Answer:
7.55 km/s
Explanation:
The force of gravity between the Earth and the Hubble Telescope corresponds to the centripetal force that keeps the telescope in uniform circular motion around the Earth:
where
is the gravitational constant
is the mass of the telescope
is the mass of the Earth
is the distance between the telescope and the Earth's centre (given by the sum of the Earth's radius, r, and the telescope altitude, h)
v = ? is the orbital velocity of the Hubble telescope
Re-arranging the equation and substituting numbers, we find the orbital velocity:
The atom in an excited state has more energy and is less stable than the atom in the ground state.
Explanation:
We'll call the radius r and the diameter d:
We also assume that the riders are at a distance r = d/2 = 7m from the center of the wheel.
The period of the wheel is 24s. The tangent velocity of the wheel (and the riders) will be: (2pi/T)*r = 0.8 m/s (circa).
It means that in 3 minutes (180 seconds) they'll run 0.8 m/s * 180s = 144m.
Hopefully I understood the question. If yes, that's the answer.
A=mgh
m=300g=0.3kg
g=9,81 m/s^2
h=10m
A=29.43J
Answer:
The correct answer to the question is
B. It always decreases
Explanation:
To solve the question, we note that the foce of gravity is given by
where
G= Gravitational constant
m₁ = mass of first object
m₂ = mass of second object
r = the distance between both objects
If the mass of one object remains unchanged while the distance to the second object and the second object’s mass are both doubled, we have
= 
Therefore the gravitational force is halved. That is it will always decrease
