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Vesna [10]
3 years ago
8

Consider a river flowing toward a lake at an average speed of 3m/s at a rate of 500 m^3/s at a location 58m above the lake surfa

ce. 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/m^3, and the acceleration due to gravity is 9.81 m/s^2.
Physics
1 answer:
mario62 [17]3 years ago
5 0

Answer:

Total power generation will be 291.465 MW

Explanation:

We have given that river is flowing at rate of  V = 500m^3/sec

Average speed v = 3 m/sec

Height h = 58 m

We know that total mechanical energy is given by

E = KE +PE

E=\frac{1}{2}mv^2+mgh

Energy per unit mass

\frac{E}{m}=\frac{1}{2}v^2+gh=\frac{1}{2}\times 3^2+9.81\times 58=583.29j/kg

Total power generation from the river is given by

P=energy\ per\ unit\ mass\times volume\ flow\ rate\times density

So P=583.29\times 500\times 1000=291.465MW

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For Al, its atomic number is 13 and its mass number is 27. How many neutrons does it have?
ira [324]

Answer:

B.14 your welcome!

Explanation:

6 0
3 years ago
The Cartesian coordinate of a point in the xy plane are (x,y)=(-3.50,-2.50)m. Find the poler coordinate of this point
Masja [62]

Answer:

The polar coordinate of P(x,y) = (-3.50\,m,-2.50\,m) is P (r,\theta) = (4.301\,m, 215.538^{\circ}).

Explanation:

Given a point in rectangular form, that is P(x,y) = (x,y), its polar form is defined by:

P(x,y) = (r,\theta) (1)

Where:

r - Norm, measured in meters.

\theta - Direction, measured in sexagesimal degrees.

The norm of the point is determined by Pythagorean Theorem:

r = \sqrt{x^{2}+y^{2}} (2)

And direction is calculated by following trigonometric relation:

\theta = \tan^{-1} \frac{y}{x} (3)

If we know that x = -3.50\,m and y = -2.50\,m, then the components of coordinates in polar form is:

r = \sqrt{(-3.50\,m)^{2}+(-2.50\,m)^{2}}

r \approx 4.301\,m

Since x < 0\,m and y < 0\,m, direction is located at 3rd Quadrant. Given that tangent function has a period of 180º, we find direction by using this formula:

\theta = 180^{\circ}+\tan^{-1} \left(\frac{-2.50\,m}{-3.50\,m} \right)

\theta \approx 215.538^{\circ}

The polar coordinate of P(x,y) = (-3.50\,m,-2.50\,m) is P (r,\theta) = (4.301\,m, 215.538^{\circ}).

5 0
3 years ago
a concrete slab 20 m long and weighing 400,000 N is supported by one pillar. A 19,600 N car is parked 8 meters from one end, whe
Elden [556K]

let the distance of pillar is "r" from one end of the slab

So here net torque must be balance with respect to pillar to be in balanced state

So here we will have

Mg(r - L/2) = mg(L/2 - 8)

here we know that

mg = 19600 N

Mg = 400,000 N

L = 20 m

from above equation we have

400,000(r - 10) = 19,600 (10 - 8)

r - 10 = 0.098

r = 10.098 m

so pillar is at distance 10.098 m from one end of the slab

3 0
3 years ago
A horizontal clothesline is tied between 2 poles, 12 meters apart. When a mass of 4 kilograms is tied to the middle of the cloth
Olegator [25]

Answer:

T=26.03 N

Explanation:

Given that

Distance between poles = 12 m

Mass of block m= 4 kg

Sag distance = 5 m

Lets take tension in the clothesline is T.

The component of tension in vertical direction will be T cosθ.

By force balancing

2 T cosθ = 40

here tan\theta =\dfrac{5}{6}

θ=39.80°

2 T cos39.8 = 40

T=26.03 N

6 0
3 years ago
A large box slides across a frictionless surface with a velocity of 12 m/s and a mass of 4
denpristay [2]
Answer= 8m/s

Because total Momentum before= total momentum after

Momentum before (p=mu)
p=(4)(12)= 48
p=2(0)=0
So total momentum before=48

Momentum after (p=mu)
Masses combined —2+4=6kg
p=6u


Mb=Ma
48=6u
u=8m/s
3 0
3 years ago
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