The new gravitational attraction will be 1/4 as much
Explanation:
The magnitude of the gravitational force between two objects is given by
where
is the gravitational constant
m1, m2 are the masses of the two objects
r is the separation between them
In this problem, the original force between the two objects is F, when they are separated by a distance r.
Later, the distance between the two objects is doubled, so the new distance is

Therefore, the new force will be

Therefore, the new force will be one-fourth as much.
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The female reproductive system is designed to carry out several functions. It produces the female egg cells necessary for reproduction, called the ova or oocytes. The system is designed to transport the ova to the site of fertilization.
The law of conservation of energy is a law of science that states that energy cannot be created or destroyed, but only changed from one form into another or transferred from one object to another.<span>
</span>
The equation
(option 3) represents the horizontal momentum of a 15 kg lab cart moving with a constant velocity, v, and that continues moving after a 2 kg object is dropped into it.
The horizontal momentum is given by:


Where:
- m₁: is the mass of the lab cart = 15 kg
- m₂: is the <em>mass </em>of the object dropped = 2 kg
: is the initial velocity of the<em> lab cart </em>
: is the <em>initial velocit</em>y of the <em>object </em>= 0 (it is dropped)
: is the final velocity of the<em> lab cart </em>
: is the <em>final velocity</em> of the <em>object </em>
Then, the horizontal momentum is:

When the object is dropped into the lab cart, the final velocity of the lab cart and the object <u>will be the same</u>, so:

Therefore, the equation
represents the horizontal momentum (option 3).
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I hope it helps you!
The time taken to hit the ground is 3.9 s, the range is 18m and the final velocity is 42.82 m/s
<h3>
Motion Under Gravity</h3>
The motion of an object under gravity is the vertical motion of the object under the influence of acceleration due to gravity.
Given that a ball is thrown horizontally from the roof of a building 75 m tall with a speed of 4.6 m/s.
a. how much later does the ball hit the ground?
The time can be calculated by considering the vertical component of the motion with the use of formula below.
h = ut + 1/2gt²
Where
- Initial velocity u = 0 ( vertical velocity )
- Acceleration due to gravity g = 9.8 m/s²
Substitute all the parameters into the formula
75 = 0 + 1/2 × 9.8 × t²
75 = 4.9t²
t² = 75/4.9
t² = 15.30
t = √15.3
t = 3.9 s
b. how far from the building will it land?
The range can be found by using the formula
R = ut
Where u = 4.6 m/s ( horizontal velocity )
R = 4.6 × 3.9
R = 18 m
c. what is the velocity of the ball just before it hits the ground?
The final velocity will be
v = u + gt
v = 4.6 + 9.8 × 3.9
v = 4.6 + 38.22
v = 42.82 m/s
Therefore, the answers are 3.9 s, 18 m and 42.82 m/s
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