During either one, the sun, moon, and Earth are lined up in the same straight line. The difference is whether the moon or the Earth is the one in the "middle".
Rubber tape is used to round sharp edges
Average speed = (total distance covered) / (time to cover the distance)
total distance covered = (4km + 2km + 1km) = 7 km
time to cover the distance = (32min + 22min + 16min) = 70 min
Average speed = (7 km) / (70 min)
Average speed = 0.1 km/minute
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):
![a=\sqrt[3]{\frac{T^{2}GM}{4\pi}^{2}}](https://tex.z-dn.net/?f=a%3D%5Csqrt%5B3%5D%7B%5Cfrac%7BT%5E%7B2%7DGM%7D%7B4%5Cpi%7D%5E%7B2%7D%7D)
(3)
![a=\sqrt[3]{\frac{(1.26(10)^{4}s)^{2}(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(5.98(10)^{24}kg)}{4\pi}^{2}}](https://tex.z-dn.net/?f=a%3D%5Csqrt%5B3%5D%7B%5Cfrac%7B%281.26%2810%29%5E%7B4%7Ds%29%5E%7B2%7D%286.674%2810%29%5E%7B-11%7D%5Cfrac%7Bm%5E%7B3%7D%7D%7Bkgs%5E%7B2%7D%7D%29%285.98%2810%29%5E%7B24%7Dkg%29%7D%7B4%5Cpi%7D%5E%7B2%7D%7D)
(4)
(5)
Substituting (5) in (2):
(6)
(7) This is the speed at which the satellite travels
The electric potential energy of the pair of charges when the second charge is at point b is 7.3 x 10⁻⁸ J.
<h3>
Electric potential energy</h3>
When work is done on a positive test charge to move it from one location to another, potential energy increases and electric potential increases.
The electric potential energy between the charges when the second charge is at point b is calculated as follows;
ΔU = -w
Ui - Uf = w
Uf = Ui - w
where;
Uf is the final potential energy
Ui is the initial potential energy
w is the work done by the force
Uf = 5.4 x 10⁻⁸ J - (-1.9 x 10⁻⁸J)
Uf = 5.4 x 10⁻⁸ J + 1.9 x 10⁻⁸ J
Uf = 7.3 x 10⁻⁸ J
Thus, the electric potential energy of the pair of charges when the second charge is at point b is 7.3 x 10⁻⁸ J.
Learn more about electric potential energy here: brainly.com/question/14306881
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