Questions & Answers

Question

Answers

A. $74$

B. $75$

C. $76$

D. $77$

Answer
Verified

i.e. $a+g$

Pressure = $h\rho \left( g+a \right)dyne/c{{m}^{2}}$

$\rho $= density of material

As given in problem that barometer is going upwards with elevator so,

Pressure = $h\rho \left( g+a \right)dyne/c{{m}^{2}}$

$\begin{align}

& P=76\times 13.6\times \left( g+a \right) \\

& p=76\times 13.6\times g \\

\end{align}$

$1+\dfrac{a}{g} > 1 cm Hg$

So, the final pressure for this will be more than $76cm$ of Hg.

Barometers and pressure altimeters (the most elementary and customary sort of altimeter) are essentially an equivalent instrument, but used for various purposes. An altimeter is meant to be used at different levels matching the corresponding air pressure to the altitude, while a barometer is kept at an equivalent level and measures subtle pressure changes caused by weather and elements of weather. The average atmospheric pressure on the earth's surface varies between $940$ and $1040hPa\left( mbar \right)$. The average atmospheric pressure at sea level is $1013hPa\left( mbar \right)$

If the elevator is “open” (a screen cage over a floor), the pressure will likely be almost unchanged relative to the surface air everywhere but during a thin layer near rock bottom , where there'll probably be an overpressure that depends on the speed (associated with the drag force) instead of the acceleration intrinsically . The pressure field above the elevator would be incredibly difficult to compute/estimate for love or money but maybe cylindrical elevators, requiring the answer of a Navier-Stokes system.

If the elevator is actually “closed” (so the pressure inside is isolated from the pressure outside) and therefore the reform the acceleration is uniform and the elevator isn’t too high and the temperature within the elevator is kept roughly constant (a lot of ifs) then the matter is comparatively simple. The pressure will rapidly attain an exponential profile from rock bottom of the elevator to the highest. The worth of the exponential constant is often computed fairly easily, IF one assumes that air is a perfect gas when estimating its constant temperature compressibility.