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Flura [38]
4 years ago
12

Two resistors, with resistances R1 and R2, are connected in series. R1 is normally distributed with mean 65 and standard deviati

on 10 , and R2 is normally distributed with mean 75 and standard deviation 5 .
a. What is the probability that R2 > R1?
Engineering
1 answer:
Sedbober [7]4 years ago
4 0

Answer:

n this question, we are asked to find the probability that  

R1 is normally distributed with mean 65  and standard deviation 10

R2 is normally distributed with mean 75  and standard deviation 5

Both resistor are connected in series.

We need to find P(R2>R1)

the we can re write as,

P(R2>R1) = P(R2-R1>R1-R1)

P(R2>R1) = P(R2-R1>0)

P(R2>R1) = P(R>0)

Where;

R = R2 - R1

Since both and are independent random variable and normally distributed, we can do the linear combinations of mean and standard deviations.

u = u2-u1

u = 75 - 65 = 10ohm

sd = √sd1² + sd2²

sd = √10²+5²

sd = √100+25 = 11.18ohm

Now we will calculate the z-score, to find  P( R>0 )

Z = ( X -u)/sd

the z score of 0 is

z = 0 - 10/11.18

z= - 0.89

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3 years ago
Describe harmful effect associated with extraction of Aluminum, Gold and Copper. Discuss each individually.
KengaRu [80]

Answer:

Harmful effect associated with extraction of Aluminum, Gold and Copper are:

During the melting of aluminium there is a released of per fluorocarbon are more harmful than carbon dioxide in the environment as they increased the level of green house gases and cause global warming. The process of transforming raw material into the aluminium are much energy intensive.

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3 0
4 years ago
Determine the complex power, apparent power, average power absorbed, reactive power, and power factor (including whether it is l
LenKa [72]

Answer:

A) complex power = apparent  power =  ( 108.253 + 62.5 i ) VA  

    active power = 108.25 watts

   reactive power = 62.5 VAR'S

    power factor = cos ∅ = cos 30° = 0.866 ( lagging )

also ; current lags voltage by 30°

B) your question is not well written hence no answer

Explanation:

A) v(t) = 100 cos (377t - 30° ) v

    Vrms = \frac{100}{\sqrt{2} } ∠ -30°

    I(t) = 2.5cos(377t- 60°) A

  Irms = \frac{2.5}{\sqrt{2} } ∠ -60°

determine complex power apparent power . ......  power factor

Note : complex power = apparent power

=( Irms ) * Vrms

=  ( \frac{2.5}{\sqrt{2} } ∠ -60° ) * ( \frac{100}{\sqrt{2} } ∠ -30° )  

= 125 ∠ 30°

= ( 108.253 + 62.5 i ) VA   ( complex power )

active power = 108.25 watts

reactive power = 62.5 VAR'S

power factor = cos ∅ = cos 30° = 0.866 ( lagging )

current lags voltage by 30°

4 0
3 years ago
Sea water with a density of 1025 kg/m3 flows steadily through a pump at 0.21 m3 /s. The pump inlet is 0.25 m in diameter. At the
myrzilka [38]

Answer:

\dot W_{pump} = 16264.922\,W\,(16.265\,kW)

Explanation:

The pump is modelled after applying Principle of Energy Conservation, whose form is:

\frac{P_{1}}{\rho\cdot g}+ \frac{v_{1}^{2}}{2\cdot g} +z_{1} + h_{pump}=\frac{P_{2}}{\rho\cdot g}+ \frac{v_{2}^{2}}{2\cdot g} +z_{2}

The head associated with the pump is cleared:

h_{pump} = \frac{P_{2}-P_{1}}{\rho\cdot g}+\frac{v_{2}^{2}-v_{1}^{2}}{2\cdot g}+(z_{2}-z_{1})

Inlet and outlet velocities are found:

v_{1} = \frac{0.21\,\frac{m^{3}}{s} }{\frac{\pi}{4}\cdot (0.25\,m)^{2} }

v_{1} \approx 4.278\,\frac{m}{s}

v_{2} = \frac{0.21\,\frac{m^{3}}{s} }{\frac{\pi}{4}\cdot (0.152\,m)^{2} }

v_{2} \approx 11.573\,\frac{m}{s}

Now, the head associated with the pump is finally computed:

h_{pump} = \frac{175\,kPa-81.326\,kPa}{(1025\,\frac{kg}{m^{3}} )\cdot (9.807\,\frac{m}{s^{2}} )} +\frac{(11.573\,\frac{m}{s} )^{2}-(4.278\,\frac{m}{s} )^{2}}{2\cdot (9.807\,\frac{m}{s^{2}} )} + 1.8\,m

h_{pump} = 7.705\,m

The power that pump adds to the fluid is:

\dot W_{pump} = \dot V \cdot \rho \cdot g \cdot h_{pump}

\dot W_{pump} = (0.21\,m^{3})\cdot (1025\,\frac{kg}{m^{3}})\cdot (9.807\,\frac{m}{s^{2}})\cdot(7.705\,m)

\dot W_{pump} = 16264.922\,W\,(16.265\,kW)

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aleksklad [387]

Answer:

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