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2 years ago

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To suck lemonade of density 1040 kg/m3 up a straw to a maximum height of 4.94 cm, what minimum gauge pressure (in atmospheres) m

ust you produce in your lungs?

2 answers:

Lady bird [3.3K]2 years ago

7 0

Answer:

The minimum gauge pressure is 0.4969 atm.

Explanation:

Given that,

Density = 1040 kg/m³

Height = 4.94 cm

We need to calculate the pressure

Using formula of pressure

$P_{g}=\rho g h$

Where, $\rho$ =density

h = height

Put the value into the formula

$P_{g}=1040\times9.8\times4.94$

$P_{g}=50348.48\ Pa$

Pressure in atmospheres

$1\ atm =101.3\ kPa$

$P_{g}=\dfrac{50348.48}{101325}$

$P_{g}=0.4969\ atm$

Hence, The minimum gauge pressure is 0.4969 atm.

m_a_m_a [10]2 years ago

3 0

Answer:

Pressure, P = 503.48 Pascals

Explanation:

Given that,

Density of lemonade, $d=1040\ kg/m^3$

Maximum height, $h=4.94\ cm=0.0494\ m$

We need to find the minimum gauge pressure produced in your lungs. The pressure exerted is given by :

$P=d\times g\times h$

$P=1040\times 9.8\times 0.0494$

P = 503.48 Pascals

So, the minimum gauge pressure is 503.48 pascals. Hence, this is the required solution.

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