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Svetach [21]
3 years ago
13

Assuming the atmospheric pressure is 1 atm at sea level, determine the atmospheric pressure at Badwater (in Death Valley, Califo

rnia) where the elevation is 86.0 m below sea level.
Physics
1 answer:
tiny-mole [99]3 years ago
4 0

Answer:

Atmospheric pressure at Badwater is 1.01022 atm

Explanation:

Data given:

1 atmospheric pressure (Pi) = 1.01 * 10^{5} Pa

Elevation (h) = 86m

gravity (g) = 9.8 m/s2

Density of air P = 1.225 kg/m3

Therefore pressure at bad water Pb = Pi + Pgh

Pb = (1.01 * 10^{5}) + (1.225 * 9.8 * 86)

Pb = (1.01 * 10^{5}) + 1032.43 = 102032 Pa

hence:

Pb = 102032 /1.01 * 10^{5} = 1.01022 atm

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The motion is over-damped when λ^2 - w^2 > 0 or when b^{2} > 0.86

The motion is critically when λ^2 - w^2 = 0 or when b^{2} = 0.86

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Explanation:

Using the newton second law

k is the spring constante

b positive damping constant

m mass attached

m\frac{d^{2} x}{dt^{2}} = - kx - b\frac{dx}{dt}

x(t) is the displacement from the equilibrium position

\frac{d^{2} x}{dt^{2}} +\frac{b}{m}\frac{dx}{dt} + \frac{k}{m}x = 0

Converting units of weights in units of mass (equation of motion)

m = \frac{W}{g} = \frac{14}{32} = 0.43 slug

From hook's law we can calculate the spring constant k

k = \frac{W}{s} = \frac{14}{2} = 7 lb/ft

If we put m and k into the DE, we get

\frac{d^{2} x}{dt^{2}} +\frac{b}{0.43}\frac{dx}{dt} + 16.28x = 0

Denoting the constants

2λ = \frac{b}{m} = \frac{b}{0.43}

λ = b/0.215

w^{2} = \frac{k}{m} = 16.28

λ^2 - w^2 = \frac{b^{2} }{0.046} - 16.28

This way,

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The motion is under-damped when λ^2 - w^2 < 0 or when b^{2} < 0.86

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