a) we can answer the first part of this by recognizing the player rises 0.76m, reaches the apex of motion, and then falls back to the ground we can ask how
long it takes to fall 0.13 m from rest: dist = 1/2 gt^2 or t=sqrt[2d/g] t=0.175
s this is the time to fall from the top; it would take the same time to travel
upward the final 0.13 m, so the total time spent in the upper 0.15 m is 2x0.175
= 0.35s
b) there are a couple of ways of finding thetime it takes to travel the bottom 0.13m first way: we can use d=1/2gt^2 twice
to solve this problem the time it takes to fall the final 0.13 m is: time it
takes to fall 0.76 m - time it takes to fall 0.63 m t = sqrt[2d/g] = 0.399 s to
fall 0.76 m, and this equation yields it takes 0.359 s to fall 0.63 m, so it
takes 0.04 s to fall the final 0.13 m. The total time spent in the lower 0.13 m
is then twice this, or 0.08s
Its C because if it is a low frequency it will not change much so it will be a longer wavelength and the higher the frequency the shorter the wavelength
Answer:
b. 0.25cm
Explanation:
You can solve this question by using the formula for the position of the fringes:

m: order of the fringes
lambda: wavelength 500nm
D: distance to the screen 5 m
d: separation of the slits 1mm=1*10^{-3}m
With the formula you can calculate the separation of two adjacent slits:

hence, the aswer is 0.25cm
Generally, the length of the line will indicate how strong the force is. If you have two opposing forces and one is higher than the other, you would draw the line of the higher force visibly longer.
Answer:
Electrification induction