<span>The force of attraction by which terrestrial bodies tend to fall toward the center of the earth strongly. If i'm wrong, correct me. otherwise i hope this helped.</span>
It goes to rest
or it goes to motion
Answer:
2.7 m/s
Explanation:
Draw a free body diagram of the ball. There are two forces:
Weight force mg pulling down
Tension force T pulling 39° above the horizontal
Sum of the forces in the y direction:
∑F = ma
T sin θ − mg = 0
T = mg / sin θ
Sum of the forces in the radial (+x) direction:
∑F = ma
T cos θ = m v² / r
Substitute:
(mg / sin θ) cos θ = m v² / r
mg / tan θ = m v² / r
g / tan θ = v² / r
v = √(gr / tan θ)
Given that r = 0.6 m and θ = 39°:
v = √(9.8 m/s² × 0.6 m / tan 39°)
v ≈ 2.7 m/s
Answer:
gravitational force and outward pressure
Explanation:
The stars are very massive stellar objects, so they have a strong gravitational force, which drives the star to contract itself, but also in stars there are nuclear reactions such as fusion of hydrogen and other elements, that releases energy and creates a pressure from the center to the star exterior, an outward pressure that goes against the gravitational force. So when a star is stable these two forces exist in equilibrium or in balance, in which the star does not collapse by gravity or disintegrate by its outward pressure.