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riadik2000 [5.3K]
3 years ago
10

A barometer accidentally contains 6.5 inches of water on top of the mercury column (so there is also water vapor instead of a va

cuum at the top of the barometer). On a day when the temperature is 70oF, the mercury column height is 28.35 inches (corrected for thermal expansion). (a) Determine the barometric pressure in psia. If the ambient temperature increased to 85oF and the barometric pressure did not change, (b) would the mercury column be longer, be shorter, or remain the same length
Physics
1 answer:
Alisiya [41]3 years ago
7 0

Answer:

(a). 14.4 lbf/in^2.

(b). 27.8 in, AS THE TEMPERATURE INCREASES, THE LENGTH OF MERCURY DECREASES.

Explanation:

So, from the question above we are given the following parameters which are going to help us in solving this particular Question;

=> The "barometer accidentally contains 6.5 inches of water on top of the mercury column (so there is also water vapor instead of a vacuum at the top of the barometer)"

=> "On a day when the temperature is 70oF, the mercury column height is 28.35 inches (corrected for thermal expansion)."

With these knowledge, let us delve right into the solution;

(a). The barometric pressure = water vapor pressure + acceleration due to gravity (ft/s^2) × water density(slug/ft^3) × {ft/12 in}^3 × [ height of mercury column + specific gravity of mercury × height of water column].

The barometric pressure= 0.363 + {(62.146) ÷ (12^3) × 390.6425}. = 14.4 lbf/in^2.

(b). { (13.55 × length of mercury) + 6.5 } × (62.15÷ 12^3) = 14.4 - 0.603.

Length of mercury = 27.8 in.

AS THE TEMPERATURE INCREASES, THE LENGTH OF MERCURY DECREASES.

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In noisy factory environments, it's possible to use a loudspeaker to cancel persistent low-frequency machine noise at the positi
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Answer:

7.39 m or 3.61 m

Explanation:

\lambda = Wavelength

f = Frequency = 90 Hz

v = Speed of sound = 340 m/s

Path difference of the two waves is given by

s_1-s_2=\frac{\lambda}{2}

Velocity of wave

v=f\lambda\\\Rightarrow \lambda=\frac{v}{f}\\\Rightarrow \lambda=\frac{340}{90}\\\Rightarrow \lambda=3.78\ m

s_1=s_2\pm\frac{\lambda}{2}\\\Rightarrow s_1=5.5\pm \frac{3.78}{2}\\\Rightarrow s_1=7.39\ m, 3.61\ m

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It mimics the real world accurately

Explanation:

Experiments conducted in the field clearly presents the real world at it is to the scientist. Hardly can any part be controlled precisely and this gives a near to perfect scenario.

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  • Most laboratory set up are driven towards a model instead of real life settings.
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Experiment brainly.com/question/5096428

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P₁ and P₂ are the initial and final pressures.

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T₁ and T₂ are the initial and final temperatures.

Given values:

P₁ = 475kPa

V₁ = 4m³, V₂ = 6.5m³

T₁ = 290K, T₂ = 277K

Substitute the terms in the equation with the given values and solve for Pf:

\frac{475*4}{290} = \frac{P₂*6.5}{277}

<h3>P₂ = 279.2kPa</h3>
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