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Nadusha1986 [10]
4 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]4 years ago
7 0

Answer:

Please look at attachment carefully.

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A turbojet aircraft flies with a velocity of 800 ft/s at an altitude where the air is at 10 psia and 20 F. The compressor has a
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Answer:

Pressure = 115.6 psia

Explanation:

Given:

v=800ft/s

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Conversion:

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Air standard assumptions:

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k= 1.4

Energy balance:

h_{1} + \frac{v_{1} ^{2} }{2} = h_{a} + \frac{v_{a} ^{2} }{2}\\

As enthalpy exerts more influence than the kinetic energy inside the engine, kinetic energy of the fluid inside the engine is negligible

hence v_{a} ^{2} = 0

h_{1} + \frac{v_{1} ^{2} }{2} = h_{a} \\h_{1} -h_{a} = - \frac{v_{1} ^{2} }{2} \\ c_{p} (T_{1} -T_{a})= - \frac{v_{1} ^{2} }{2} \\(T_{1} -T_{a}) = - \frac{v_{1} ^{2} }{2c_{p} }\\ T_{a}=T_{1} +  \frac{v_{1} ^{2} }{2c_{p} }

T_{1} = 20+460 = 480°R

T_{a}  =480+  \frac{(800)(800}{2(0.240)(25037}= 533.25°R

Pressure at the inlet of compressor at isentropic condition

P_{a } =P_{1}(\frac{T_{a} }{T_{1} }) ^{k/(k-1)}

P_{a} = (10)(\frac{533.25}{480}) ^{1.4/(1.4-1)}= 14.45 psia

P_{2}= 8P_{a} = 8(14.45) = 115.6 psia

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Technician B is correct because torque is a force of an object.
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What material resources and intellectual resources were used in self driving cars?
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