6 3/7 * 1 5/9
45/7 * 14/9
630/63
10
The spring has been stretched 0.701 m
Explanation:
The elastic potential energy of a spring is the potential energy stored in the spring due to its compression/stretching. It is calculated as

where
k is the spring constant
x is the elongation of the spring with respect to its equilibrium position
For the spring in this problem, we have:
E = 84.08 J (potential energy)
k = 342.25 N/m (spring constant)
Therefore, its elongation is:

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Answer:
Explanation:
According to the <u>Third Kepler’s Law</u> of Planetary motion:
(1)
Where;:
is the period of the satellite
is the Gravitational Constant and its value is
is the mass of the Earth
is the semimajor axis of the orbit the satllite describes around the Earth (as we know it is a circular orbit, the semimajor axis is equal to the radius of the orbit).
On the other hand, the orbital velocity
is given by:
(2)
Now, from (1) we can find
, in order to substitute this value in (2):
(3)
(4)
(5)
Substituting (5) in (2):
(6)
(7) This is the speed at which the satellite travels