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Brrunno [24]
3 years ago
13

The tidal bulge on the side of the earth opposite the moon is due to _______ .

Physics
2 answers:
butalik [34]3 years ago
8 0
Due to the moon's gravitational force and inertias counterbalance.
Zepler [3.9K]3 years ago
5 0

Answer:

Due to the moon's gravitational force and inertias counterbalance.

:

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Why would an object float in water, but sink in rubbing alcohol?
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Read 2 more answers
Suppose the rocket in the Example was initially on a circular orbit around Earth with a period of 1.6 days. Hint (a) What is its
ruslelena [56]

Answer:

a

The orbital speed is v= 2.6*10^{3} m/s

b

The escape velocity of the rocket is  v_e= 3.72 *10^3 m/s

Explanation:

Generally angular velocity is mathematically represented as

            w = \frac{2 \pi}{T}

Where T is the period which is given as 1.6 days = 1.6 *24 *60*60 = 138240 sec

       Substituting the value

         w = \frac{2 \pi}{138240}

             = 4.54*10^ {-5} rad /sec

At the point when the rocket is on a circular orbit  

   The gravitational force =  centripetal force and this can be mathematically represented as

              \frac{GMm}{r^2} = mr w^2

Where  G is the universal gravitational constant with a value  G = 6.67*10^{-11}

            M is the mass of the earth with a constant value of M = 5.98*10^{24}kg

            r is the distance between earth and circular orbit where the rocke is found

               Making r the subject

                     r = \sqrt[3]{\frac{GM}{w^2} }

                        = \sqrt[3]{\frac{6.67*10^{-11} * 5.98*10^{24}}{(4.45*10^{-5})^2} }

                        = 5.78 *10^7 m

The orbital speed is represented mathematically as

                   v=wr

Substituting value

                  v= (5.78*10^7)(4.54*10^{-5})

                     v= 2.6*10^{3} m/s    

The escape velocity is mathematically represented as

                            v_e = \sqrt{\frac{2GM}{r} }

Substituting values

                             = \sqrt{\frac{2(6.67*10^{-11})(5.98*10^{24})}{5.78*10^7} }

                             v_e= 3.72 *10^3 m/s

7 0
3 years ago
A ball of mass m falls vertically, hits the floor with a speed ui , and rebounds with a speed u f . what is the magnitude of the
Sholpan [36]

Answer:

Change of momentum = M (Vf - (-Vi))   where V represents the scalar speeds of the ball or

I = M (ui + uf)  and I is the impulse ΔM V = I     Force = Change in Momentum

5 0
2 years ago
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