When they say use energy, you want to use
Total energy = potential energy + kinetic energy = mgh + 1/2mv²
I assume you mean 200 g ball,
so, we know the total distance traveled is going to be 13 - 1.3 = 11.7 m
If the ball just makes it to the top ( 13 m ) , then the ball will stop moving and the kinetic energy will be 0,
therefore, the potential energy at the top will be the total energy of the system = mgh
from this, we say that mgh = 1/2mv² solve for v
<span>
v = sqrt (2gh) = 15.2 m/s </span>
Answer:
true
Explanation:
the sun warms the atmosphere and warms the air which drives our weather
Answer:
No.
Explanation:
The force that two particle experience is inversely proportional to the sqare of the distance, this is:
for a distance D
If we move them so that D is doubled:
= 
Then the force they experience is one fourth of the original.
We actually don't need to know how far he/she is standing from the net, as we know that the ball reaches its maximum height (vertex) at the net. At the vertex, it's vertical velocity is 0, since it has stopped moving up and is about to come back down, and its displacement is 0.33m. So we use v² = u² + 2as (neat trick I discovered just then for typing the squared sign: hold down alt and type 0178 on ur numpad wtih numlock on!!!) ANYWAY....... We apply v² = u² + 2as in the y direction only. Ignore x direction.
IN Y DIRECTION: v² = u² + 2as 0 = u² - 2gh u = √(2gh) (Sub in values at the very end)
So that will be the velocity in the y direction only. But we're given the angle at which the ball is hit (3° to the horizontal). So to find the velocity (sum of the velocity in x and y direction on impact) we can use: sin 3° = opposite/hypotenuse = (velocity in y direction only) / (velocity) So rearranging, velocity = (velocity in y direction only) / sin 3° = √(2gh)/sin 3° = (√(2 x 9.8 x 0.33)) / sin 3° = 49 m/s at 3° to the horizontal (2 sig figs)
Water that flows across the surface is called a;
Runoff
That's when rain has saturated the ground to the point it cant hold anymore and it runs over the surface.
