m/s<br>80 kg m/s<br>200 kg m/s​
o-na [289]

Answer: Here this will help you..

Explanation:

1 kg-m/s to kilogram-force meter/second = 1 kilogram-force meter/second

5 kg-m/s to kilogram-force meter/second = 5 kilogram-force meter/second

10 kg-m/s to kilogram-force meter/second = 10 kilogram-force meter/second

20 kg-m/s to kilogram-force meter/second = 20 kilogram-force meter/second

30 kg-m/s to kilogram-force meter/second = 30 kilogram-force meter/second

40 kg-m/s to kilogram-force meter/second = 40 kilogram-force meter/second

50 kg-m/s to kilogram-force meter/second = 50 kilogram-force meter/second

75 kg-m/s to kilogram-force meter/second = 75 kilogram-force meter/second

100 kg-m/s to kilogram-force meter/second = 100 kilogram-force meter/second

8 0
3 years ago
A sound wave traveling through dry air has a frequency of 15 Hz, a
Korolek [52]

Answer:

Option B

Explanation:

Speed of a wave is denoted by:

v=fλ

where f is the frequency which is unchanged 15Hz and λ is the new wavelength which is 28m

v=fλ

v=15(28)\\v=420m/s

3 0
3 years ago
Read 2 more answers
Need help ??? Please
Yanka [14]
Is there information in the previous question which relates to this one?
8 0
3 years ago
Block A of mass M is on a horizontal surface of negligible friction. An identical block B is attached to block A by a light stri
miv72 [106K]

Answer:

T’= 4/3 T  

The new tension is 4/3 = 1.33 of the previous tension the answer e

Explanation:

For this problem let's use Newton's second law applied to each body

Body A

X axis

      T = m_A a

Axis y

     N- W_A = 0

Body B

Vertical axis

     W_B - T = m_B a

In the reference system we have selected the direction to the right as positive, therefore the downward movement is also positive. The acceleration of the two bodies must be the same so that the rope cannot tension

We write the equations

    T = m_A a

    W_B –T = M_B a

We solve this system of equations

     m_B g = (m_A + m_B) a

    a = m_B / (m_A + m_B) g

In this initial case

     m_A = M

     m_B = M

     a = M / (1 + 1) M g

     a = ½ g

Let's find the tension

    T = m_A a

    T = M ½ g

    T = ½ M g

Now we change the mass of the second block

    m_B = 2M

    a = 2M / (1 + 2) M g

    a = 2/3 g

We seek tension for this case

    T’= m_A a

    T’= M 2/3 g

   

Let's look for the relationship between the tensions of the two cases

   T’/ T = 2/3 M g / (½ M g)

   T’/ T = 4/3

   T’= 4/3 T

The new tension is 4/3 = 1.33 of the previous tension the answer  e

4 0
3 years ago
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