Answer:
8.800s
Explanation:
When the performer swings, she oscillates in SHM about Lo of the string with time period To = 8.90s.
First, determine the original length Lo, where for a SHM the time period is related to length and the gravitational acceleration by the equation
T = 2π×√(Lo/g)..... (1)
Let's make Lo the subject of the formulae
Lo = gTo^2/4π^2 ..... (2)
Let's put our values into equation (2) to get Lo
Lo = gTo^2/4π^2
= (9.8m/s^2)(8.90s)^2
------------------------------
4π^2
= 19.663m
Second instant, when she rise by 44.0cm, so the length Lo will be reduced by 44.0cm and the final length will be
L = Lo - (0.44m)
= 19.663m - 0.44m
= 19.223m
Now let use the value of L into equation (1) to get the period T after raising
T = 2π×√(L/g)
= 2π×√(19.223m/9.8m/s^2)
= 8.800s
Things too small to be viewed by a light microscope can be viewed with an Electron Microscope
Hope this helps!
Answer:
it's called transpiration when stomata opens and release extra water
<span>In
the desert food web shown below, which of the following best describes
the transfer of energy between the lubber grasshopper and the kangaroo
rat?
</span>
<span>C) About 10 percent of the kangaroo rat's energy transfers to the lubber grasshopper.
</span>
