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padilas [110]
2 years ago
13

What must be the distance between two 0.800 kg balls if the magnitude of the gravitational between them is equal to that in samp

le problem C?
1.25 x10^{22} kg 1.20 x 10^{6} m
Physics
1 answer:
marta [7]2 years ago
4 0

The distance between the two balls at the given force is 0.7 m.

The given parameters;

  • <em>mass of each ball, m = 0.8 kg</em>
  • <em>gravitational force between the balls from sample problem C, F = 8.92 x 10⁻¹¹ N</em>

The distance between the balls is calculated by applying Coulomb's law as shown below;

F = \frac{Gm^2}{r^2} \\\\r^2 = \frac{Gm^2}{F} \\\\r = \sqrt{\frac{Gm^2}{F} } \\\\r = \sqrt{\frac{(6.67\times 10^{-11}) \times (0.8)^2}{8.92 \times 10^{-11}} } \\\\r = 0.7 \ m

Thus, the distance between the two balls is 0.7 m.

Learn more about Coulomb's law here: brainly.com/question/14270204

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Answer:

The value is  M  =  42.3 \  kg

Explanation:

From the question we are told that

    The first  position of the fulcrum  is  x = 49.7 cm

    The mass  attached is m  =  41.5 \  g

    The position of the attachment is  x_1 =  28.5 \  cm  

    The second position of the fulcrum is  x_2 = 39.2 \  cm

Generally the sum of clockwise torque =  sum of anti - clockwise torque

So  

       CWT  =  m (x_2 - x_1)

Here CWT  stands for clockwise torque

       ACWT  =  M  ( x - x_2)

So

      m (x_2 - x_1) =    M  ( x - x_2)

=>   41.5  (39.2 -  28.5 ) =    M  ( 49.7  -39.2 )

=>    M  =  42.3 \  kg

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If the moon had twice the diameter with same mass and orbital distance from earth, then the high tides on Earth would be practically the same.

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As the tides occurring in the earth is mostly due to the gravitational force of the moon acting on the earth. The distance between the earth and the moon mainly influences high tides. So, tidel force can be termed as directly proportionate to the gravitational forces acting between moon and earth.

As the gravitational forces act on Earth due to the moon will be directly proportionate to the product of masses of Earth and Moon and inversely proportionate to the distance squared (separation between Earth and Moon), due to the universal law of gravitation.

Thus, it can be stated that the diameter of the moon has no role to play in the high tides. As it is stated that the mass and distance almost remained the same, and there is only a change in the diameter of the moon, the high tides will not be exhibiting much change. Thus, on earth the high tides would be practically the same on increasing the diameter of the Moon.

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