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RSB [31]
3 years ago
14

Consider a river flowing toward a lake at an average speed of 3 m/s at a rate of 550 m3/s at a location 58 m above the lake surf

ace. Determine the total mechanical energy of the river water per unit mass (in kJ/kg) and the power generation potential of the entire river at that location (in MW). The density of water is 1000 kg/m3, and the acceleration due to gravity is 9.81 m/s2. The total mechanical energy of the river per unit mass is kJ/kg. The power generation potential of the entire river at that location is MW..
Physics
1 answer:
Vladimir [108]3 years ago
4 0

Answer:

1. 0.574 kJ/kg

2. 315.7 MW

Explanation:

1. The mechanical energy per unit mass of the river is given by:

E_{m} = E_{k} + E_{p}

E_{m} = \frac{1}{2}v^{2} + gh

Where:

Ek is the kinetic energy

Ep is the potential energy

v is the speed of the river = 3 m/s

g is the gravity = 9.81 m/s²

h is the height = 58 m

E_{m} = \frac{1}{2}(3 m/s)^{2} + 9.81 m/s^{2}*58 m = 0.574 kJ/Kg

Hence, the total mechanical energy of the river is 0.574 kJ/kg.

2. The power generation potential on the river is:

P = m(t)E_{m} = \rho*V(t)*E_{m} = 1000 kg/m^{3}*550 m^{3}/s*0.574 kJ/kg = 315.7 MW

Therefore, the power generation potential of the entire river is 315.7 MW.

I hope it helps you!

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Sergei uses a lever to lift a heavy rock. He obtains a 2.5m lever and places the fulcrum 0.7m from the rock. What is the ideal m
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Answer:

m = 3.57

Explanation:

Given that,

Sergei uses a lever to lift a heavy rock. He obtains a 2.5m lever and places the fulcrum 0.7m from the rock.

We need to find the ideal mechanical advantage of Sergei's lever.

It is equal to the ratio of resistance arm to the effort arm. In terms of length it is given by :

m=\dfrac{d_2}{d_1}\\\\m=\dfrac{2.5}{0.7}\\\\=3.57

So, the ideal mechanical advantage of the lever is 3.57.

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A wire has a current density of 6.25 × 10 6 A / m 2 6.25×106 A/m2 . If the cross-sectional area of the wire is 1.79 mm 2 1.79 mm
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Answer:17.44A

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A 0.150-kg glider is moving to the right with a speed of 0.80 m/s on a frictionless, horizontal air track. The glider has a head
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Answer:

The final speed of 0.150-kg glider after collision is 3.2 m/s to the left

The final speed of 0.300-kg glider after collision is 0.20 m/s to the left

Explanation:

Given;

mass of glider moving to the right, m₁ = 0.150-kg

mass of glider moving to the left, m₂ = 0.300-kg

initial speed of glider moving to the right, u₁ = 0.80 m/s

initial speed of glider moving to the left, u₂ = 2.20 m/s

Apply the principle of conservation of linear momentum;

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

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Again, their relative velocity after collision is given as;

u₁ - u₂ = v₂ - v₁

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3 = v₂ - v₁

v₂ =  v₁ + 3  ------------equation (ii)

Substitute v₂ in equation (i)

0.15v₁ + 0.3v₂ = - 0.54

0.15v₁ + 0.3(v₁ + 3 ) = - 0.54

0.15v₁ + 0.3v₁ + 0.9 = - 0.54

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0.45v₁  = -1.44

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v₁ = - 3.2 m/s

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v₂ = -3.2 + 3

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Therefore, the final speed of 0.300-kg glider after collision is 0.20 m/s to the left

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