Physics, 28.12.2019 04:31 jjbossgt

Suppose the moon had twice the diameter but the same mass and same orbital distance from earth. in that case, the high tides on earth would be :

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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.

Explanation:

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.

Answer from: Quest

the unit of the measure = $\frac{j}{kgk}$

explanation:

the property that is quantified by the energy needed per of material to raise the temperature by one.

si unit of energy = joule (j)

si unit of temperature = kelvin (k)

si unit of mass of material = kilogram (kg)

the property that is quantified by the energy needed per of material to raise the temperature by one is called specific heat.

the unit of specific heat is $\frac{j}{kgk}$

Answer from: Quest

W= fd
where f is an applied force and d is the distance over which it is applied. when a parachuting skydiver is at constant velocity, he is experiencing "terminal velocity", the point at which his acceleration down is canceled out by air resistance. that means that the resisting force of the air is equal to his weight, but in the opposite direction. with this, we can solve for the work done by the force.
w = fd
w = (92.0)(9.81 m/s^2) (325)
w= 2.93 x 10^5 j
the value for work in this case is positive because it is pushing up on the skydiver, who has a downward force of gravity.
therefore the answer is 2.93 x 10^5 j

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Suppose the moon had twice the diameter but the same mass and same orbital distance from earth. in t...

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