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Lerok [7]
3 years ago
11

A Carnot heat engine uses a hot reservoir consisting of a large amount of boiling water and a cold reservoir consisting of a lar

ge tub of ice and water. In 5 minutes of operation of the engine, the heat rejected by the engine melts a mass of ice equal to 5.90×10−2 kg . Throughout this problem use Lf=3.34×105J/kg for the heat of fusion for water. Part A During this time, how much work W is performed by the engin
Physics
1 answer:
JulsSmile [24]3 years ago
4 0

Answer:

W = 7.214 k J

Explanation:

given,

time of operation = 5 minutes

mass of ice melt = 5.90 × 10⁻² kg

L_f = 3.34 × 10⁵ J/kg

Work done = ?

We know work done

     |W| = |Q_h| - |Q_c|

Q_c = - m L_f

Q_c= - 5.90\times 10^{-2}\times 3.34 \times 10^5

Q_c = -19.706 kJ

Q_h = Q_c \dfrac{T_H}{T_C}

T_H is the heat from hot reservoir = 373 K

T_C is the energy released to cold reservoir = 273 K

Q_h = -19.706\times \dfrac{373}{273}

Q_h =-26.92 kJ

now,

     |W| = |-26.92| - |-19.706|

              W = 7.214 k J

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Answer:

A) The event horizon, singularity, and the chute located between the two.

6 0
3 years ago
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If you have lifted 10 pounds 2 feet, you have done ___ foot-pounds of work.
nasty-shy [4]
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          (10 pounds)(2 feet)
Work = 20 foot-pounds of work

hope this helps :)


8 0
3 years ago
What did Rutherford’s model of the atom include that Thomson’s model did not have?
Ksivusya [100]
Rutherford overturned Thomson's model in 1911 with his well-known gold foil experiment in which he demonstrated that the atom has a tiny and heavy nucleus. Rutherford designed an experiment to use the alpha particles emitted by a radioactive element as probes to the unseen world of atomic structure.
3 0
3 years ago
A purse at radius 2.30 m and a wallet at radius 3.45 m travel in uniform circular motion on the floor of a merry-go-round as the
ivolga24 [154]

Answer:

The acceleration of the wallet is 3\hat{i}+6\hat{j}

Explanation:

Given that,

Radius of purse r= 2.30 m

Radius of wallet r'= 3.45 m

Acceleration of the purse a=2\hat{i}+4.00\hat{j}

We need to calculate the acceleration of the wallet

Using formula of acceleration

a=r\omega^2

Both the purse and wallet have same angular velocity

\omega=\omega'

\sqrt{\dfrac{a}{r}}=\sqrt{\dfrac{a'}{r'}}

\dfrac{a}{r}=\dfrac{a'}{r'}

\dfrac{a'}{a}=\dfrac{r'}{r}

\dfrac{a'}{a}=\dfrac{3.45}{2.30}

\dfrac{a'}{a}=\dfrac{3}{2}

a'=\dfrac{3}{2}\times(2\hat{i}+4.00\hat{j})

a'=3\hat{i}+6\hat{j}

Hence, The acceleration of the wallet is 3\hat{i}+6\hat{j}

4 0
3 years ago
1) On the way to the moon, the Apollo astro-
kramer
(1) You must find the point of equilibrium between the two forces,

<span>G * <span><span><span>MT</span><span>ms / </span></span><span>(R−x)^2 </span></span>= G * <span><span><span>ML</span><span>ms / </span></span><span>x^2
MT / (R-x)^2 = ML / x^2

So,

x = R * sqrt(ML * MT) - ML / (MT - ML)
R = is the distance between Earth and Moon.

</span></span></span>The result should be,
x = 3.83 * 10^7m
from the center of the Moon, and 

R - x = 3.46*10^8 m
from the center of the Earth.


(2) As the distance from the center of the Earth is the number we found before,
d = R - x = 3.46*10^8m
The acceleration at this point is
g = G * MT / d^2
g = 3.33*10^-3 m/s^2
6 0
3 years ago
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