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.
To learn more about the ocean bulge refer;
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Answer:
Acceleration = 5.77 m/s²
Distance cover in 13 seconds = 487.56 meter
Explanation:
Given:
Final velocity of mobile device = 75 m/s
initial velocity of mobile device = 0 m/s
Time taken = 13 seconds
Find:
Acceleration
Distance cover in 13 seconds
Computation:
v = u + at
75 = 0 + (a)(13)
13a = 75
a = 5.77
Acceleration = 5.77 m/s²
s = ut + (1/2)(a)(t²)
s = (0)(t) + (1/2)(5.77)(13²)
Distance cover in 13 seconds = 487.56 meter
Answer:
See answers below
Explanation:
a.
F = mg,
15.5 N = m(9.8 m/s²)
m = 1.58 kg
b.
Fnet = Applied force - resistance,
Fnet = 18 N - 4.30 N,
Fnet = 13.70 N
Fnet = ma
13.70 N = (1.58 kg)a
a = 8.67 m/s²
For the free body diagram, draw a box with an upward arrow labeled 15.5 N, a downward label labeled 15.5 N, a right label labeled 18 N, and a left label labeled 4.30 N.
Answer:
(a) 490 N on earth
(b) 80 N on earth
(c) 45.4545 kg on earth
(d) 270.27 kg on moon
Explanation:
We have given 1 kg = 9.8 N = 2.2 lbs on earth
And 1 kg = 1.6 N = 0.37 lbs on moon
(a) We have given mass of the person m = 50 kg
As it is given that 1 kg = 9.8 N
So 50 kg = 50×9.8 =490 N
(b) Mass of the person on moon = 50 kg
As it is given that on moon 1 kg = 1.6 N
So 50 kg = 50×1.6 = 80 N
(c) We have given that weight of the person on the earth = 100 lbs
As it is given that 1 kg = 2.2 lbs on earth
So 100 lbs = 45.4545 kg
(d) We have given weight of the person on moon = 100 lbs
As it is given that 1 kg = 0.37 lbs
So 100 lbs 
