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2 years ago

6

One of the dangers of tornados and hurricanes is the rapid drop in air pressure that is associated with such storms. Assume that

the air pressure inside of a sealed house is 1.02 atm when a hurricane hits. The hurricane rapidly decreases the external air pressure to 0.910 atm. What net outward force is exerted on a square window of the house that is 2.03 m on each side?

1 answer:

Rufina [12.5K]2 years ago

5 0

Answer:

45930.52N

Explanation:

Net force = (internal pressure - external pressure)× area of window

Net force = (1.02 - 0.910)atm × 2.03m × 2.03m = 0.11atm × 4.1209m^2 = 0.11 × 101325N/m^2 × 4.1209m^2 = 45930.52N

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Explanation:

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calculate the time rate of change in air density during expiration. Assume that the lung has a total volume of 6000mL, the diame

kipiarov [429]

Answer:

The time rate of change in air density during expiration is 0.01003kg/m³-s

Explanation:

Given that,

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diameter of the trachea is 18mm = 0.018m

Velocity v = 20cm/s = 0.20m/s

dv /dt = -100mL/s (volume rate decrease)

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Area for trachea =

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0 - p × Area for trachea =

$\frac{d}{dt} (pv)=v\frac{ds}{dt} + p\frac{dv}{dt}$

$-1.225\times2.5434\times10^-^4\times0.20=6\times10^-^3\frac{ds}{dt} +1.225(-1\times10^-^4)$

$-1.225\times2.5434\times10^-^4\times0.20=6\times10^-^3\frac{ds}{dt} +1.225(-1\times10^-^4)$

⇒ $-0.623133\times10^-^4+1.225\times10^-^4=6\times10^-^3\frac{ds}{dt}$

$\frac{ds}{dt} = \frac{0.6018\times10^-^4}{6\times10^-^3} \\\\= 0.01003kg/m^3-s$

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2 years ago

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A uniform 1500-kg beam, 20.0 m long, supports a 15,000-kg printing press

Ghella [55]

Answer:

$\mathbf{F_1=4.41*10^4\ N}$

$\mathbf{F_2 = 1.176*10^5 \ N}$

Explanation:

The missing image of the figure slide is attached in below.

However, from the model, it is obvious that it is in equilibrium.

As a result, the relation of the force and the torque is said to be zero.

i.e.

$\sum F = 0$ and $\sum \tau = 0$

From the image, expressing the forces through the y-axis, we have:

$F_1+F_2 = W_B + W_P \\ \\ \implies 9.8(1500+15000) \\ \\ \implies \mathtt{1.617\times 10^5 \ N}$

Also, let the force $F_1$ be the pivot and computing the torque to determine $F_2$ :

Then:

$F_1(0)+F_2(20.0) = 10.0W_B + 15.0W_P$

$F_2 = \dfrac{((10*1500)+(15*15000))*9.8}{20.0}$

$F_2 = 117600 \ N$

$\mathbf{F_2 = 1.176*10^5 \ N}$

For the force equation:

$F_1+F_2=1.617*10^5 \ N;$

where:

$F_2 = 1.176*10^5 \ N$

Then:

$F_1+1.176*10^5 \ N=1.617*10^5 \ N$

$F_1=1.617*10^5 \ N-1.176*10^5 \ N$

$F_1=44100\ N$

$\mathbf{F_1=4.41*10^4\ N}$

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2 years ago

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