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3 years ago

13

In a heat engine, Qh and Qc represents high temperature substance and low temperature substance respectively. Which of the follo

wing expressions is correct for the efficiency of the heat engine?

1 answer:

ElenaW [278]3 years ago

6 0

The efficiency of a heat engine is given by the expression $eff =\frac{Q_h-Q_c}{Q_h}$

The efficiency of a heat engine is the ratio of the work done by the engine to the heat given as the input to the engine.

$eff =\frac{W}{Q_h}$

The heat engine absorbs Q h from the hot reservoir , performs a work <em>W</em> on the absorbed heat and rejects Qc to the cold reservoir.

Therefore, the work done is given by,

$W=Q_h - Q_c$

Thus the efficiency is given by, $eff =\frac{Q_h-Q_c}{Q_h}$

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ou are looking for a new tennis partner. Which of these people would most likely demonstrate good sportsmanship based on the inf

Dennis_Churaev [7]

<h2><u>Answer:</u></h2>

As you are looking for a new tennis partner. People should keep in mind that they should go for the one who most likely demonstrate good sportsmanship

Luis, when you pursue the principles in tennis, you realize when to talk up, you don't blast a racquet or shout, holler.

Whatever it is following the principles and being respectful it the most ideal approach.

7 0

3 years ago

Which is an example of a covalent bond?

AleksAgata [21]

Hello there.

Question: <span>Which is an example of a covalent bond?

Answer: Water and Diamonds are good examples of covalent bonds.

Hope This Helps You!
Good Luck Studying ^-^</span>

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4 years ago

An open-end mercury manometer is connected to a low-pressure pipeline that supplies a gas to a laboratory. Because paint was spi

Svetllana [295]

Answer:

a

$P_G = 14.03 \ psig$

b

$h_m = 0.148 \ m$

Explanation:

From the question we are told that

The pressure of the manometer when there is no gas flow is $P_{m} = 15.5 \ psig = 15.5 * 6894.76 = 106868.78 \ N/m^2$

The level of mercury is $h = 950 \ mm = 0.950 \ m$

The drop in the mercury level at the visible arm is $d = 39.0 = 0.039 \ m$

Generally when there is no gas flow the pressure of the manometer is equal to the gauge pressure which is mathematically represented as

$P_g = P_m = g * \delta h * \rho$

Here $\rho$ is the density of mercury with value $\rho = 13.6 *10^{3} kg/m^3$

and $\delta h$ is the difference in the level of gas in arm one and two

So

$\delta h = \frac{106868.78}{ 13.6 *10^{3} * 9.8 }$

$\delta h = 0.802 \ m$

Generally the height of the mercury at the arm connected to the pipe is mathematically represented as

$h_m = 0.950 - 0.802$

=> $h_m = 0.148 \ m$

Generally from manometry principle we have that

$P_G + \rho * g * d - \rho * g * [h - (h_m + d)] = 0$

Here $P_G$ is the pressure of the gas

$P_G +13.6 *10^{3} * 9.8 * 0.039 - 13.6 *10^{3} * 9.8 * [0.950 - (0.148 + 0.039)] = 0$

$P_G = 9.6724 04 *10^{4} \ N/m^2$

converting to psig

$P_G = \frac{ 9.6724 04 *10^{4} }{6894.76}$

$P_G = 14.03 \ psig$

6 0

3 years ago

Compare and contrast Ursa Major and the Milky Way

Natali [406]

One thing thats the same is there both universes.

One thing diffrent is that in the milky way there is proven life on earth but ursa major is proven without life.

7 0

3 years ago

Ice skaters often end their performances with spin turns, where they spin very fast about their center of mass with their arms f

vampirchik [111]

Answer:

$\large \boxed{\text{30 rev/s}}$

Explanation:

This question is based on the Law of Conservation of Angular Momentum.

Angular momentum (L) equals the moment of inertia (I) times the angular speed (ω).

L = Iω

If momentum is conserved,

I₁ω₁ = I₂ω₂

Data:

I₁ = 3.5 kg·m²s⁻¹

ω₁ = 6.0 rev·s⁻¹

I₂ = 0.70 kg·m²s⁻¹

Calculation:

$\begin{array}{rcl}I_{1}\omega_{1} &= &I_{2}\omega_{2}\\\text{3.5 kg$\cdot$m$^{2}$}\times \text{6.0 rev/s} &= &\text{0.70 kg$\cdot$m$^{2}$}\times\omega_{2}\\\text{21 rev/s} &= &0.70\omega_{2}\\\omega_{2} & = & \dfrac{\text{21 rev/s}}{0.70}\\\\&=&\textbf{30 rev/s}\\\end{array}\\\text{The skater's final rotational speed is $\large \boxed{\textbf{30 rev/s}}$}$

8 0

4 years ago