Answer:
c)wind
Explanation:
Wind from the given choices will have the greatest amount of kinetic energy.
Kinetic energy is the energy due to motion of a body. It is different from the energy at rest in a body.
- Wind is air in motion.
- Wind energy is a form of kinetic energy in motion.
A book on a table, a slice of pizza and a person at the top of the stairs are all at rest and will possess potential energy.
Answer:
First of all the formula is F= uR,( force= static friction× reaction)
mass= 5+25=30
F= 50
R= mg(30×10)=300
u= ?
F=UR
u= F/R
u= 50/300=0.17N
600/3 = 200
the slope is 200m/min
OR
600/ (3/60) =
600 x 60/3 =
600 x 20 = 12,000 meters per hour
Answer:
The bending of light as it passes from one medium to another is called refraction. The angle and wavelength at which the light enters a substance and the density of that substance determine how much the light is refracted. The bending occurs because light travels more slowly in a denser medium.
hope this helped :))
Answer:
The ball would have landed 3.31m farther if the downward angle were 6.0° instead.
Explanation:
In order to solve this problem we must first start by doing a drawing that will represent the situation. (See picture attached).
We can see in the picture that the least the angle the farther the ball will go. So we need to find the A and B position to determine how farther the second shot would go. Let's start with point A.
So, first we need to determine the components of the velocity of the ball, like this:






we pick the positive one, so it takes 0.317s for the ball to hit on point A.
so now we can find the distance from the net to point A with this time. We can find it like this:



Once we found the distance between the net and point A, we can similarly find the distance between the net and point B:







t= -0.9159s or t=0.468s
we pick the positive one, so it takes 0.468s for the ball to hit on point B.
so now we can find the distance from the net to point B with this time. We can find it like this:



So once we got the two distances we can now find the difference between them:

so the ball would have landed 3.31m farther if the downward angle were 6.0° instead.