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larisa [96]
2 years ago
6

Consider a river flowing toward a lake at an average velocity of 3 m/s at a rate of 510 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:
tangare [24]2 years ago
4 0

The mechanical energy of the river water per unit mass is 887.4 J/kg

The power generated is 452.574 MW.

Given:

Average velocity (v) = 3 m/s

Rate = 510 m³/s

Height (h)  = 90 m

We know, that mechanical energy is the sum of potential energy and kinetic energy.

So,

E = \frac{1}{2}×m×v² + m×g×h                    

and energy per mass unit is

E/m =  \frac{1}{2}×v² + g×h

E/m =  \frac{1}{2}×3² + 9.81×90

E/m = 887.4 J/kg

So, mechanical energy per unit mass is 887.4 J/kg.

Power generated is expressed as;

Power generated = energy per unit mass ×rate×density

Density of water = 1000 kg/m³

Power generated = 887.4× 510× 1000

Power generated = 452574000 W

So, the power generated is 452.574 MW.

Learn more about mechanical energy here:

brainly.com/question/15191390

#SPJ4

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Answer: Force F will be one-sixteenth of the new force when the charges are doubled and distance halved

Explanation:

Let the charges be q1 and q2 and the distance between the charges be 'd'

Mathematical representation of coulombs law will be;

F1=kq1q2/d²...(1)

Where k is the electrostatic constant.

If q1 and q2 is doubled and the distance halved, we will have;

F2 = k(2q1)(2q2)/(d/2)²

F2 = 4kq1q2/(d²/4)

F2 = 16kq1q2/d²...(2)

Dividing equation 1 by 2

F1/F2 = kq1q2/d² ÷ 16kq1q2/d²

F1/F2 = kq1q2/d² × d²/16kq1q2

F1/F2 = 1/16

F1 = 1/16F2

This shows that the force F will be one-sixteenth of the new force when the charges are doubled and distance halved

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3 years ago
Suppose the ring rotates once every 4.30 ss . If a rider's mass is 58.0 kgkg , with how much force does the ring push on her at
Stells [14]

Answer:

422.36 N

Explanation:

given,

time of rotation = 4.30 s

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v = \dfrac{2\pi\times 8}{4.30}

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now, Force does the ring push on her at the top

- N - m g = \dfrac{-mv^2}{R}

N + m g = \dfrac{mv^2}{R}

N = \dfrac{mv^2}{R}- m g

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The force exerted by the ring to push her is equal to 422.36 N.

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Consider the Uniform Circular Motion Gizmo configured as shown. Notice that, under the current settings, |a|=0.50m/s2. What chan
Eddi Din [679]

To increase the centripetal acceleration to 2.00 m/s^2, you can double the speed or decrease the radius by 1/4

Explanation:

An object is said to be in uniform circular motion when it is moving at a constant speed in a circular path.

The acceleration of an object in uniform circular motion is called centripetal acceleration, and it is given by

a=\frac{v^2}{r}

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so, the acceleration has increased by a factor 4 again.

Learn more about centripetal acceleration:

brainly.com/question/2562955

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#LearnwithBrainly

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