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3 years ago

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A dam is used to block the passage of a river and to generate electricity. Approximately 5.73 × 104 kg of water fall each second

through a height of 19.6 m. If 85 % of the gravitational potential energy of the water were converted to electrical energy, how much power would be generated?

1 answer:

Step2247 [10]3 years ago

4 0

Answer:

9.36 MW

Explanation:

Let g = 9.81 m/s2. Then the gravitational energy of that much water at a height of 19.6 m is as following

$E_P = mgh = 5.73*10^4*9.81*19.6 = 11017414.8 J$ per second or 11017414.8 W

If only 85% of this energy is converted to electrical energy then the power generated would be

11017414.8 * 85% = 9364802.58 W or 9.36 MW

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evablogger [386]

Answer:

a)n= 3.125 x $10^{19$ electrons.

b)J= 1.515 x $10^{6$ A/m²

c) $V_{d$ =1.114 x $10^{4$ m/s

d) see explanation

Explanation:

Current 'I' = 5A =>5C/s

diameter 'd'= 2.05 x $10^{-3$ m

radius 'r' = d/2 => 1.025 x $10^{-3$ m

no. of electrons 'n'= 8.5 x $10^{28}$

a) the amount of electrons pass through the light bulb each second can be determined by:

I= Q/t

Q= I x t => 5 x 1

Q= 5C

As we know that: Q= ne

where e is the charge of electron i.e 1.6 x $10^{-19$ C

n= Q/e => 5/ 1.6 x $10^{-19$

n= 3.125 x $10^{19$ electrons.

b) the current density 'J' in the wire is given by

J= I/A => I/πr²

J= 5 / (3.14 x (1.025x $10^{-3$ )²)

J= 1.515 x $10^{6$ A/m²

c) The typical speed' $V_{d$ ' of an electron is given by:

$V_{d$ = $\frac{J}{n|q|}$

=1.515 x $10^{6$ / 8.5 x $10^{28}$ x |-1.6 x $10^{-19$ |

$V_{d$ =1.114 x $10^{4$ m/s

d) According to these equations,

J= I/A

$V_{d$ = $\frac{J}{n|q|}$ = $\frac{I}{nA|q|}$

If you were to use wire of twice the diameter, the current density and drift speed will change

Increase in the diameter increase the cross sectional area and decreases the current density as it has inverse relation.

Also drift velocity will decrease as it is inversely proportional to the area

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Volgvan

Answer:

4.1 m

Explanation:

10 kW = 10000 W

20mi/h = 20*1.6 km/mi = 32 km/h = 32 * 1000 (m/km) *(1/3600) hr/s = 8.89 m/s

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$P = \frac{1}{2} \rho v^3 A C_p$

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$r = \sqrt{16.79} = 4.1 m$

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2 years ago

For a moon orbiting its planet, rp is the shortest distance between the moon and its planet andra is the longest distance betwee

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Explanation:

The formula to calculate the eccentricity $e$ of an ellipse is (assuming the moon's orbit in the shape of an ellipse):

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Where:

$r_{a}$ is the apoapsis (the longest distance between the moon and its planet)

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Then:

$e=\frac{r_{a}-0.27 r_{a}}{r_{a}+0.27 r_{a}}$

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podryga [215]

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Explanation:

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