Complete Question
An isolated charged soap bubble of radius R0 = 7.45 cm is at a potential of V0=307.0 volts. V0=307.0 volts. If the bubble shrinks to a radius that is 19.0%19.0% of the initial radius, by how much does its electrostatic potential energy ????U change? Assume that the charge on the bubble is spread evenly over the surface, and that the total charge on the bubble r
Answer:
The difference is 
Explanation:
From the question we are told that
The radius of the soap bubble is 
The potential of the soap bubble is 
The new radius of the soap bubble is 
The initial electric potential is mathematically represented as
The final electric potential is mathematically represented as
The initial potential is mathematically represented as

The final potential is mathematically represented as

Now

substituting values

=> 
So
Therefore
where k is the coulomb's constant with value 
substituting values

Answer:
2.36 x 10^5 kg
Explanation:
radius of hose, r = 0.017 m
radius of underground pipe, R = 0.088 m
number of hoses, n = 3
velocity of water in underground pipe, V = 2.7 m/s
Let v is the velocity of water in each hose.
According to the equation of continuity
A x V = n x a x v
π R² x V = n x π x r² x v
0.088 x 0.088 x 2.7 = 3 x 0.017 x 0.017 x v
v = 24.12 m/s
(a) Amount of water poured onto a fire in one hour by all the three hoses
= n x a x v x density of water x time
= 3 x 3.14 x 0.017 x 0.017 x 24.12 x 1000 x 3600
= 2.36 x 10^5 kg
Thus, the amount of water poured onto the fire in one hour is 2.36 x 10^5 kg.
Needs mass to be reasonably solvable, sorry
Answer:
ΔD = 2.29 10⁻⁵ m
Explanation:
This is a problem of thermal expansion, if the temperature changes are not very large we can use the relation
ΔA = 2α A ΔT
the area is
A = π r² = π D² / 4
we substitute
ΔA = 2α π D² ΔT/4
as they do not indicate the initial temperature, we assume that ΔT = 75ºC
α = 1.7 10⁻⁵ ºC⁻¹
we calculate
ΔA = 2 1.7 10⁻⁵ pi (1.8 10⁻²) ² 75/4
ΔA = 6.49 10⁻⁷ m²
by definition
ΔA = A_f- A₀
A_f = ΔA + A₀
A_f = 6.49 10⁻⁷ + π (1.8 10⁻²)² / 4
A_f = 6.49 10⁻⁷ + 2.544 10⁻⁴
A_f = 2,551 10⁻⁴ m²
the area is
A_f = π D_f² / 4
A_f =
D_f =
D_f = 1.80229 10⁻² m
the change in diameter is
ΔD = D_f - D₀
ΔD = (1.80229 - 1.8) 10⁻² m
ΔD = 0.00229 10⁻² m
ΔD = 2.29 10⁻⁵ m