To solve this problem we will use the related concepts in Newtonian laws that describe the force of gravitational attraction. We will use the given value and then we will obtain the proportion of the new force depending on the Radius. From there we will observe how much the force of attraction increases in the new distance.
Planet gravitational force



Distance between planet and star

Gravitational force is

Applying the new distance,


Replacing with the previous force,

Replacing our values


Therefore the magnitude of the force on the star due to the planet is 
This is an example of the Newton`s Second Law:
F = m * a
a = F / m
F = 8 N, m = 2 kg.
a = 8 N : 2 kg
Answer:
a = 4 m/s²
Answer:
So I never really knew you
God, I really tried to
Blindsided, addicted
Felt we could really do this
But really I was foolish
Hindsight, it's obvious
Explanation:
Answer:
274N 0.41
Explanation:
As he is sliding down in a constant speed then the force that accelerates him (weight) and the force that slows his down (friction) are equal.
then
<em>friction=mass x gravity x sin(21)</em>
Fr=78kg x 9.8m/s2 x sin(21)=274N
<em>friction= coefficient of kinetic friction x normal force of from the slope</em>
Fr= u x 78kg x 9.8m/s2 x cos(21)=274N
Fr= u x 78kg x 9.8m/s2 x cos(21)=274Nu=274/677=0.41