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ipn
2 months ago
6

A hydroelectric dam holds back a lake of surface area 3.0×106m2 that has vertical sides below the water level. The water level i

n the lake is 150 m above the base of the dam. When the water passes through turbines at the base of the dam, its mechanical energy is converted to electrical energy with 90% efficiency.
(a) If gravitational potential energy is taken to be zero at the base of the dam, how much energy is stored in the top meter of the water in the lake? The density of water is 1000kg/m3
(b) What volume of water must pass through the dam to produce 1000 kilowatt-hours of electrical energy? What distance does the level of water in the lake fall when this much water passes through the dam?
Physics
1 answer:
Softa [3K]2 months ago
8 0

Answer:

4.41 × 10¹² J, 2.72 × 10³ m³, 0.907 × 10 ⁻³ m

Explanation:

Gravitational potential energy is given by the formula: mgh

with m being the mass in kg, g reflecting the gravitational acceleration in m/s², and h indicating the distance above the dam's base.

The mass of the surface water amounts to: density of water × volume of water × 1 m = 1000 kg/m³ × 3.0 × 10⁶ m² × 1 m = 3 × 10⁹ kg

This leads to a total gravitational potential energy of 3 × 10⁹ kg × 9.81 m/s² × 150 m = 4.41 × 10¹² J

b) to find out how much water must flow through the dam to yield 1000 kW-hrs:

1,000 kW-hr = 3.6 × 10⁹ J

Given that the dam has a conversion efficiency of 90% for mechanical to electrical energy:

The necessary gravitational potential energy is 3.6 × 10⁹ J / 0.9 = 4 × 10⁹ J

The water mass needed is calculated using Energy required / gh = 4 × 10⁹ J / (9.81 m/s² × 150 m) = 2.718 × 10⁶ kg

To find the volume, we use density = mass/volume which gives us volume = mass/density = 2.718 × 10⁶ kg / (1000 kg/m³) = 2.72 × 10³ m³

The reduction in the lake's water level equals volume/area = 2.72 × 10³ m³ / (3.0 × 10⁶ m²) = 0.907 × 10 ⁻³ m

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1 month ago
An air hockey game has a puck of mass 30 grams and a diameter of 100 mm. The air film under the puck is 0.1 mm thick. Calculate
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Answer:

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Explanation:

initially given information

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diameter = 100 mm = 0.1 m

thickness = 0.1 mm = 1 ×10^{-4} m

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to determine

time needed for the puck to reduce its speed by 10%

solution

we note that velocity changes from 0 to v

assuming initial velocity = v

therefore final velocity = 0.9v

implying a change in velocity is du = v

and clearance dy = h

shear stress acting on the surface is expressed as

= µ \frac{du}{dy}

therefore

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substituting the values

= 1.75 ×10^{-5} × \frac{v}{10^{-4}}

= 0.175 v

the area between the air and puck is given by

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area = \frac{\pi }{4} 0.1^{2}

area = 7.85 × \frac{v}{10^{-3}} m²

thus, the force on the puck can be represented as

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force = 0.175 v × 7.85 × 10^{-3}

force = 1.374 × 10^{-3} v

now applying Newton's second law

force = mass × acceleration

- force = mass \frac{dv}{dt}

- 1.374 × 10^{-3} v = 0.03 \frac{0.9v - v }{t}

solving for t = \frac{0.1 v * 0.03}{1.37*10^{-3} v}

the time needed after impact for the puck is 2.18 seconds

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2 months ago
At a location where the acceleration due to gravity is 9.807 m/s2, the atmospheric pressure is 9.891 × 104 Pa. A barometer at th
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Answer:

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Explanation:

The measurement of pressure is indicated as p=\rho gh where p denotes the pressure, \rho signifies density, and h represents height

Given values include pressure p=9.891\times 10^4\ Pa, gravity's acceleration g=9.9870\ m/sec^2, and height =1.163 m

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Answer:

She exerts a force of 40 N.

Explanation:

The fact that the ring remains stationary indicates that the forces are in equilibrium.

Let’s denote Jo's force as x.

The equation to consider is

140 = x + 100

x = 40

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1 month ago
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