Wood frogs have this adaptation where they accumulate urea in their bodies and convert their liver glycogen to glucose to act as cryoprotectants. This prevents the formation of ice crystals in their bodies that could cause damage cells during freezing in winter.
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
Answer:
They both have the same angular speed.
Explanation:
The mathematical formula for angular speed is:

where
is angular speed,
is a constant, and
is the period (the time it takes the marry-go-round to complete a lap).
What we can see from the formula is that, since the
does not change its value, the angular speed depends only on the period T.
In this case for both the children closer to the outher edge and for the children closer to the center, the time to complete a lap is the same, because the time does not depend on where they are sitting in the marry go round. This means that the period for both is the same.
Thus, since the period for both is the same, the angular speed given by
will also be the same
There’s no picture and what’s the length?