Ocean bulges on Earth would be bigger if the Moon had twice as much mass and yet orbited the planet at the same distance. Option B is correct.
<h3>What is ocean bludge?</h3>
The fluid and moveable ocean water are drawn towards the moon by the gravitational attraction between the moon and the Earth.
The ocean nearest to the moon experiences a bulge as a result, and as the Earth rotates, the affected seas' locations shift.
The Moon's bulges in the oceans would be larger if it had twice the mass and orbited Earth at the same distance.
Hence option B is corect.
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High energy waves have Gamma rays
Answer:
The answer is 39.
Explanation:
The atomic number refers to the number of protons and the atomic mass is the sum of the protons and neutrons. So, you would just do 70 - 31 and that gets you 39.
There is no need for tangential acceleration when moving in a circle at a constant speed.
<h3>What is centripetal acceleration?</h3>
centripetal acceleration refers to the speed at which a body moves through a circle. Due to the fact that velocity is a vector quantity (i.e., it has both a magnitude, the speed, and a direction), when a body travels in a circle, its direction is constantly changing, which causes a change in velocity, which results in an acceleration.
<h3>Which is an example of centripetal acceleration?</h3>
Centripetal acceleration occurs when you spin a ball on a string above your head. A car experiences centripetal acceleration when it is being driven in a circle. Additionally, a satellite in orbit around the Earth experiences centripetal acceleration.
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Answer: 0.01 m
Explanation: The formulae for capillarity rise or fall is given below as
h = (2T×cosθ)/rpg
Where θ = angle mercury made with glass = 50°
T = surface tension = 0.51 N/m
g = acceleration due gravity = 9.8 m/s²
r = radius of tube = 0.5mm = 0.0005m
p = density of mercury.
h = height of rise or fall
From the question, specific gravity of density = 13.3
Where specific gravity = density of mercury/ density of water, where density of water = 1000 kg/m³
Hence density of mercury = 13.3×1000 = 13,300 kg/m³.
By substituting parameters, we have that
h = 2×0.51×cos 50/0.0005×9.8×13,300
h = 0.6556/65.17
h = 0.01 m