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)
Answer:
i do belive its C
Explanation:
i remeber this question from somewhere also it makes the most sense
Here in this case we can use work energy theorem
As per work energy theorem
Work done by all forces = Change in kinetic Energy of the object
Total kinetic energy of the solid sphere is ZERO initially as it is given at rest.
Final total kinetic energy is sum of rotational kinetic energy and translational kinetic energy

also we know that


Now kinetic energy is given by





Now by work energy theorem
Work done = 10500 - 0 = 10500 J
So in the above case work done on sphere is 10500 J
Answer:
X-rays go all the way through the body, but ultraviolet rays do not.
Explanation:
An x-ray will show inside the body, but uv light isn't strong enough to go all the way through the body.
Answer:
Both warming up and cooling down or not important