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3 years ago

The picture shown shows a practical application of

Physics

2 answers:

kati45 [8]3 years ago

3 0

Answer:

You'd have to show the picture..

Ulleksa [173]3 years ago

3 0

<em>Nuclear power</em> is used thoughout the world safely and efficiently. It makes up approximately 20% of the US production of electricity.

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When two substance mixed together what does not indicate that a chemical reaction has

nevsk [136]

Answer:

When two substances are mixed together, a color change is observed. This observation is not enough evidence that there is a chemical reaction happening because color changes are not indicators of chemical reaction:

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3 years ago

Which axis is the IV on a graph?

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3 years ago

Sarah bikes 15 miles in 70 minutes. Which is her average speed per minute?

valentina_108 [34]

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4 years ago

A bowling ball with a mass of 9 kg is moving at a speed of 4 m/s. What is its

Ilia_Sergeevich [38]

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3 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:

$P_G = 14.03 \ psig$

$h_m = 0.148 \ m$

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

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

$\delta h = 0.802 \ m$

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