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3 years ago

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Charge of uniform surface density (4.0 nC/m2 ) is distributed on a spherical surface (radius = 2.0 cm). What is the total electr

ic flux through a concentric spherical surface with a radius of 4.0 cm?

1 answer:

Sonja [21]3 years ago

3 0

To solve this problem we will use the values of the density on surface, which relates the total load and the area of the surface. From there we will get the total charge, which will allow us to find the electric flow.

Surface density is defined as,

$\mu= \frac{Q_{tot}}{A_{surface}}$

Where,

$A_{surface} = 4\pi r^2$

Replacing,

$4*10^{-9} = \frac{Q_{tot}}{(4\pi 0.02^2)}$

$Q_{tot}= 20.11*10^{-2} C$

At the same time electric flux can be defined as,

$\Phi = \frac{Qtot}{\epsilon_0}$

Here,

$\epsilon_0 =$ Permittivity vacuum constant

Replacing,

$\Phi =\frac{(20.11 * 10-12 )}{(8.85 * 10-12)}$

$\Phi =2.27N \cdot m^2 \cdot C^{-1}$

Therefore the total electric flux through the concentric spherical surface is $2.27Nm^2/C$

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3 years ago

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3 years ago

Refrigerant R-12 is used in a Carnot refrigerator

Sphinxa [80]

Answer:

Heat transferred from the refrigerated space = 95.93 kJ/kg

Work required = 18.45 kJ/kg

Coefficient of performance = 3.61

Quality at the beginning of the heat addition cycle = 0.37

Explanation:

From figure

$Q_H$ is heat rejection process

$Q_L$ is heat transferred from the refrigerated space

$T_H$ is high temperature = 50 °C + 273 = 323 K

$T_L$ is low temperature = -20 °C + 273 = 253 K

$W_{net}$ is net work of the cycle (the difference between compressor's work and turbine's work)

Coefficient of performance of a Carnot refrigerator $(COP_{ref})$ is calculated as

$COP_{ref} = \frac{T_L}{T_H - T_L}$

$COP_{ref} = \frac{253 K}{323 K - 253 K}$

$COP_{ref} = 3.61$

From figure it can be seen that heat rejection is latent heat of vaporisation of R-12 at 50 °C. From table

$Q_H = 122.5 kJ/kg$

From coefficient of performance definition

$COP_{ref} = \frac{Q_L}{Q_H - Q_L}$

$Q_H \times COP_{ref} = (COP_{ref} + 1) \times Q_L$

$Q_L = \frac{Q_H \times COP_{ref}}{(COP_{ref} + 1)}$

$Q_L = \frac{122.5 kJ/kg \times 3.61}{(3.61 + 1)}$

$Q_L = 95.93 kJ/kg$

Energy balance gives

$W_{net} = Q_H - Q_L$

$W_{net} = 122.5 kJ/kg - 95.93 kJ/kg$

$W_{net} = 26.57 kJ/kg$

Vapor quality at the beginning of the heat addition cycle is calculated as (f and g refer to saturated liquid and saturated gas respectively)

$x = \frac{s_1 - s_f}{s_g - s_f}$

From figure

$s_1 = s_4 = 1.165 kJ/(K kg)$

Replacing with table values

$x = \frac{1.165 kJ/(K \, kg) - 0.9305 kJ/(K \, kg)}{1.571 kJ/(K \, kg) - 0.9305 kJ/(K \, kg)}$

$x = 0.37$

Quality can be computed by other properties, for example, specific enthalpy. Rearrenging quality equation we get

$h_1 = h_f + x \times (h_g - h_f)$

$h_1 = 181.6 kJ/kg + 0.37 \times 162.1 kJ/kg$

$h_1 = 241.58 kJ/kg$

By energy balance, $W_{t}$ turbine's work is

$W_{t} = |h_1 - h_4|$

$W_{t} = |241.58 kJ/kg - 249.7 kJ/kg|$

$W_{t} = 8.12 kJ/kg$

Finally, $W_{c}$ compressor's work is

$W_{c} = W_{net} + W_{t}$

$W_{c} = 26.57 kJ/kg + 8.12 kJ/kg$

$W_{c} = 34.69 kJ/kg$

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3 years ago

Practice Problem: Large-Particle CompositesThe mechanical properties of a metal may be improved by incorporating fine particles

trasher [3.6K]

Answer: (a). Ec(μ) = 165.6 GPa

(b). Ec(∝) = 83.09 GPa

Explanation:

this is quite straightforward, so we will go step by step.

from the data we have that,

Moduli of elasticity of the metal -(Em) is 60 Gpa

Moduli of elasticity of oxide is (Ep) is 380 Gpa

volume Vp = 33% = 0.33

(a). To solve the upper bound-modulus of the elasticity is calculate thus;

Ec (μ) = EmVm + EpVp ----------------(1)

where E rep the modulus of elasticity

v rep the volume fraction

c rep the composite

Vm = 100% - Vp

Vm = 100% - 33% = 67%

Vm = 0.67

substituting the valus of Em, Vm, Ep, Vp from equation (1) we have;

Ec(μ) = (60×0.67) + (380×0.33)

Ec(μ) = 40.2 + 125.4 = 165.6 GPa

Ec(μ) = 165.6 GPa

(b). The lower bound modulus of elasticity can be calculated thus;

Ec(∝) = EmVp / EpVm + EmVp -------------- (2)

substituting values Em,Vm,Ep,Vp.

Ec(∝) = 60×30 / (380×0.67) + (60 ×0.33)

Ec(∝) = 22800 / 254.6 + 19.8 = 83.09 GPa

Ec(∝) = 83.09 GPa

cheers i hope this helps!!!!

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4 years ago