Answer:
Honestly i think the answer is B
Explanation:
Answer:
500ms times 2 would be when the ball reaches the max horizontal distance.
Then to find the angle, use the formula of time to reach max height t = u sin theta / g . With t being the max height time 500ms, u being 10m/s
For initial vertical velocity just use u sin theta.
W=F*D
83J=F*14
83/14=F
5.92N
Answer:

Explanation:
The formula for the single-slit diffraction is

where
y is the distance of the n-minimum from the centre of the diffraction pattern
D is the distance of the screen from the slit
d is the width of the slit
is the wavelength of the light
In this problem,


, with n=2 (this is the distance of the 2nd-order minimum from the central maximum)
Solving the formula for d, we find:
