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mixer [17]
3 years ago
15

A sphere of radius R contains charge Q spread uniformly throughout its volume. Find an expression for the electrostatic energy c

ontained within the sphere itself.
Physics
1 answer:
tensa zangetsu [6.8K]3 years ago
7 0

Answer:

E = \frac{3kQ^2}{5R}

Explanation:

Let the sphere is uniformly charge to radius "r" and due to this charged sphere the electric potential on its surface is given as

V = \frac{kq}{r}

now we can say that

q = \frac{Q}{\frac{4}{3}\pi R^3} (\frac{4}{3}\pi r^3)

q = \frac{Qr^3}{R^3}

now electric potential is given as

V = \frac{k\frac{Qr^3}{R^3}}{r}

V = \frac{kQr^2}{R^3}

now work done to bring a small charge from infinite to the surface of this sphere is given as

dW = V dq

dW = \frac{kQr^2}{R^3} dq

here we know that

dq = \frac{3Qr^2dr}{R^3}

now the total energy of the sphere is given as

E = \int dW

E = \int_0^R  \frac{kQr^2}{R^3} (\frac{3Qr^2dr}{R^3})

E = \frac{3kQ^2}{R^6} (\frac{R^5}{5} - 0)

E = \frac{3kQ^2}{5R}

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A hot (70°C) lump of metal has a mass of 250 g and a specific heat of 0.25 cal/g⋅°C. John drops the metal into a 500-g calorimet
Gnom [1K]

Answer:

d. 37 °C

Explanation:

m_{m} = mass of lump of metal = 250 g

c_{m} = specific heat of lump of metal  = 0.25 cal/g°C

T_{mi} = Initial temperature of lump of metal = 70 °C

m_{w} = mass of water = 75 g

c_{w} = specific heat of water = 1 cal/g°C

T_{wi} = Initial temperature of water = 20 °C

m_{c} = mass of calorimeter  = 500 g

c_{c} = specific heat of calorimeter = 0.10 cal/g°C

T_{ci} = Initial temperature of calorimeter = 20 °C

T_{f} = Final equilibrium temperature

Using conservation of heat

Heat lost by lump of metal = heat gained by water + heat gained by calorimeter

m_{m} c_{m} (T_{mi} - T_{f}) = m_{w} c_{w} (T_{f} - T_{wi}) +  m_{c} c_{c} (T_{f} - T_{ci}) \\(250) (0.25) (70 - T_{f} ) = (75) (1) (T_{f} - 20) + (500) (0.10) (T_{f} - 20)\\T_{f} = 37 C

6 0
3 years ago
Physics {deceleration}
seropon [69]
Using kinematic equation, v^2 - u^2 = 2as. 5^2 - 3^2 = 2a x 16. a = 0.5m/s^2. So particle will deaccelerate at 0.5m/s^2. ( v = final velocity, u= initial velocity, a= acceleration, s= displacement.)
7 0
3 years ago
A sample of monatomic ideal gas occupies 5.00 L at atmospheric pressure and 300 K (point A). It is warmed at constant volume to
leonid [27]

Answer:

(a) 0.203 moles

(b) 900 K

(c) 900 K

(d) 15 L

(e) A → B, W = 0, Q = Eint = 1,518.91596 J

B → C, W = Q ≈ 1668.69974 J Eint = 0 J

C → A, Q = -2,531.5266 J, W = -1,013.25 J, Eint = -1,518.91596 J

(g) ∑Q = 656.089 J, ∑W =  655.449 J, ∑Eint = 0 J

Explanation:

At point A

The volume of the gas, V₁ = 5.00 L

The pressure of the gas, P₁ = 1 atm

The temperature of the gas, T₁ = 300 K

At point B

The volume of the gas, V₂ = V₁ = 5.00 L

The pressure of the gas, P₂ = 3.00 atm

The temperature of the gas, T₂ = Not given

At point C

The volume of the gas, V₃ = Not given

The pressure of the gas, P₃ = 1 atm

The temperature of the gas, T₂ = T₃ = 300 K

(a) The ideal gas equation is given as follows;

P·V = n·R·T

Where;

P = The pressure of the gas

V = The volume of the gas

n = The number of moles present

R = The universal gas constant = 0.08205 L·atm·mol⁻¹·K⁻¹

n = PV/(R·T)

∴ The number of moles, n = 1 × 5/(0.08205 × 300) ≈ 0.203 moles

The number of moles in the sample, n ≈ 0.203 moles

(b) The process from points A to B is a constant volume process, therefore, we have, by Gay-Lussac's law;

P₁/T₁ = P₂/T₂

∴ T₂ = P₂·T₁/P₁

From which we get;

T₂ = 3.0 atm. × 300 K/(1.00 atm.) = 900 K

The temperature at point B, T₂ = 900 K

(c) The process from points B to C is a constant temperature process, therefore, T₃ = T₂ = 900 K

(d) For a constant temperature process, according to Boyle's law, we have;

P₂·V₂ = P₃·V₃

V₃ = P₂·V₂/P₃

∴ V₃ = 3.00 atm. × 5.00 L/(1.00 atm.) = 15 L

The volume at point C, V₃ = 15 L

(e) The process A → B, which is a constant volume process, can be carried out in a vessel with a fixed volume

The process B → C, which is a constant temperature process, can be carried out in an insulated adjustable vessel

The process C → A, which is a constant pressure process, can be carried out in an adjustable vessel with a fixed amount of force applied to the piston

(f) For A → B, W = 0,

Q = Eint = n·cv·(T₂ - T₁)

Cv for monoatomic gas = 3/2·R

∴ Q = 0.203 moles × 3/2×0.08205 L·atm·mol⁻¹·K⁻¹×(900 K - 300 K) = 1,518.91596 J

Q = Eint = 1,518.91596 J

For B → C, we have a constant temperature process

Q = n·R·T₂·㏑(V₃/V₂)

∴ Q = 0.203 moles × 0.08205 L·atm/(mol·K) × 900 K × ln(15 L/5.00 L) ≈ 1668.69974 J

Eint = 0

Q = W ≈ 1668.69974 J

For C → A, we have a constant pressure process

Q = n·Cp·(T₁ - T₃)

∴ Q = 0.203 moles × (5/2) × 0.08205 L·atm/(mol·K) × (300 K - 900 K) = -2,531.5266 J

Q = -2,531.5266 J

W = P·(V₂ - V₁)

∴ W = 1.00 atm × (5.00 L - 15.00 L) = -1,013.25 J

W = -1,013.25 J

Eint = n·Cv·(T₁ - T₃)

Eint = 0.203 moles × (3/2) × 0.08205 L·atm/(mol·K) × (300 K - 900 K) = -1,518.91596 J

Eint = -1,518.91596 J

(g) ∑Q = 1,518.91596 J + 1668.69974 J - 2,531.5266 J = 656.089 J

∑W = 0 + 1668.69974 J -1,013.25 J = 655.449 J

∑Eint = 1,518.91596 J + 0 -1,518.91596 J = 0 J

5 0
3 years ago
What happens to a liquid beyond its certain temperature
Natalka [10]

Answer: what do you mean ''beyond certain temperature'', Do you mean ask in cooling the liquid down or heating the liquid up?

Explanation:

7 0
3 years ago
Two air track carts move along an air track towards each other. Cart A has a mass of 450 g and moves toward the right with a spe
Blizzard [7]

Answer:

The right answer is D) the total momentum of the system is 0.047 kg · m/s toward the right.

Explanation:

Hi there!

The total momentum of the system is given by the sum of the momentum vectors of each cart. The momentum is calculated as follows:

p = m · v

Where:

p = momentum.

m = mass.

v = velocity.

Then, the momentum of the system will be the momentum of cart A plus the momentum of cart B (let´s consider the right as the positive direction):

mA · vA + mB · Vb

0.450 kg · 0.850 m/s + 0.300 kg · (- 1.12 m/s) = 0.047 kg · m/s

The right answer is D) the total momentum of the system is 0.047 kg · m/s toward the right.

6 0
3 years ago
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