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marshall27 [118]
2 years ago
10

1.A river flowing steadily at a rate of 240 m3/s is considered for hydroelectric power generation. It is determined that a dam c

an be built to collect water and release it from an elevation difference of 50 m to generate power. Determine how much power can be generated from this river water after the dam is filled
Physics
1 answer:
SIZIF [17.4K]2 years ago
4 0

Answer:

the power that can be generated by the river is 117.6 MW

Explanation:

Given that;

Volume flow rate of river v = 240 m³/s

Height above the lake surface a h = 50 m

Amount of power can be generated from this river water after the dam is filled = ?

Now the collected water in the dam contains potential energy which is used for the power generation,

hence, total mechanical energy is due to potential energy alone.

E_{mech} = m(gh)

first we determine the mass flow rate of the fluid m

m = p×v

where p is density ( 1000 kg/m³

so we substitute

m = 1000kg/m³ × 240 m³/s

m = 240000 kg/s

so we plug in our values into ( E_{mech} = m(gh) kJ/kg )

E_{mech} = 240000 × 9.8 × 50

E_{mech} = 117600000 W

E_{mech} = 117.6 MW

Therefore, the power that can be generated by the river is 117.6 MW

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Inessa05 [86]

Explanation:

The given data is as follows.

    Inner wheel Radius = 20 m,

   Distance between left and right wheel = 1.5m,

Let us assume speed of drive shaft is N rpm.

Formula to calculate angular velocity is as follows.

    Angular velocity of automobile = w = \frac{V}{R}

where,   V = linear velocity of automobile m/min,

              R = turning radius from automobile center in meter

In the given case, angular velocity remains same for inner and outer wheel but there is change in linear velocity of inner wheel and outer wheel.

Now, we assume that

         u = linear velocity of inner wheel

and,   u' = linear velocity of outer wheel.

Formula for angular velocity of inner wheel w = ,

Formula for angular velocity of outer wheel w =

Now, for inner wheels

                   w =

                      = \frac{u}{(R - d)}

                  u = V \times \frac{(R - d)}{R}

                    = V \times (1 - \frac{d}{R})

If radius of wheel is r it will cover  distance in one min.

Since, velocity of wheel is u it will cover distance u in unit time(min)

Thus,             u = 2\pi rn = V \times (1 - \frac{d}{R})

Now, rotation per minute of inner wheel is calculated as follows.

         n = \frac{V}{2 \pi r \times (1 - \frac{d}{R})}

            = \frac{V}{2 \pi r \times (1 - \frac{0.75}{20})} (since 2d = 1.5m given, d = 0.75m),

             = \frac{V}{r} \times 0.1532

So, rotation per minute of outer wheel; n' =  

                   = \frac{V}{2 \pi r \times (1 + \frac{0.75}{20})}

                   = \frac{V}{r} \times 0.1651

5 0
3 years ago
A vertical spring has a spring constant of 2900 N/m. The spring is compressed 80 cm and a 8 kg spider is placed on the spring. T
Serga [27]

Answer:

a)  k_{e} = 928 J , b)U = -62.7 J , c) K = 0 , d) Y = 11.0367 m,  e)  v = 15.23 m / s  

Explanation:

To solve this exercise we will use the concepts of mechanical energy.

a) The elastic potential energy is

      k_{e} = ½ k x²

      k_{e} = ½ 2900 0.80²

      k_{e} = 928 J

b) place the origin at the point of the uncompressed spring, the spider's potential energy

     U = m h and

     U = 8 9.8 (-0.80)

     U = -62.7 J

c) Before releasing the spring the spider is still, so its true speed and therefore the kinetic energy also

      K = ½ m v²

      K = 0

d) write the energy at two points, maximum compression and maximum height

     Em₀ = ke = ½ m x²

     E_{mf} = mg y

     Emo = E_{mf}

     ½ k x² = m g y

     y = ½ k x² / m g

     y = ½ 2900 0.8² / (8 9.8)

     y = 11.8367 m

As zero was placed for the spring without stretching the height from that reference is

     Y = y- 0.80

     Y = 11.8367 -0.80

     Y = 11.0367 m

Bonus

Energy for maximum compression and uncompressed spring

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     E_{mf}= ½ m v²

     Emo = E_{mf}

     Emo = ½ m v²

      v =√ 2Emo / m

     v = √ (2 928/8)

     v = 15.23 m / s

8 0
3 years ago
How are energy time and power related physics?
olganol [36]
Power is the rate of energy. Mathematically, it is

Power (p) = Energy(E) / Time(t)

Hope this helps!
6 0
3 years ago
The box leaves position x=0 with speed v0. The box is slowed by a constant frictional force until it comes to rest at position x
Diano4ka-milaya [45]

Answer:

Ff=m\times \dfrac{V_o^2}{2X_1}

Explanation:

Given that

At X=0 V=Vo

At X=X1  V=0

As we know that friction force is always try to oppose the motion of an object. It means that it provide acceleration in the negative direction.

We know that

V^2=U^2-2aS

0=V_o^2-2a X_1

a=\dfrac{V_o^2}{2X_1}

So the friction force on the box

Ff= m x a

Ff=m\times \dfrac{V_o^2}{2X_1}

Where m is the mass of the box.

4 0
3 years ago
Brianna pushes a 20 kg box with a force of 50N to achieve an acceleration of 2.5 m/s/s. In order to push a 30 kg box at the same
Elanso [62]

Answer:

She applies 75N of force

Explanation:

F=ma

50=20(2.5)

F=30(2.5)

F=75 N

5 0
2 years ago
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