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:

When something is hit harder just like when sound is turned up the waves become higher and more frequent like a zig zag more so then wavy.
[two waves] pass a point [every second]... The answer is in the question (B)
Answer:
x(t) = -3sin2t
Explanation:
Given that
Spring force of, W = 720 N
Extension of the spring, s = 4 m
Attached mass to the spring, m = 45 kg
Velocity of, v = 6 m/s
The proper calculation is attached via the image below.
Final solution is x(t) = -3.sin2t