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)
Tsunamis can be a few kilometres
Answer:
The charge on the droplet is .
Yes, quantization of charge is obeyed within experimental error.
Explanation:
Given that,
Radius = 1.6μm
Electric field = 46 N/C
Density of oil = 0.085 g/cm³
We need to calculate the charge on the droplet
Using formula of force
Put the value into the formula
We need to calculate the quantization of charge
Using formula of quantization
Put the value into the formula
Yes, quantization of charge is obeyed within experimental error.
Hence, The charge on the droplet is .
Yes, quantization of charge is obeyed within experimental error.
Answer:
want
Explanation:
because it is not actually needed for the existence of life at that particular time