It’s important because if you were let’s say training to see how far you go in 10sec you need to record your data to see how much you are improving in that skill or something
(D) The gravitational force between the astronaut and the asteroid.
Reason :
All the other forces given in the options, except (D), doesn't account for the motion of the astronaut. They are the forces that act between nucleons or atoms and neither of them accounts for an objects motion.
Answer:
a) 5 N b) 225 N c) 5 N
Explanation:
a) Per Coulomb's Law the repulsive force between 2 equal sign charges, is directly proportional to the product of the charges, and inversely proportional to the square of the distance between them, acting along the line that joins the charges, as follows:
F₁₂ = K Q₁ Q₂ / r₁₂²
So, if we make Q1 = Q1/5, the net effect will be to reduce the force in the same factor, i.e. F₁₂ = 25 N / 5 = 5 N
b) If we reduce the distance, from r, to r/3, as the factor is squared, the net effect will be to increase the force in a factor equal to 3² = 9.
So, we will have F₁₂ = 9. 25 N = 225 N
c) If we make Q2 = 5Q2, the force would be increased 5 times, but if at the same , we increase the distance 5 times, as the factor is squared, the net factor will be 5/25 = 1/5, so we will have:
F₁₂ = 25 N .1/5 = 5 N
Answer:
The average speed of the earth in its orbit is 
Explanation:
The average distance between the Earth and the Sun is 
.
The average speed of the earth in its orbit can be found by the next equation :
(1)
Where r is the radius and T is the period.
In this case, the orbit of the Earth can be considered as a circle
(
) instead of an ellipse.
It takes 1 year to the Earth to make one revolution around the Sun. Therefore, its period will be 365.25 days.
Notice that to express the period in terms of seconds, the following is needed:
⇒ 
Then, equation 1 can be used:
