Answer:
56
Explanation:
I just want the points to be completely honest with you.
After reading this whole question, I feel like I've already
earned 5 points !
-- Two satellites at the same distance, different masses:
The forces of gravity between two objects are directly
proportional to the product of the objects' masses. In
other words, the gravitational forces between the Earth
and an object on its surface are proportional to the mass of
the object. In other words, people with more mass weigh more
on the Earth, and the Earth weighs more on them.
If the satellites are both at the same distance from Earth,
then the Earth pulls on the one with more mass with greater
force, and also the one with more mass pulls on the Earth
with greater force.
-- Two satellites with the same mass, at different distances:
The forces of gravity between two objects are inversely
proportional to the square of the distance between them.
In other words, the gravitational
forces between the Earth
and an object are inversely proportional
to the square of
the distance between the object and the center of the Earth.
If
the satellites both have the same mass, then the Earth
pulls on the nearer one with greater force, and also the
nearer one pulls on the Earth with greater force.
-- Resistor in a circuit when the voltage changes:
The resistance depends on how the resistor was manufactured.
Its resistance is marked on it, and doesn't change. It remains
the same whether the voltage changes, the current changes,
the time of day changes, the cost of oil changes, etc.
If you increase the voltage in the circuit where that resistor is
installed, the current through the resistor increases. If the current
remains constant, then you can be sure that somebody snuck over
to your circuit when you weren't looking, and they either installed
another resistor in series with the original one to make the total
resistance bigger, or else they snipped the original one out of the
circuit and quickly connected one with more resistance in its place.
Explanation:
We'll need two equations.
v² = v₀² + 2a(x - x₀)
where v is the final velocity, v₀ is the initial velocity, a is the acceleration, x is the final position, and x₀ is the initial position.
x = x₀ + ½ (v + v₀)t
where t is time.
Given:
v = 47.5 m/s
v₀ = 34.3 m/s
x - x₀ = 40100 m
Find: a and t
(47.5)² = (34.3)² + 2a(40100)
a = 0.0135 m/s²
40100 = ½ (47.5 + 34.3)t
t = 980 s
Answer:
9 Brainly hahaha ............huh
In physics, "work<span>" is when a force applied to an object moves the object in the same direction as the force. If someone pushes against a wall, no </span>work<span> is done on the system. It is calculated as follows:
Work = Force x distance
Work = 25 N x 4 meters
Work = 100 N.m</span>
