Answer:
5.1*10^3 J/m^3
Explanation:
Using E = q/A*eo
And
q =75*10^-6 C
A = 0.25
eo = 8.85*10^-12
Energy density = 1/2*eo*(E^2) = 1/2*eo*(q/A*eo)^2 = [q^2] / [2*(A^2)*eo]
= [(75*10^-6)^2] / [2*(0.25)^2*8.85*10^-12]
= 5.1*10^3 J/m^3
density of water = 
velocity of flow = 
radius of pipe = 
Height of second floor = 
Now we can use here Bernuoli's Equation to find the speed of water flow at second floor



Now in order to find the radius of pipe we can use equation of continuity



So radius of pipe at second floor is 0.034 meter
Answer:
310 meters
Explanation:
Given:
v₀ = 0 m/s
t = 8.0 s
a = -9.8 m/s²
Find: Δy
Δy = v₀ t + ½ at²
Δy = (0 m/s) (8.0 s) + ½ (-9.8 m/s²) (8.0 s)²
Δy = -313.6
Rounded to two significant figures, the object fell 310 meters.
Yes it is. Uh huh, uh huh, shore enuff. Mmm hmm. Yeah yeah yeah. Yah Mon ! Indubitably.
There's no such thing as "an unbalanced force".
If all of the forces acting on an object all add up to zero, then we say that
<span>the group </span>of forces is balanced. When that happens, the group of forces
has the same effect on the object as if there were no forces on it at all.
An example:
Two people with exactly equal strength are having a tug-of-war. They pull
with equal force in opposite directions. Each person is sweating and straining,
grunting and groaning, and exerting tremendous force. But their forces add up
to zero, and the rope goes nowhere. The <u>group</u> of forces on the rope is balanced.
On the other hand, if one of the offensive linemen is pulling on one end of
the rope, and one of the cheerleaders is pulling on the other end, then their
forces don't add up to zero, because even though they're opposite, they're
not equal. The <u>group</u> of forces is <u>unbalanced</u>, and the rope moves.
A group of forces is either balanced or unbalanced. A single force isn't.