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Tasya [4]
3 years ago
8

Consider a spherical capacitor with radius of the inner conducting sphere a and the outer shell b. The outer shell is grounded (

i.e., it is at zero potential). The charges are +Q and −Q. a A B +Q −Q b What is the magnitude of electric field at in the region between the sphere and the outer

Physics
1 answer:
AleksAgata [21]3 years ago
6 0

Answer:

Explanation:

The application of Gauss's law is used in the derivation as shown with detailed step by step in the attached file.

The potential difference on this spherical capacitor is ΔV = Va - Vb = kQ/a - kQ/b = kQ(1/a - 1/b)

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Answer:

The thermosphere

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3 years ago
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The British gold sovereign coin is an alloy of gold and copper having a total mass of 7.988 g, and is 22-karat gold 24 x (mass o
matrenka [14]

Answers:

(a) 0.0073kg

(b) Volume gold: 3.79(10)^{-7}m^{3}, Volume cupper: 7.6(10)^{-8}m^{3}

(c) 17633.554kg/m^{3}

Explanation:

<h2>(a) Mass of gold </h2><h2 />

We are told the total mass M of the coin, which is an alloy  of gold and copper is:

M=m_{gold}+m_{copper}=7.988g=0.007988kg   (1)

Where  m_{gold} is the mass of gold and m_{copper} is the mass of copper.

In addition we know it is a 22-karat gold and the relation between the number of karats K and mass is:

K=24\frac{m_{gold}}{M}   (2)

Finding {m_{gold}:

m_{gold}=\frac{22}{24}M   (3)

m_{gold}=\frac{22}{24}(0.007988kg)   (4)

m_{gold}=0.0073kg   (5)  This is the mass of gold in the coin

<h2>(b) Volume of gold and cupper</h2><h2 />

The density \rho of an object is given by:

\rho=\frac{mass}{volume}

If we want to find the volume, this expression changes to: volume=\frac{mass}{\rho}

For gold, its volume V_{gold} will be a relation between its mass m_{gold}  (found in (5)) and its density \rho_{gold}=19.30g/cm^{3}=19300kg/m^{3}:

V_{gold}=\frac{m_{gold}}{\rho_{gold}}   (6)

V_{gold}=\frac{0.0073kg}{19300kg/m^{3}}   (7)

V_{gold}=3.79(10)^{-7}m^{3}   (8)  Volume of gold in the coin

For copper, its volume V_{copper} will be a relation between its mass m_{copper}  and its density \rho_{copper}=8.96g/cm^{3}=8960kg/m^{3}:

V_{copper}=\frac{m_{copper}}{\rho_{copper}}   (9)

The mass of copper can be found by isolating m_{copper} from (1):

M=m_{gold}+m_{copper}  

m_{copper}=M-m_{gold}  (10)

Knowing the mass of gold found in (5):

m_{copper}=0.007988kg-0.0073kg=0.000688kg  (11)

Now we can find the volume of copper:

V_{copper}=\frac{0.000688kg}{8960kg/m^{3}}   (12)

V_{copper}=7.6(10)^{-8}m^{3}   (13)  Volume of copper in the coin

<h2>(c) Density of the sovereign coin</h2><h2 />

Remembering density is a relation between mass and volume, in the case of the coin the density \rho_{coin will be a relation between its total mass M and its total volume V:

\rho_{coin}=\frac{M}{V} (14)

Knowing the total volume of the coin is:

V=V_{gold}+V_{copper}=3.79(10)^{-7}m^{3}+7.6(10)^{-8}m^{3}=4.53(10)^{-7}m^{3} (15)

\rho_{coin}=\frac{0.007988kg}{4.53(10)^{-7}m^{3}} (16)

Finally:

\rho_{coin}=17633.554kg/m^{3}} (17)  This is the total density of the British sovereign coin

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3 years ago
A system gains 767 kJ of heat, resulting in a change in internal energy of the system equal to +151 kJ. How much work is done?
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Answer:

The work done on the system is -616 kJ

Explanation:

Given;

Quantity of heat absorbed by the system, Q = 767 kJ

change in the internal energy of the system, ΔU = +151 kJ

Apply the first law of thermodynamics;

ΔU = W + Q

Where;

ΔU  is the change in internal energy

W is the work done

Q is the heat gained

W = ΔU  - Q

W = 151 - 767

W = -616 kJ (The negative sign indicates that the work is done on the system)

Therefore, the work done on the system is -616 kJ

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