Physics, asked by Yacika, 1 year ago

Very urgent solve fast

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Answered by abi2790
1
ρ = 1.295 kg/m³

decrease in the pressure of air = ρ g h = 1.295 * g * 107  Pa

ρ g h = 13.6 * 10³ * g * H

H = decrease in barometric height of mercury
    = ρ g h / (13.6*10³ * g) = 10.188 mm

pressure on mercury column in barometer = weight of air column /area of cross section row * g * h of air = row * g * H of mercury H of mercury = air density * height of air / density of mercury.

Hope this helped.
Answered by samuelpaul
0
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