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blagie [28]
3 years ago
6

Consider a river flowing toward a lake at an average velocity of 3 m/s at a rate of 500 m3/s at a location 90 m above the lake s

urface. Determine the total mechanical energy of the river water per unit mass and the power generation potential of the entire river at that location.
Physics
1 answer:
WINSTONCH [101]3 years ago
7 0

Answer: a) 5 x 10^5 kg/s b) 444 MW

Explanation:

Kinetic energy per unit mass Ke is

Ke = V^2 / 2

Ke = 3^2 / 2 = 4.5 J/kg = 0.0045 kJ/kg

Now potential energy per unit mass Pe is

Pe = g x z = 9.8 x 90 = 882.9 J/kg = 0.8829 kJ/kg

The total mechanical energy of the River per unit mass e = Ke + Pe = 0.0045 + 0.8829 = 0.88744 J/kg

M = P x V = 1000 x 500 = 5 x 10^5 kg/s

b) power generation potential of the entire river at the location Wmax = Emech = M x Emech = 500,000 x 0.88744 = 444,000kW = 444MW

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A train locomotive is pulling two cars of the same mass behind it. Determine the ratio of the tension in the coupling (think of
Anna007 [38]

Answer:

The ratio is  \frac{F_{T1}}{F_{T2}}  =  2

Explanation:

The diagram for this question is shown on the first uploaded image

Here we are assume the acceleration of the train is a

which makes the acceleration of each car a

From the question we are told that

      Considering the second car

 The force causing it s movement  is mathematically represented as

       F_{T2} =  ma

 Considering the first car

 The force causing it s movement  is mathematically represented as

      F  = F_{T1} -F_{T2} = ma

=>   F_{T1} -ma  = ma

=>   F_{T1} =  2 ma

=> \frac{F_{T1}}{ma}  =  2

=> \frac{F_{T1}}{F_{T2}}  =  2

7 0
3 years ago
Voltage needed to raise current to 3.75a using 20,20,200 resistor set
Varvara68 [4.7K]

<u>Answer:</u> The voltage needed is 35.7 V

<u>Explanation:</u>

Assuming that the resistors are arranged in parallel combination.

For the resistors arranged in parallel combination:

\frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}

We are given:

R_1=20\Omega\\R_2=20\Omega\\R_3=200\Omega

Using above equation, we get:

\frac{1}{R}=\frac{1}{20}+\frac{1}{20}+\frac{1}{200}\\\\\frac{1}{R}=\frac{10+10+1}{200}\\\\R=\frac{200}{21}=9.52\Omega

Calculating the voltage by using Ohm's law:

V=IR         .....(1)

where,

V = voltage applied

I = Current = 3.75 A

R = Resistance = 9.52\Omega

Putting values in equation 1, we get:

V=3.75\times 9.52\\\\V=35.7V

Hence, the voltage needed is 35.7 V

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The distance between Jupiter and the sun is 5.2 AU.

According to Kepler's third law, the square of the period of revolution of planets is proportional to the cube of their mean distances from the sun. From this; T^2 = r^3.

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