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sashaice [31]
2 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]2 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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kotegsom [21]

Answer:

The amount of heat required is H_t =  1.37 *10^{6} \ J

Explanation:

From the question we are told that

The mass of water is m_w  =  20 \ ounce = 20 * 28.3495 = 5.7 *10^2 g

The temperature of the water before drinking is T_w  =  3.8 ^oC

The temperature of the body is T_b  =  36.6^oC

Generally the amount of heat required to move the water from its former temperature to the body temperature is

H=  m_w  *  c_w * \Delta T

Here c_w is the specific heat of water with value c_w = 4.18 J/g^oC

So

H=   5.7 *10^2 * 4.18 * (36.6 - 3.8)

=> H= 7.8 *10^{4} \  J

Generally the no of mole of sweat present mass of water is

n = \frac{m_w}{Z_s}

Here Z_w is the molar mass of sweat with value

Z_w =  18.015 g/mol

=> n = \frac{5.7 *10^2}{18.015}

=> n = 31.6 \  moles

Generally the heat required to vaporize the number of moles of the sweat is mathematically represented as

H_v  =  n  *  L_v

Here L_v is the latent heat of vaporization with value L_v  = 7 *10^{3} J/mol

=> H_v  =  31.6 * 7 *10^{3}

=> H_v  = 1.29 *10^{6} \  J

Generally the overall amount of heat energy required is

H_t =  H +  H_v

=> H_t =  7.8 *10^{4} +  1.29 *10^{6}

=> H_t =  1.37 *10^{6} \ J

4 0
2 years ago
5. A 1-kg car and a 2-kg car are both released from the top of the same hill and roll down a frictionless track. At the bottom o
grigory [225]

The cars will have equal speeds and the 2 kg car will have greater kinetic energy.

7 0
2 years ago
Michael Phelps needs to swim at an average speed of 2.00 m/s in order to set a new world record in the 200 m freestyle. If he sw
Natasha_Volkova [10]

Answer:

Explanation:

Given

average speed of Phelps v_{avg}=2\ m/s

total distance d=200\ m

he swims first 100 m at an average speed if 1.8 m/s

so time taken is t_1=\frac{100}{1.8}=55.55\ s

suppose t_2 is the time taken to swim remaining half

average velocity is v_{avg}=\frac{displacement}{total\ time}

v_{avg}=\frac{100+100}{55.55+t_2}

t_2+55.55=\frac{200}{2}=100

t_2=44.45\ s

so velocity in the second half is

v_2=\frac{100}{45.45}

v_2=2.19\approx 2.2\ m/s                                      

3 0
2 years ago
To a stationary observer, a man jogs east at 2.5 m/s and a woman jogs west at 1.5 m/s. from the woman's frame of reference, what
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miss Akunina [59]

Answer:

The airplane should release the parcel 6.7*10^2 m before reaching the island

Explanation:

The height of the plane is y_0=100m, and its speed is v=150 m/s

When an object moves horizontally in free air (no friction), the equation for the y measured with respect to ground is

y=y_0 - \frac{gt^2}{2}    [1]

And the distance X is

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Being t the time elapsed since the release of the parcel

If we isolate t from the equation [1] and replace it in equation [2] we get

X = V . \sqrt{\frac{2y_0}{g}}

Using the given values:

x = 150 m/s  \sqrt{\frac{2\times 100m}{9.8 m/sec^2}}

x = 6.7*10^2 m

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2 years ago
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