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sashaice [31]
3 years ago
14

When the Glen Canyon hydroelectric power plant in Arizona is running at capacity, 690 m3 of water flows through the dam each sec

ond. The water is released 220 m below the top of the reservoir.
Part A

If the generators that the dam employs are 90% efficient, what is the maximum possible electric power output?

Express your answer with the appropriate units.
Physics
1 answer:
bixtya [17]3 years ago
3 0

Answer:

1340.2MW

Explanation:

Hi!

To solve this problem follow the steps below!

1 finds the maximum maximum power, using the hydraulic power equation which is the product of the flow rate by height by the specific weight of fluid

W=αhQ

α=specific weight for water =9.81KN/m^3

h=height=220m

Q=flow=690m^3/s

W=(690)(220)(9.81)=1489158Kw=1489.16MW

2. Taking into account that the generator has a 90% efficiency, Find the real power by multiplying the ideal power by the efficiency of the electric generator

Wr=(0.9)(1489.16MW)=1340.2MW

the maximum possible electric power output is 1340.2MW

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katen-ka-za [31]

The initial potential energy of the wagon containing gold boxes will enable

it roll down the hill when cut loose.

The Lone Ranger and Tonto have approximately <u>5.1 seconds</u>.

Reasons:

Mass wagon and gold = 166 kg

Location of the wagon = 77 meters up the hill

Slope of the hill = 8°

Location of the rangers = 41 meters from the canyon

Mass of Lone Ranger, m₁ = 65 kg

Mass of Tonto m₂ = 66 kg

Solution;

Height of the wagon above the level ground, h = 77 m × sin(8°) ≈ 10.72 m

Potential energy = m·g·h

Where;

g = Acceleration due to gravity ≈ 9.81 m/s²

Potential energy of wagon, P.E. ≈ 166 × 9.81 × 10.72 = 17457.0912

Potential energy of wagon, P.E. ≈ 17457.0912 J

By energy conservation, P.E. = K.E.

K.E. = \mathbf{\dfrac{1}{2} \cdot m \cdot v^2}

Where;

v = The velocity of the wagon a the bottom of the cliff

Therefore;

\dfrac{1}{2} \times 166 \times v^2 = 17457.0912

v = \sqrt{\dfrac{17457.0912}{\dfrac{1}{2} \times 166} } \approx 14.5

Velocity of the wagon, v ≈ 14.5 m/s

Momentum = Mass, m × Velocity, v

Initial momentum of wagon = m·v

Final momentum of wagon and ranger = (m + m₁ + m₂)·v'

By conservation of momentum, we have;

m·v = (m + m₁ + m₂)·v'

\therefore v' = \mathbf{ \dfrac{m \cdot v}{(m + m_1 + m_2)  }}

Which gives;

\therefore v' = \dfrac{166 \times 14.5}{(166 + 65 + 66)  } \approx 8.1

The velocity of the wagon after the Ranger and Tonto drop in, v' ≈ 8.1 m/s

Time = \dfrac{Distance}{Velocity}

\mathrm{The \ time \ the\ Lone \  Ranger \  and  \ Tonto \  have,  \ t} = \dfrac{41 \, m}{8.1 \, m/s} \approx 5.1 \, s

The Lone Range and Tonto have approximately <u>5.1 seconds</u> to grab the

gold and jump out of the wagon before the wagon heads over the cliff.

Learn more here:

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5 0
2 years ago
A 2kg block of which material would require 450 joules of thermal energy to increase its temperature by 1 degree Celsius?
12345 [234]

The block is made of A) Tin, as its specific heat capacity is 0.225 J/(g^{\circ}C)

Explanation:

When an amount of energy Q is supplied to a sample of material of mass m, the temperature of the material increases by \Delta T, according to the following equation :

Q=mC_s \Delta T

where  C_s is the specific heat capacity of the material.

In this problem, we have:

m = 2 kg = 2000 g is the mass of the unknown material

Q = 450 J is the amount of energy supplied to the block

\Delta T = 1^{\circ}C is the change in temperature of the material

Solving the equation for C_s, we can find the specific heat capacity of the unknown sample:

C_s = \frac{Q}{m \Delta T}=\frac{450}{(2000)(1)}=0.225 J/(g^{\circ}C)

And by comparing with tabular values, we can find that this value is approximately the specific heat capacity of tin.

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7 0
3 years ago
A race car has a centripetal acceleration of 15.625 m/s2 as it goes around a curve. If the curve is a circle with radius 40 m, w
myrzilka [38]
The centripetal acceleration is given by
a_c =  \frac{v^2}{r}
where v is the tangential speed and r the radius of the circular orbit.

For the car in this problem, a_c = 15.625 m/s^2 and r=40 m, so we can re-arrange the previous equation to find the velocity of the car:
v= \sqrt{a_c r}= \sqrt{(15.625 m/s^2)(40 m)}=25 m/s
8 0
3 years ago
Alan rents a private jet for the weekend. He flies 400 km south to New York, then flies 700 km west to Chicago, then 1200 km to
Ratling [72]

Answer:

Total distance = 400+700+1200= 2300km

Explanation:

the resultant of d 1st right angle triangle + 1200

= 806.2 + 1200 = 2006.2km

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3 years ago
An unusually high tide is called a _______ tide.
Monica [59]
 Proxigean Spring <span>Tide</span>
8 0
3 years ago
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