In this question, you're determining the time (t) taken for an object to fall from a distance (d).
The equation to represent this is:
Time equals the square root of 2 times the distance divided by the gravitational force of earth.
In equation from it looks like this (there isn't an icon to represent square root so just pretend like there's a square root there):
t = 2d/g (square-rooted)
d = 8,848m and g = 9.8m/s
Now plug in the information we have:
t = 2 x 8,848m/9.8m/s (square-rooted)
The first step is to multiply 2 times 8,848m:
t = 17,696m/9.8m/s (square-rooted)
Now divide 9.8m/s by 17,696m (note that the two m's (meters) cancels out leaving you with only s (seconds):
t = 1805.72s (square-rooted)
Now for the last step, find the square root of the remaining number:
t = 42.5s
So the time it takes the ball to drop from the height (distance) of 8,848 meters, and falling with the gravitational pull of 9.8 meters per second is 42.5 seconds.
I hope this helps :)
W=20 e(-kt)
A. Rearranging gives k= -(ln(w/20)/t
Substituting w= 10 and solving gives k=0.014
B. Using W=20e(-kt). After 0 hours, W=20. After 24 hours, W=14.29g. After 1 week (24x7=168h) W=1.9g
C. Rearranging gives t=-(ln(10/20)/k. Substituting w=1 and solving gives t=214 hours.
D. Differentiating gives dW/ dt = -20ke(-kt). Solving for t=100 gives dW/dt = 0.07g/h. Solving for t=1000 gives 0.0000002g/h
E. dW/dt = -20ke(-kt). But W=20e(-kt) so dW/dt = -kW
<span>speed = wavelength x frequency
speed = 0.4m X 10 Hz
speed = 4 m/s</span>
Protons, neutrons, and electrons<span> are the three main subatomic particles found in an atom.</span>
