The Moon is 3.8 108 m from Earth and has a mass of 7.34 1022 kg. 5.97 1024 kg is the mass of the Earth.
<h3>What kind of gravitational pull does the moon have on the planet?</h3>
On the surface of the Moon, the acceleration caused by gravity around 1.625 m/s2 which is 16.6% greater than on the surface of the Earth 0.166.
<h3>What does the Earth's center's gravitational pull feel like?</h3>
Gravity is zero if you are in the centre of the earth since everything around you is pulling "up" (up is the only direction).
<h3>Where is the Earth's and the moon's gravitational centre?</h3>
It is around 1700 kilometres below Earth's surface.
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Answer: 0.6m
Explanation:
Given that:
force = 4.5 N
Work done = 2.7J
Distance moved by the book = ?
Since work is done when force is applied on an object over a distance, apply the formula:
work = force x distance
2.7J = 4.5N x distance
Distance = (2.7J / 4.5N)
Distance = 0.6 m
Thus, the book was moved 0.6 metres far
Answer:
The resultant velocity is <u>169.71 km/h at angle of 45° measured clockwise with the x-axis</u> or the east-west line.
Explanation:
Considering west direction along negative x-axis and north direction along positive y-axis
Given:
The car travels at a speed of 120 km/h in the west direction.
The car then travels at the same speed in the north direction.
Now, considering the given directions, the velocities are given as:
Velocity in west direction is,
Velocity in north direction is,
Now, since are perpendicular to each other, their resultant magnitude is given as:
Plug in the given values and solve for the magnitude of the resultant.This gives,
Let the angle made by the resultant be 'x' degree with the east-west line or the x-axis.
So, the direction is given as:
Therefore, the resultant velocity is 169.71 km/h at angle of 45° measured clockwise with the x-axis or the east-west line.
The vertical velocity of the projectile upon returning to its original is 17. 74 m/s
<h3>
How to determine the vertical velocity</h3>
Using the formula:
Vertical velocity component , Vy = V * sin(α)
Where
V = initial velocity = 36. 6 m/s
α = angle of projectile = 29°
Substitute into the formula
Vy = 36. 6 * sin ( 29°)
Vy = 36. 6 * 0. 4848
Vy = 17. 74 m/s
Thus, the vertical velocity of the projectile upon returning to its original is 17. 74 m/s
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