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Lilit [14]
3 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]3 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]3 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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A disk with a rotational inertia of 5.0 kg. m2 and a radius of 0.25 m rotates on a frictionless fixed axis perpendicu-
diamong [38]

Answer:1.6 rad/s

Explanation:

Given

moment of Inertia  of disk I=5 kg-m^2

radius of disc r=0.25 m

Force F=8 N

Torque T=I\alpha =F\cdot r

5\times \alpha =8\times 0.25

\alpha =0.4 rad/s^2

using

\theta =\omega _0\times t+\frac{\alpha t^2}{2}

\pi =0+\frac{0.4t^2}{2}

2\pi =0.4t^2

t^2=5\pi

t=\sqrt{5\pi }

t=3.96 s

\omega =\omega _0+\alpha t

\omega =0+0.4\times 3.96

\omega =1.58 rad/s\approx 1.6 rad/s

                         

3 0
4 years ago
What process marks the birth of a star?
Misha Larkins [42]

Answer:

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Explanation:

7 0
4 years ago
The force applied while lifting a box would be ?
V125BC [204]
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3 0
3 years ago
The mass of the Sun is 2 × 1030 kg, and the distance between Neptune and the Sun is 30 AU. What is the orbital period of Neptune
Veronika [31]
Kepler's third law states that, for a planet orbiting around the Sun, the ratio between the cube of the radius of the orbit and the square of the orbital period is a constant:
\frac{r^3}{T^2}= \frac{GM}{4 \pi^2} (1)
where
r is the radius of the orbit
T is the period
G is the gravitational constant
M is the mass of the Sun

Let's convert the radius of the orbit (the distance between the Sun and Neptune) from AU to meters. We know that 1 AU corresponds to 150 million km, so
1 AU = 1.5 \cdot 10^{11} m
so the radius of the orbit is
r=30 AU = 30 \cdot 1.5 \cdot 10^{11} m=4.5 \cdot 10^{12} m

And if we re-arrange the equation (1), we can find the orbital period of Neptune:
T=\sqrt{ \frac{4 \pi^2}{GM} r^3} =  \sqrt{ \frac{4 \pi^2}{(6.67 \cdot 10^{-11} m^3 kg^{-1} s^{-2} )(2\cdot 10^{30} kg)}(4.5 \cdot 10^{12} m)^3 }= 5.2 \cdot 10^9 s

We can convert this value into years, to have a more meaningful number. To do that we must divide by 60 (number of seconds in 1 minute) by 60 (number of minutes in 1 hour) by 24 (number of hours in 1 day) by 365 (number of days in 1 year), and we get
T=5.2 \cdot 10^9 s /(60 \cdot 60 \cdot 24 \cdot 365)=165 years
3 0
3 years ago
Help meeeeeeeeeeee<br>what is the rheostat and what its ruleeee??​
oee [108]

Answer:

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Explanation:

6 0
3 years ago
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