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Nadusha1986 [10]
3 years ago
9

A Gaussian random voltage X volts is input to a half-wave rectifier and the output voltage is Y = Xu (X) Volts were u (x) is the

unit step function. Assume X has mean 0 V and variance σ^2 V^2. The output voltage Y is then applied across a (nonrandom) resistance of R ohms. The answers below should be expressed in terms of the Φ or Q functions or in closed form (no integrals).
(a) Find the probability that the current which flows through the resistor exceeds 1 Amp.
(b) Find the probability that the power which is dissipated in the resistor exceeds 1 watt.
(c) Find the mean and variance of the current which flows through the resistor.
(d) Find the mean and variance of the power which is dissipated in the resistor.

Engineering
1 answer:
adelina 88 [10]3 years ago
7 0

Answer:

Please look at attachment carefully.

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Calculate the number of vacancies per cubic meter for some metal, M, at 783°C. The energy for vacancy formation is 0.95 eV/atom,
djyliett [7]

Answer:

Following are the solution to this question:

Explanation:

The number of vacancies by the cubic meter is determined.  

N_V =N exp(\frac{Q_v}{kT})

      = \frac{N_A \rho}{A} exp (\frac{Q_v}{kT})

      =  \frac{6.022 \times 10^{23} \times 6.10}{43.41} \exp(\frac{-0.95}{8.62\times 10^{-5} \times (783+273)})\\\\=  \frac{36.7342 \times 10^{23}}{43.41} \exp(\frac{-0.95}{0.0313626})\\\\=  0.846215158 \times 10^{23} \exp(-30.290856)\\\\

      =1.57 \times 10^{25} \ cm^{-3}

7 0
3 years ago
8. What are used by the project architect to depict different building systems and to show how they correlate to one anothe
grigory [225]

Explanation:

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6 0
2 years ago
Define volume flow rate Q of air flowing in a duct of area A with average velocity V
Shalnov [3]

Answer:

The volume flow rate of air is Q=A\times V

Explanation:

A random duct is shown in the below attached figure

The volume flow rate is defined as the volume of fluid that passes a section in unit amount of time

Now by definition of velocity we can see that 'v' m/s means that in 1 second the flow occupies a length of 'v' meters

From the attached figure we can see that

The volume of the prism that the flow occupies in 1 second equals

Volume=Area\times V=A\times V

Hence the volume flow rate is Q=V\times A

3 0
2 years ago
Technician A says independent shops are not affiliated with vehicle manufacturers, but it is easy for technicians who work in th
KatRina [158]

Answer:

b

Explanation:

i did it yeater dayajsbs

8 0
3 years ago
Oil with a density of 800 kg/m3 is pumped from a pressure of 0.6 bar to a pressure of 1.4 bar, and the outlet is 3 m above the i
Naddik [55]

Answer:

23.3808 kW

20.7088 kW

Explanation:

ρ = Density of oil = 800 kg/m³

P₁ = Initial Pressure = 0.6 bar

P₂ = Final Pressure = 1.4 bar

Q = Volumetric flow rate = 0.2 m³/s

A₁ = Area of inlet = 0.06 m²

A₂ = Area of outlet = 0.03 m²

Velocity through inlet = V₁ = Q/A₁ = 0.2/0.06 = 3.33 m/s

Velocity through outlet = V₂ = Q/A₂ = 0.2/0.03 = 6.67 m/s

Height between inlet and outlet = z₂ - z₁ = 3m

Temperature to remains constant and neglecting any heat transfer we use Bernoulli's equation

\frac {P_1}{\rho g}+\frac{V_1^2}{2g}+z_1+h=\frac {P_2}{\rho g}+\frac{V_2^2}{2g}+z_2\\\Rightarrow h=\frac{P_2-P_1}{\rho g}+\frac{V_2^2-V_1^2}{2g}+z_2-z_1\\\Rightarrow h=\frac{(1.4-0.6)\times 10^5}{800\times 9.81}+\frac{6.67_2^2-3.33^2}{2\times 9.81}+3\\\Rightarrow h=14.896\ m

Work done by pump

W_{p}=\rho gQh\\\Rightarrow W_{p}=800\times 9.81\times 0.2\times 14.896\\\Rightarrow W_{p}=23380.8\ W

∴ Power input to the pump 23.3808 kW

Now neglecting kinetic energy

h=\frac{P_2-P_1}{\rho g}+z_2-z_1\\\Righarrow h=\frac{(1.4-0.6)\times 10^5}{800\times 9.81}+3\\\Righarrow h=13.19\ m\\

Work done by pump

W_{p}=\rho gQh\\\Rightarrow W_{p}=800\times 9.81\times 0.2\times 13.193\\\Rightarrow W_{p}=20708.8\ W

∴ Power input to the pump 20.7088 kW

6 0
3 years ago
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