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Gala2k [10]
3 years ago
10

Would it be true that if you double the distance of an astronaut from a planet, the gravitational pull between them would be hal

f as strong?​
Physics
2 answers:
velikii [3]3 years ago
5 0

Answer:

Yes

Explanation:

Newton's law of universal gravitation is usually stated that every particle attracts every other particle in the universe with a force which is directly proportional to the product of their masses(m1 and m2) and inversely proportional to the square of the distance between their centers(r).

F = Gm1m2/r²

This is a general physical law derived from

empirical observations by what Isaac Newton called inductive reasoning.

when distance is doubled the gravitational force will be reduced by quarter not half.

WARRIOR [948]3 years ago
5 0
Yes, that would be the case, because the gravity is pulling you towards the planet, so doubling the the distance cuts the gravitational pull between them in half.
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Jake is in chemistry class. He makes a list of the chemicals his instructor described and the properties of each.
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Carbon: bonds with other atoms through strong shared electrical bonds 
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6 0
3 years ago
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When the voltage and current have _____ polarities in a pure capacitive circuit, the capacitor is discharging and the energy is
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Answer:opposite

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3 years ago
Adding or removing thermal energy to or from a substance does not always cause its temperature
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Change in thermal energy not always cause it's temperature change. It is the situation when water reaches either at 0 C or 100 C then thermal energy doesn't cause change in temperature instead it changes the state of matter.

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3 years ago
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A person of mass 70 kg stands at the center of a rotating merry-go-round platform of radius 2.9 m and moment of inertia 900 kg⋅m
Cloud [144]

Explanation:

It is given that,

Mass of person, m = 70 kg

Radius of merry go round, r = 2.9 m

The moment of inertia, I_1=900\ kg.m^2

Initial angular velocity of the platform, \omega=0.95\ rad/s

Part A,

Let \omega_2 is the angular velocity when the person reaches the edge. We need to find it. It can be calculated using the conservation of angular momentum as :

I_1\omega_1=I_2\omega_2

Here, I_2=I_1+mr^2

I_1\omega_1=(I_1+mr^2)\omega_2

900\times 0.95=(900+70\times (2.9)^2)\omega_2

Solving the above equation, we get the value as :

\omega_2=0.574\ rad/s

Part B,

The initial rotational kinetic energy is given by :

k_i=\dfrac{1}{2}I_1\omega_1^2

k_i=\dfrac{1}{2}\times 900\times (0.95)^2

k_i=406.12\ rad/s

The final rotational kinetic energy is given by :

k_f=\dfrac{1}{2}(I_1+mr^2)\omega_1^2

k_f=\dfrac{1}{2}\times (900+70\times (2.9)^2)(0.574)^2

k_f=245.24\ rad/s

Hence, this is the required solution.

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3 years ago
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