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Lilit [14]
4 years ago
11

A town is exploring the option of generating hydroelectric power from a stream. What is the maximum available power in a waterfa

ll that has a flow rate of 200 kg/s dropping over a 15 m change in height?
Physics
2 answers:
tester [92]4 years ago
8 0
The power is equal to energy transferred per unit time.
So, here power will be due to potential energy of the water kept at a certain height. And water is flowing down at the given flow rate.

The flow rate of water is 200kg/s
And the height given is : 15m

So, we calculate the potential energy as: mgh
Where m is the mass,
g is acc due to gravity = 9.8m/s
h is the height

So, potential energy = 200kg*9.8*15
= 29,400 J

Therefore, the maximum power generated is 29400 J/sec = 29400 Watts
dybincka [34]4 years ago
6 0
To answer this we calculate the potential energy of 200kg of water at 15m height. We know that this amount of energy is available every second and power is energy expressed per unit time (often per second).

Ep = mass x gravity x height
Ep = 200kg x 9.8 x 15 = 29,400 Joules or 29.4 mega joules (MJ)

Energy = Power x Time so Power = Energy / Time
1 Watt of power equals 1 Joule of energy per second
So we don't actually need to do a conversion, the waterfall is providing 29.4MJ of energy per second, which equates to 29.4 kilowatts (kW) of available power.
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In the first stage of a two-stage Carnot engine, energy is absorbed as heat Q1 at temperature T1 = 500 K, work W1 is done, and e
Nataly [62]

Answer:

Efficiency = 52%

Explanation:

Given:

First stage

heat absorbed, Q₁ at temperature T₁ = 500 K

Heat released, Q₂ at temperature T₂ = 430 K

and the work done is W₁

Second stage

Heat released, Q₂ at temperature T₂ = 430 K

Heat released, Q₃ at temperature T₃ = 240 K

and the work done is W₂

Total work done, W = W₁ + W₂

Now,

The efficiency is given as:

\eta=\frac{\textup{Total\ work\ done}}{\textup{Energy\ provided}}

or

Work done = change in heat

thus,

W₁ = Q₁ - Q₂

W₂ = Q₂ - Q₃

Thus,

\eta=\frac{(Q_1-Q_2)\ +\ (Q_2-Q_3)}{Q_1}}

or

\eta=1-\frac{(Q_1-Q_3)}{Q_1}}

or

\eta=1-\frac{(Q_3)}{Q_1}}

also,

\frac{Q_1}{T_1}=\frac{Q_2}{T_2}=\frac{Q_3}{T_3}

or

\frac{T_3}{T_1}=\frac{Q_3}{Q_1}

thus,

\eta=1-\frac{(T_3)}{T_1}}

thus,

\eta=1-\frac{(240\ K)}{500\ K}}

or

\eta=0.52

or

Efficiency = 52%

8 0
4 years ago
the very high voltage needed to create a spark across the spark plug is produced at the a. transformer's primary winding. b. tra
Karo-lina-s [1.5K]
I think the correct answer from the choices listed above is option B. The very high voltage needed to create a spark across the spark plug is produced at the  transformer's secondary winding. <span>The secondary coil is engulfed by a powerful and changing magnetic field. This field induces a current in the coils -- a very high-voltage current.</span>
5 0
3 years ago
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Find the magnitude of the emf induced in the rod when it is moving toward the right with a speed 7.50 m/s.
Flauer [41]

Answer:

3 volts

Explanation:

It is given that,

Magnetic field, B = 0.8 T

Length of a conducting rod, l = 50 cm = 0.5 m

Velocity of the conducting rod, v = 7.5 m/s

We need to find the magnitude of the emf induced in the rod when it is moving toward the right. When a rod is moved in a magnetic field, an emf is induced in it and it is given by :

\epsilon=Blv

Putting all the values,

\epsilon=0.8\times 0.5\times 7.5\\\\\epsilon=3\ V

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6 0
4 years ago
Define oxidation number
Delicious77 [7]

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8 0
3 years ago
The volume flow rate of blood leaving the heart to circulate throughout the body is about 5 L/min for a person at rest. All this
skelet666 [1.2K]

Answer:

n=2.9\times 10^9

A=1.88\times 10^{-8}\ m^2

Explanation:

Given that

Q= 5 L/min

1 L = 10⁻³ m³/s

1 min = 60 s

Q=0.083 x 10⁻³ m³/s

d= 6 μm

v= 1 mm/s

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q = A v

A=\dfrac{\pi}{4}d^2

A=\dfrac{\pi}{4}\times (6\times 10^{-6})^2\ m^2

A=2.8 x 10⁻¹¹ m²

v= 1 x 10⁻³  m/s

q= 2.8 x 10⁻¹⁴  m³/s

Lets take total number of tube is n

Q= n q

n=Q/q

n=\dfrac{0.083\times 10^{-3} }{ 2.8\times 10^{-14}}

n=2.9\times 10^9

Surface  area A

A= π d L

A=\pi \times 6\times 10^{-6}\times 10^{-3}\ m^2

A=1.88\times 10^{-8}\ m^2

7 0
3 years ago
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