Answer:
If the distance is doubled, the force of gravity between the two bodies is one-fourth as strong as before
Explanation:
The force of gravity between two bodies depends on the mass and distance. But we will focus on distance since that's what the question asks
Therefore, the force of gravity decreases as distance between the bodies increases.
Answer:
The answer to the question is as follows
The acceleration due to gravity for low for orbit is 9.231 m/s²
Explanation:
The gravitational force is given as
Where 
= Gravitational force
G = Gravitational constant = 6.67×10⁻¹¹
m₁ = mEarth = mass of Earth = 6×10²⁴ kg
m₂ = The other mass which is acted upon by and = 1 kg
rEarth = The distance between the two masses = 6.40 x 10⁶ m
therefore at a height of 400 km above the erth we have
r = 400 + rEarth = 400 + 6.40 x 10⁶ m = 6.80 x 10⁶ m
and = 
= 9.231 N
Therefore the acceleration due to gravity = /mass
9.231/1 or 9.231 m/s²
Therefore the acceleration due to gravity at 400 kn above the Earth's surface is 9.231 m/s²
Answer:
There are six main components, or parts, of weather. They are <u>temperature, atmospheric pressure, wind, humidity, precipitation, and cloudiness</u>. Together, these components describe the weather at any given time. These changing components, along with the knowledge of atmospheric processes, help meteorologists—scientists who study weather—forecast what the weather will be in the near future.
Answer:
Explanation:
For the first case , the expression for electrostatic force can be given by the following .
F = K x 8Q x 2Q / r² where k is a constant .
F = K 16 Q² / r²
When they touch , some charge is neutralized . Net charge remaining
= 8Q - 2 Q = 6 Q
Charge on each sphere = 6Q/2 = 3 Q .
Force between them
F₁ = k 3Q x 3 Q / r² = k 9 Q² / r²
F₁ / F = 9 / 16
F₁ = 9 F / 16 .
Answer:
4.25 m/s
Explanation:
Force, F = 22 N
Time, t = 0.029 s
mass, m = 0.15 kg
initial velocity of the cue ball, u = 0
Let v be the final velocity of the cue ball.
Use newton's second law
Force = rate of change on momentum
F = m (v - u) / t
22 = 0.15 ( v - 0) / 0.029
v = 4.25 m/s
Thus, the velocity of cue ball after being struck is 4.25 m/s.
