Answer:
4.14 m
Explanation:
In the last leg of the journey the ball covers 2 m in 2ms or 0.2 s .
Let in this last leg , u be the initial velocity.
s = ut + 1/2 g t²
2 = .2 u + .5 x 9.8 x .04
u = 9.02 m /s .
Let v be the final velocity in this leg
v² = u² + 2 g s
v² = (9.02)² + 2 x 9.8 x 2
= 81.36 +39.2
v = 10.97 m / s
Now consider the whole height from where the ball dropped . Let it be h.
Initial velocity u = 0
v² = u² +2gh
(10.97 )² = 2 x 9.8 h
h = 6.14 m
Height from window
= 6.14 - 2m
= 4.14 m
I might have did mistake with calculations but this is how you should do.
Answer: the external agent must do work equal to -1.3 × 10⁻⁸ J
Explanation:
Given that;
Mass M1 = 7.0 kg
r = 3.0/2 m = 1.5 m
Mass M2 = 21 kg
we know that G = 6.67 × 10⁻¹¹ N.m²/kg²
work done by an external agent W = -2GM2M1 / r
so we substitute
W = (-2 × 6.67 × 10⁻¹¹ × 21 × 7) / 1.5
W = -1.96098 × 10⁻⁸ / 1.5
W = -1.3 × 10⁻⁸ J
Therefore the external agent must do work equal to -1.3 × 10⁻⁸ J
They are fused in the core of the star due to great pressures and temperatures. They are made all the way through iron. At that point the star dies. If it is a really large star it will become a supernova when it dies, creating all of the elements beyond iron as well, but only in its death. No star can create anything beyond iron in its life cycle
