Milan1025

14.11.2021 •

English

How does the long-line free verse used by both Whitman and Ginsberg reflect their rebellious attitudes

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Ответ:

emmiles2005

12.06.2022

English

1.25 cm/day

Explanation:

An air thickness , (l) = 0.15 cm

Air Temperature =

$(T_a)=20^0C = (20+273)K\\(T_a)=293K$

Mass Diffusion coefficient (D) = 0.25cm^2/sec

If the air pressure $(P_a) = 0.5 P_{sat}$

We are to determine how fast will the water (H_2O) level drop in a day.

From the property of air at T = 20° C

$P_{sat} = 2.34$ from saturated water properties.

The mass flow of (H_2O) can be calculated as:

$H_2O = \frac{D}{\phi} \delta C$

where:

$\delta C = \frac{P_{sat}*P_a}{RT }$

R(constant) = 8.314 kJ/mol.K

$\delta C = \frac{2.34*0.5}{8.314*293 }$

$\delta C = 4.803*10^{-4}\\ \delta C =0.48*10^{-3} mol/m^3\\ \delta C = 0.48*10^{-6} mol/cm^3$

Since 1 mole = 18 cm ³ of water

$0.48*10^{-6}mol/cm^3$ will be: $(0.48*10^{-6}mol/cm^3 *18)cm^3/cm^3$

$\delta C = 8.64 * 10^{-6}$

Again:

$H_2O = \frac{D}{\phi} \delta C$

$= \frac{0.25}{0.15}*8.69*10^{-6} \frac{cm^2/sec}{cm}$

$=1.4481*10^{-5} \frac{cm^2/sec}{cm}$

Converting the above value to cm/day: we have:

$1.448*10^{-5}*3600*24\frac{cm}{s}*\frac{s}{yr}*\frac{yr}{day}$

= 1.25 cm/day

∴ the rate at which the water level drop in a day = 1.25 cm/day

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