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:
n_cladding = 1.4764
Explanation:
We are told that θ_max = 5 °
Thus;
θ_max + θ_c = 90°
θ_c = 90° - θ_max
θ_c = 90° - 5°
θ_c = 85°
Now, critical angle is given by;
θ_c = sin^(-1) (n_cladding/n_core)
sin θ_c = (n_cladding/n_core)
n_cladding = (n_core) × sin θ_c
Plugging in the relevant values, we have;
n_cladding = 1.482 × sin 85
n_cladding = 1.4764
Answer:
F = 37.8 × 10^(6) N
Explanation:
The charges are 0.06 C and 0.07 C.
Thus;
Charge 1; q1 = 0.06 C
Charge 2; q2 = 0.07 C
Distance between them; r = 3 m
Formula for the force in between them is;
F = kq1•q2/r²
Where k is a constant = 9 × 10^(9) N.m²/C²
Thus;
F = (9 × 10^(9) × 0.06 × 0.07)/3²
F = 37.8 × 10^(6) N
Answer:
Insulation helps to prevent that transfer of heat.
