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3 years ago

10

Two nuclear reactors provide 3200 MW of power.If the transmission system loses 5.1% of the energy produced,how much power from t

hese two reactors would customers receive ?

1 answer:

Annette [7]3 years ago

6 0

The power that the customers receive is 163.2 MW

Explanation:

The efficiency of a system is given by

$\eta = \frac{P_{out}}{P_{in}}$

where

$P_{out}$ is the power in output

$P_{in}$ is the power in input

For the nuclear reactors in this problem, we have:

$P_{in} = 3200 MW$ (power generated by the reactors)

$\eta=0.051$ (efficiency is 5.1%)

Therefore, the power that reaches the customers is:

$P_{out} = \eta P_{in} = (0.051)(3200)=163.2 MW$

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Answer:

Saved Energy ?

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3 years ago

An astronaut in a space craft looks out the window and sees an asteroid move pas a backward direction at 68 mph relative to the

Annette [7]

Answer: -194 mph

Explanation:

Taking into account the <u>Sun as the center </u>(origin, point zero) <u>of the reference system</u>, the velocity of the spacecraft relative to the Sun $V_{R-S}$ is:

$V_{R-S}=126mph$ Note it is <u>positive</u> because the spacecraft is moving <u>away</u> from the Sun

Taking into account the <u>spacecraft as the center of another reference system</u>, the velocity of the asteroid relative to the spacecraft $V_{A-R}$ is:

$V_{A-R}=-68mph$ Note it is <u>negative</u> because the asteroid is moving<u> towards</u> the spacracft.

Now, the velocity of the asteroid relative to the Sun $V_{A-S}$ is:

$V_{A-S}=V_{A-R}-V_{R-S}$

$V_{A-S}=-68mph-126mph$

Finally:

$V_{A-S}=-194mph$ This is the velocity of the asteroid relative to the Sun and its negative sign indicates it is <u>moving towards the Sun</u>.

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3 years ago

Supposed an object is weighted with a spring balance, first in air then while totally immersed in water.The readings on the bala

Sedbober [7]

Is An object completely submerged in a fluid displaces its own volume of fluid ". This is: D

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3 years ago

Calculate the energy transferred by an appliance using mains electricity (230V) if the charge is 150C. Give your answer in kiloj

Otrada [13]

The energy transferred by the appliance using mains electricity is 17.3 KJ

<h3>Data obtained from the question </h3>

Potential difference (V) = 230V
Charge (Q) = 150 C

Energy (E) =?

<h3>How to determine the energy transferred </h3>

The energy transferred can be obtained as follow:

E = ½QV

E = ½ × 150 × 230

E = 75 × 230

E = 17250 J

Divide by 1000 to express in kilojoules

E = 17250 / 1000

E = 17.3 KJ

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2 years ago

Two charges are located in the x–y plane. If q1 = -2.90 nC and is located at x = 0.00 m, y = 0.840 m and the second charge has m

Lunna [17]

Answer:

Epx= - 21.4N/C

Epy= 19.84N/C

Explanation:

Electric field theory

The electric field at a point P due to a point charge is calculated as follows:

E= k*q/r²

E= Electric field in N/C

q = charge in Newtons (N)

k= electric constant in N*m²/C²

r= distance from load q to point P in meters (m)

Equivalences

1nC= 10⁻⁹C

known data

q₁=-2.9nC=-2.9 *10⁻⁹C

q₂=5nC=5 *10⁻⁹C

r₁=0.840m

$r_{2} =\sqrt{1^{2} +0.8^{2} } =\sqrt{1.64}$

$sin\beta =\frac{0.8}{\sqrt{1.64} } =0.6246$

$cos\beta =\frac{1}{\sqrt{1.64} } =0.7808$

Calculation of the electric field at point P due to q1

Ep₁x=0

$Ep_{1y} =\frac{k*q_{1} }{r_{1}^{2} } =\frac{8.99*10^{9}*2.9*10^{-9} }{0.84^{2} } =36.95\frac{N}{C}$

Calculation of the electric field at point P due to q2

$Ep_{2x} =-\frac{k*q_{2} *cos\beta }{r_{2}^{2} } =-\frac{8.99*10^{9}*5*10^{-9} *0.7808 }{(\sqrt{1.64})^{2} } =-21.4\frac{N}{C}$

$Ep_{2y} =-\frac{k*q_{2} *sin\beta }{r_{2}^{2} } =-\frac{8.99*10^{9}*5*10^{-9} *0.6242 }{(\sqrt{1.64})^{2} } =-17.11\frac{N}{C}$

Calculation of the electric field at point P(0,0) due to q1 and q2

Epx= Ep₁x+ Ep₂x==0 - 21.4N/C =- 21.4N/C

Epy= Ep₁y+ Ep₂y=36.95 N/C-17.11N =19.84N/C

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3 years ago