Hey Dave... you need to learn and I want you to improve I believe it to be at the bottom of the hill. Please read your siht homie
First I’ll show you this standard derivation using conservation of energy:
Pi=Kf,
mgh = 1/2 m v^2,
V = sqrt(2gh)
P is initial potential energy, K is final kinetic, m is mass of object, h is height from stopping point, v is final velocity.
In this case the height difference for the hill is 2-0.5=1.5 m. Thus the ball is moving at sqrt(2(10)(1.5))=
5.477 m/s.
Lol what???? i don’t understand
Answer:
simple
Explanation:
<h3>CONCAVE MIRRORS AND LENSES</h3>
<h3>f= negative</h3>
<h3>CONVEX MIRRORS AND LENSES</h3><h3 /><h3>f= positive</h3>
<h3>PLEASE FOLLOW ME AND MARK IT BRAINLIEST</h3>
