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Daniel [21]
3 years ago
11

Please help me i dont understand it This subject is Earth science

Physics
1 answer:
Anton [14]3 years ago
6 0
Here are your answers

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A capacitor is formed from two concentric spherical conducting shells separated by vacuum. The inner sphere has radius 11.0 cm ,
viktelen [127]
Part A)
First of all, let's convert the radii of the inner and the outer sphere:
r_A = 11.0 cm = 0.110 m
r_B = 16.5 cm=0.165 m
The capacitance of a spherical capacitor which consist of two shells with radius rA and rB is
C=4 \pi \epsilon _0  \frac{r_A r_B}{r_B- r_A}=4\pi(8.85 \cdot 10^{-12}C^2m^{-2}N^{-1}) \frac{(0.110m)(0.165m)}{0.165m-0.110m}=
=3.67\cdot 10^{-11}F

Then, from the usual relationship between capacitance and voltage, we can find the charge Q on each sphere of the capacitor:
Q=CV=(3.67\cdot 10^{-11}F)(100 V)=3.67\cdot 10^{-9}C

Now, we can find the electric field at any point r located between the two spheres, by using Gauss theorem:
E\cdot (4 \pi r^2) =  \frac{Q}{\epsilon _0}
from which
E(r) =  \frac{Q}{4 \pi \epsilon_0 r^2}
In part A of the problem, we want to find the electric field at r=11.1 cm=0.111 m. Substituting this number into the previous formula, we get
E(0.111m)=2680 N/C

And so, the energy density at r=0.111 m is
U= \frac{1}{2} \epsilon _0 E^2 =  \frac{1}{2} (8.85\cdot 10^{-12}C^2m^{-2}N^{-1})(2680 N/C)^2=3.17 \cdot 10^{-5}J/m^3

Part B) The solution of this part is the same as part A), since we already know the charge of the capacitor: Q=3.67 \cdot 10^{-9}C. We just need to calculate the electric field E at a different value of r: r=16.4 cm=0.164 m, so
E(0.164 m)= \frac{Q}{4 \pi \epsilon_0 r^2}=1228 N/C

And therefore, the energy density at this distance from the center is
U= \frac{1}{2}\epsilon_0 E^2 =  \frac{1}{2} (8.85\cdot 10^{-12}C^2m^{-2}N^{-1})(1228 N/C)^2=6.68 \cdot 10^{-6}J/m^3
8 0
3 years ago
A 1.0-m-diameter vat of liquid is 2.0 m deep. The pressure at the bottom of the vat is 1.3 atm. What is the mass of the liquid i
gladu [14]

Answer:

The mass of the liquid = 10538 kg

Explanation:

The pressure in a liquid is

P = ρgh ......................... Equation 1

ρ = P/gh ...................... Equation 2

Where P = pressure, ρ = density, g = acceleration due to gravity, h = height.

Given: p = 1.3 atm, h = 2.0 m, g = 9.81 m/s²

If,  1 atm = 1.013×10⁵ N/m²

Then, P = 1.3×1.013×10⁵ N/m² = 1.3169×10⁵ N/m²

Substituting in equation 2,

ρ  =  1.3169×10⁵/(9.81×2)

ρ =   1.3169×10⁵/19.62

 ρ = 6712.03 kg/m³.

But Density,

ρ  = m/v  

m =  ρ × v........................ Equation 3

Where m = mass of the liquid, v = volume of the liquid in the vat

v = πd²h/4, where d = diameter = 1.0 m, h = 2.0 m.

v = 3.14(1)²×2/4

v = 1.57 m³ also,  ρ =  6712.03 kg/m³.

Substituting into equation 3

m = 1.57×6712.03

m = 10537.887 kg

m ≈ 10538 kg.

Thus the mass of the liquid = 10538 kg

3 0
3 years ago
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