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3 years ago

3

An often-overlooked consequence of power generation is "thermal pollution" from excess heat deposited in the envi- ronment. The

heating can be considerable. Here are some typical numbers: A nuclear power plant generates 1.0 GW of electric power with an operating efficiency of 35%. The reactor has a single-pass cooling system; water from a nearby river is brought in, warmed, and returned to the river. When operating at full power, the plant takes in 90 m3 of water second, comparable to the flow of a small river. If the river water comes in at 10°C, at what temperature does it emerge?

1 answer:

weeeeeb [17]3 years ago

8 0

Answer:

Explanation:

1 GW = 10⁹ W

35% of 10⁹ W

= 35 x 10⁷W

Q = 35 x 10⁷J

Mass of water

= volume x density

= 90 x 10³ kg

= 9 x 10⁴ kg

specific heat s = 4.142 x 10³ J / kg

Now

Q = ms Δ T

Δ T is rise in temp

Δ T = Q / ms

= 35 x 10⁷ / (9 x 10⁴ x4.142 x 10³)

= .94°C

So the emerging water will have temperature of

10.94 °C.

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satela [25.4K]

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(1 joule-hour/sec) · (3600 sec/hour) = 3600 joules

So 1 watt-hour = 3,600 joules

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3 years ago

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Art [367]

Answer:

poet

Explanation:

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3 years ago

Alex is asked to move two boxes of books in contact with each other and resting on a rough floor. He decides to move them at the

Over [174]

Answer:

Part a)

$a= 0.32 m/s^2$

Part b)

$F_c = 3.6 N$

Part c)

$F_c = 5.5 N$

Explanation:

Part a)

As we know that the friction force on two boxes is given as

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$F_f = 0.02(10.6 + 7)9.81$

$F_f = 3.45 N$

Now we know by Newton's II law

$F_{net} = ma$

so we have

$F_p - F_f = (m_a + m_b) a$

$9.1 - 3.45 = (10.6 + 7) a$

$a = \frac{5.65}{17.6}$

$a= 0.32 m/s^2$

Part b)

For block B we know that net force on it will push it forward with same acceleration so we have

$F_c - F_f = m_b a$

$F_c = \mu m_b g + m_b a$

$F_c = 0.02(7)(9.8) + 7(0.32)$

$F_c = 3.6 N$

Part c)

If Alex push from other side then also the acceleration will be same

So for box B we can say that Net force is given as

$F_p - F_f - F_c = m_b a$

$9.1 - 0.02(7)(9.8) - F_c = 7(0.32)$

$F_c = 9.1 - 0.02 (7)(9.8) - 7(0.32)$

$F_c = 5.5 N$

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3 years ago

The distance covered by a car at a time, t is given by x = 20t + 6t4, calculate

Anni [7]

Answer:

(i) v = 44 m/s

(ii) a = 72 m/s^2

Explanation:

You have the following equation for the potion of a car:

$x=20t+6t^4$

(i) The instantaneous velocity is the derivative of x in time:

$\frac{dx}{dt}=20+(6)(4)t^3=20+24t^3$

for t = 1 is:

$v(t=1)=\frac{dx}{dt}=20+24(1)^3=44m/s$

(ii) The instantaneous acceleration is the derivative of the velocity:

$\frac{dv}{dt}=24(3)t^2=72t^2$

for t = 1

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8 0

4 years ago

Vector sum of 15 m, East and 9 m, West is 24 m, East. True False

Firlakuza [10]

Answer:

False

Explanation:

Because when you go through east

( +x axis ) then you go to west ( -x axis )

You will subtract -9 from +15

it's become +6

( I talk about the displacement not distance) ( West = - East )

I hope that it's a clear ") .

7 0

3 years ago

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