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3 years ago

A barometer reads 780 mm Hg. Mercury has a density of 1.36 x 10^4 kg /m^3.

2 answers:

7 0

The pressure of the atmosphere, when a barometer reads 780 mm Hg. Mercury which a density of 1.36 x 10^4 kg /m^3 is B 1.1 x 10^5 N/m^2

This problem can be solved using the formula below

P = dgh................. Equation 1

Where P = Pressure of the atmosphere, d = density of the mercury, h = height of the mercury, g = acceleration due to gravity.

From the question,

Given: d = 1.36×10⁴ kg/m³, h = 780 mm = 0.78 m,

Constant: g = 10 m/s²

Substitute these values into equation 1

P = (1.36×10⁴)(10)(0.78)

P = 10.608×10⁴ N/m²

P ≈ 1.1×10⁵ N/m²

Hence the right answer is B. 1.1×10⁵ N/m²

Learn more about Pressure here: brainly.com/question/23603188

frutty [35]3 years ago

5 0

Answer:

Explanation:

Pressure = density × g × height

$p \: = (1.36 \times {10}^{4} )(10)(780 \times {10}^{ - 3} ) \\ p = 1.1 \times {10}^{5}$

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3 years ago

Town A lies 15 km north of town B. Town C lies 10 km west of town A. A small plane flies directly from town B to town C. What is

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3 years ago

A mole of ideal gas expands at T=27 °C. The pressure changes from 20 atm to 1 atm. What’s the work that the gas has done and wha

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Answer:

The work made by the gas is 7475.69 joules
The heat absorbed is 7475.69 joules

Explanation:

We know that the differential work made by the gas its defined as:

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We can solve this by integration:

$\Delta W = \int\limits_{s_1}^{s_2}\,dW = \int\limits_{v_1}^{v_2} P \ dv$

but, first, we need to find the dependence of Pressure with Volume. For this, we can use the ideal gas law

$P \ V = \ n \ R \ T$

$P = \frac{\ n \ R \ T}{V}$

This give us

$\int\limits_{v_1}^{v_2} P \ dv = \int\limits_{v_1}^{v_2} \frac{\ n \ R \ T}{V} \ dv$

As n, R and T are constants

$\int\limits_{v_1}^{v_2} P \ dv = \ n \ R \ T \int\limits_{v_1}^{v_2} \frac{1}{V} \ dv$

$\Delta W= \ n \ R \ T \left [ ln (V) \right ]^{v_2}_{v_1}$

$\Delta W = \ n \ R \ T ( ln (v_2) - ln (v_1 )$

$\Delta W = \ n \ R \ T ln (\frac{v_2}{v_1})$

But the volume is:

$V = \frac{\ n \ R \ T}{P}$

$\Delta W = \ n \ R \ T ln(\frac{\frac{\ n \ R \ T}{P_2}}{\frac{\ n \ R \ T}{P_1}} )$

$\Delta W = \ n \ R \ T ln(\frac{P_1}{P_2})$

Now, lets use the value from the problem.

The temperature its:

$T = 27 \° C = 300.15 \ K$

The ideal gas constant:

$R = 8.314 \frac{m^3 \ Pa}{K \ mol}$

So:

$\Delta W = \ 1 mol \ 8.314 \frac{m^3 \ Pa}{K \ mol} \ 300.15 \ K ln (\frac{20 atm}{1 atm})$

$\Delta W = 7475.69 joules$

We know that, for an ideal gas, the energy is:

$E= c_v n R T$

where $c_v$ its the internal energy of the gas. As the temperature its constant, we know that the gas must have the energy is constant.

By the first law of thermodynamics, we know

$\Delta E = \Delta Q - \Delta W$

where $\Delta W$ is the Work made by the gas (please, be careful with this sign convention, its not always the same.)

So:

$\Delta E = 0$

$\Delta Q = \Delta W$

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3 years ago

An airplane dropped a flare from a height of 2860 feet above a lake. How many seconds did it take for the flare to reach the wat

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Answer: 13.2 seconds.

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