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nikklg [1K]
3 years ago
15

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 :
Physics
1 answer:
Inessa [10]3 years ago
5 0

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.

<u>Explanation:</u>

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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A 0.350 kg block at -27.5°C is added to 0.217 kg of water at 25.0°C. They come to equilibrium at 16.4°C. What is the specific he
Gekata [30.6K]

The specific heat capacity of the block is 508J/kg^{\circ}C

Explanation:

As the block is placed into the water, heat energy is transferred from the water (which is at higher temperature) to the block (which is at lower temperature), until the block and the water are in thermal equilibrium (= same temperature).

Therefore, we can write:

Q_{water}=Q_{block}

Where

Q_{water}=m_w C_w (T_w-T_{eq}) is the heat energy released by the water, where

m_w = 0.217 kg is the mass of the water

C_w = 4186 J/kg^{\circ}C is the water heat specific capacity

T_w = 25.0^{\circ} is the initial temperature of the water

T_{eq}=16.4^{\circ} is the temperature at equilibrium

Substituting,

Q_{water}=(0.217)(4186)(25.0-16.4)=7812 J

Now we can write the heat energy absorbed by the block as

Q_{block}=m_b C_b(T_{eq}-T_b)

where

m_b=0.350 kg is the mass of the block

C_b is the specific heat capacity of the block

T_b = -27.5^{\circ} is the initial temperature of the block

And solving for C_b,

C_b=\frac{Q_{block}}{m_b(T_{eq}-T_b)}=\frac{7812}{(0.350)(16.4-(-27.5))}=508J/kg^{\circ}C

Learn more about specific heat capacity:

brainly.com/question/3032746

brainly.com/question/4759369

#LearnwithBrainly

6 0
3 years ago
Raising 100 grams of water from 40 to 60 °C (the specific heat capacity of water is 1
faust18 [17]

Heat in a system can be calculated by multiplying the given mass to the specific heat capacity of the substance and the temperature difference. It is expressed as follows:<span>

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<span><span>
</span></span>

<span><span>Hope this answers the question. Have a nice day.</span></span>

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3 years ago
What is the magnetic force on a proton that is moving at 4.5 x 10^7 m/s to the right through a magnetic field that is 1.6 T and
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The magnetic force on a charged particle is given as: F = qvBsinФ
B is the magnetic field.
q is the charge on the particle.
v is the velocity of the charged particle.
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⇒ F = qvb sin Ф

Here, Ф = 90 degree.
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