Answer:
M = 0.730*m
V = 0.663*v
Explanation:
Data Given:
Conservation of Momentum:
Energy Balance:
Substitute Eq 2 into Eq 1
Using Eq 1
Since the main objective of this experiment is to determine the effect of porosity on seedling growth that should be the only independent variable. In short, that is the only variable that should be different to ensure fair testing.
The answer should be B:
he plants seedlings in soils with different levels of porosity and equal levels of permeability.
Permeability is not what needs to be tested. If it changes, you may not be able to determine whether it was the porosity or permeability that cause changes.
Answer:
<h2>1.5 ohms</h2>
Explanation:
Power is expressed as P = V²/R
R = resistance
V = supplied voltage
Given P = 600W and V = 30V
R = V²/P
R = 30²/600
R = 900/600
R = 1.5ohms
magnitude of its resistance is 1.5ohms
<span>f2 = f0/4
The gravity from the planet can be modeled as a point source at the center of the planet with all of the planet's mass concentrated at that point. So the initial condition for f0 has the satellite at a distance of 2r, where r equals the planet's radius.
The expression for the force of gravity is
F = G*m1*m2/r^2
where
F = Force
G = Gravitational constant
m1,m2 = masses involved
r = distance between center of masses.
Now for f2, the satellite has an altitude of 3r and when you add in the planet's radius, the distance from the center of the planet is now 4r. When you compare that to the original distance of 2r, that will show you that the satellite is now twice as far from the center of the planet as it was when it started. So let's compare the gravitational attraction, before and after.
f0 = G*m1*m2/r^2
f2 = G*m1*m2/(2r)^2
f2/f0 = (G*m1*m2/(2r)^2) / (G*m1*m2/r^2)
The Gm m1, and m2 terms cancel, so
f2/f0 = (1/(2r)^2) / (1/r^2)
f2/f0 = (1/4r^2) / (1/r^2)
And the r^2 terms cancel, so
f2/f0 = (1/4) / (1/1)
f2/f0 = (1/4) / 1
f2/f0 = 1/4
f2 = f0*1/4
f2 = f0/4
So the gravitational force on the satellite after tripling it's altitude is one fourth the original force.</span>
