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jeyben [28]
4 years ago
13

assuming 100% efficient energy conversion how much water stored behind a 50 centimeter high hydroelectric dam would be required

to charged the battery ​
Physics
1 answer:
Nataly [62]4 years ago
4 0

Answer:

The amount of water that will power a battery with that rating = 7.35 m³

Explanation:

The power rating for the battery is missing from the question.

Complete Question

Assuming 100% efficient energy conversion how much water stored behind a 50 centimeter high hydroelectric dam would be required to charged the battery with power rating, 12 V, 50 Ampere-minutes

Solution

Potential energy possessed by water at that height = mgH

m = mass of the water = ρV

ρ = density of water = 1000 kg/m³

V = volume of water = ?

g = acceleration due to gravity = 9.8 m/s²

H = height of water = 50 cm = 0.5 m

Potential energy = ρVgH = 1000 × V × 9.8 × 0.5 = (4900V) J

Energy of the battery = qV

q = 50 A.h = 50 × 60 = 3,000 C

V = 12 V

qV = 3,000 × 12 = 36,000 J

Energy = 36,000 J

At a 100% conversion rate, the energy of the water totally powers the battery

(4900V) = (36,000)

4900V = 36,000

V = (36,000/4900)

V = 7.35 m³

Hope this Helps!!!

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Andreas93 [3]

Answer:

The difference in the amount of energy transferred by the two bulbs is 1200 J.

Explanation:

The energy transferred by the two lightbulbs can be calculated with the given equation:

E = P*t

Where P is the power and t is the time

For the 60 W lightbulb:

E_{60} = P*t = 60 W*30 s = 1800 J

For the 100 W lightbulb:

E_{100} = P*t = 100 W*30 s = 3000 J

Hence, the difference in the amount of energy transferred is:

E_{t} = E_{100} - E_{60} = 3000 J - 1800 J = 1200 J

Therefore, the difference in the amount of energy transferred by the two bulbs is 1200 J.

I hope it helps you!

4 0
3 years ago
Which statement accurately describes the relationship between air pressure, air density, or altitude?​
8_murik_8 [283]

Answer:

The answer is: dense air exerts more pressure than less dense air.

Pressure is due to the weight of the above column of the substance which is a force resulting from its weight that is the product of its mass and gravitational pull. The denser the air and the grater the column, the larger its mass and hence the more the pressure it exerts.

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Explanation:

8 0
4 years ago
Read 2 more answers
A 0.13 g plastic bead is charged by the addition of 1.0Ã1010 excess electrons. what electric field eâ (strength) will cause the
slavikrds [6]
The working equation for this is: E = F/Q, where E is the strength of electric field, F is the force and Q is the charge. The force is equal to:

F = mg = (0.13/1000 kg)(9.81 m/s²) = 1.2753×10⁻³ N

The charge of he excess electrons is equal to:
Q = (-1.6021766208×10⁻¹⁹ C/electron)(1×10¹⁰ electrons)
Q = -1.6021766208×10⁻⁹ C

E = 1.2753×10⁻³ N/-1.6021766208×10⁻⁹ C
E = -795,979.66 N/C
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koban [17]
Yes everything is made of cells because without cells our being can't stay together.
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Two point charges, q1 = 2.0 × 10−7 C and q2 = −6.0 × 10−8 C, are held 25.0 cm apart. (a) What is the electric field at a point 5
AlladinOne [14]

Answer:

a)432000\frac{N}{C}\hat{i}

b)-6.92\times10^{-14}N\hat{i}

Explanation:

a)

The magnitude of the electric field generated by a charged particle at a distance r is:

E= k\frac{|Q|}{r^{2}}

With Q the charge of the particle and k the constant ()

So, the electric field generated by q1 knowing that the point 5.0 cm apart the negative charge is 25.0cm-5.0cm=20.0 cm=0.2m apart the positive charge is:

E_1= k\frac{|q1|}{r_1^{2}} =(9.0\times10^{9}\,\frac{Nm^{2}}{C^{2}})\frac{|2.0\times10^{-7}|}{(0.2)^{2}}

E_1= 45000\frac{N}{C}

and the electric field generated by q2:

E_2= k\frac{|q2|}{r_2^{2}} =(9.0\times10^{9}\,\frac{Nm^{2}}{C^{2}})\frac{|-6.0\times10^{-8}|}{(0.05)^{2}}

E_2=216000\frac{N}{C}

Those are the magnitudes of the electric field, but electric field is a vector quantity, so the direction is important. Electric field generated by negative particles points towards the charge and electric field generated by positive particles points away the particle. So, if we define positive direction towards negative particle (x-axis):

\overrightarrow{E_2}=+216000\frac{N}{C}\hat{i}

\overrightarrow{E_1}= +45000\frac{N}{C}\hat{i}

Vector quantities satisfy superposition principle, this is \overrightarrow{E}=\overrightarrow{E_1}+\overrightarrow{E_2}, with E the total electric field.

\overrightarrow{E}=(216000+45000)\hat{i}=432000\frac{N}{C}\hat{i}

b) The force is:

\overrightarrow{F}=e*\overrightarrow{E},

with q the charge of an electron

\overrightarrow{F}=(-1.61\times10^{-19})*(432000)\hat{i}=-6.92\times10^{-14}N\hat{i}

8 0
3 years ago
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