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Ainat [17]
2 years ago
14

the water behind hoover dam in nevada is 206 m higher than the colorado river below it. at what rate must water pass through the

hydraulic turbines of this dam to produce 100 mw of power if the turbines are 100 percent efficient
Physics
1 answer:
Nitella [24]2 years ago
5 0

Answer:

the required mass flow rate is 49484.37 kg/s

Explanation:

Given the data in the question;

we first determine the relation for mass flow rate of water that passes through the turbine;

so the relation for net work on the turbine due to the change in potential energy considering 100% efficiency is;

W_{net} = m ( Δ P.E )

so we substitute (gh) for ( Δ P.E );

W_{net} = m (gh)

m = W_{net} / gh

so we substitute our given values into the equation

m = 100 MW / ( 9.81 m/s²) × 206 m

m = ( 100 MW × 10⁶W/MW) / ( 9.81 m/s²) × 206 m

m = 10 × 10⁷ / 2020.86

m = 49484.37 kg/s

Therefore, the required mass flow rate is 49484.37 kg/s

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The driver of a car slams on the brakes when he sees a tree blocking the road. The car slows uniformly with acceleration of −5.2
sergejj [24]

Answer:

The car strikes the tree with a final speed of 4.165 m/s

The acceleration need to be of -5.19 m/seg2 to avoid collision by 0.5m

Explanation:

First we need to calculate the initial speed V_{0}

x=V_{0} *t+\frac{1}{2} *a*(t^{2} )\\62.5m=V_{0} *4.15s+\frac{1}{2} *-5.25\frac{m}{s^{2} } *(4.15^{2} )\\V_{0}=25.953\frac{m}{s}

Once we have the initial speed, we can isolate the final speed from following equation:

V_{f} =V_{0} +a*t  V_{f}= 4.165 \frac{m}{s}  

Then we can calculate the aceleration where the car stops 0.5 m before striking the tree.

To do that, we replace 62 m in the first formula, as follows:

x=V_{0} *t+\frac{1}{2} *a*(t^{2} )\\62m=25.953\frac{m}{s}*4.15s+\frac{1}{2} *-a\frac{m}{s^{2} } *(4.15^{2} )\\a=-5.19\frac{m}{s^{2} }

3 0
2 years ago
Read 2 more answers
The air in a pipe resonates at 150 Hz and 750 Hz, one of these resonances being the fundamental. If the pipe is open at both end
Xelga [282]

Answer:

Explanation:

Two frequencies with magnitude 150 Hz and 750 Hz are given

For Pipe open at both sides

fundamental frequency is 150 Hz as it is smaller

frequency  of pipe is given by

f=\frac{nv}{2L}

where L=length of Pipe

v=velocity of sound

f=150\ Hz for n=1

and f=750 is for n=5

thus there are three resonance frequencies for n=2,3 and 4

For Pipe closed at one end

frequency is given by

f=\frac{(2n+1)}{4L}\cdot v

for n=0

f_1=\frac{v}{4L}

f_1=150\ Hz

for n=2

f_2=\frac{5v}{4L}

Thus there is one additional resonance corresponding to n=1 , between f_1 and f_2

8 0
2 years ago
A segment of wire of total length 2.0 m is formed into a circular loop having 5.0 turns. If the wire carries a 1.2-A current, de
docker41 [41]

Answer:

Magnetic field at the center of the loop B=5.89\times 10^{-5}\ T.

Explanation:

It is given that total length of wire is 2 m and number of circular loop is 5 turns.

Therefore ,

5\times ( 2\pi r)=2 \ m .\\\\r=\dfrac{1}{5 \pi}=0.064\ m.

We know , magnetic field at the center of loop is given by :

B=N\dfrac{\mu_o i}{2r}

Putting all values in above equation we get :

B=5\times \dfrac{4\pi\times 10^{-7}\times 1.2}{2\times 0.064}\\\\B=5.89\times 10^{-5}\ T.

Hence , this is the required solution.

8 0
2 years ago
A 1-kg mass is dropped from a third floor window. The acceleration of the mass is found to be 8 m/s2. What is the average force
Paha777 [63]
Summary:
m=1kg
a=8 m/s^2
g= 9,8 m/s^2
F(ar)=?


I hope to help you

7 0
2 years ago
An astronomy student, for her PhD, really needs to estimate the age of a cluster of stars. Which of the following would be part
LiRa [457]

Answer:

A. plot an H-R diagram for the stars in the cluster.

Explanation:

A star cluster can be defined as a constellation of stars, due to gravitational force, which has the same origin.

The astronomy student would have to plot an H-R diagram for the stars in the cluster and determine the age of the cluster by observing the turn-off point. The turn-off is majorly as a result of gradual depletion of the source of energy of the star. Thus, it projects off the constellation.

4 0
2 years ago
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