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attashe74 [19]
3 years ago
9

Assuming that the density of a person is the same as the density of water, what buoyant force does the atmosphere exert on a per

son whose mass is 70 kg?
Physics
1 answer:
Charra [1.4K]3 years ago
4 0

Answer:

the buoyant force F will be 0.84 N

Explanation:

Assuming that a person has the density of water , then its volume could be calculated through

V= m/ρ , where m= mass of the person , ρ= density of water

then

V= m/ρ =  70 kg/ 1000 kg/m³ = 0.07 m³

then the buoyant force will be equal to the mass of displaced air in the volume V.

if we assume that the density of air in the atmosphere is ρa=1.225 kg/m³ then the buoyant force F will be

F= ρa*V*g = 0.07 m³*1.225 kg/m³ *9.8 m/s² =0.84 N

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A typical ceiling fan running at high speed has an airflow of about 1.85 ✕ 103 ft3/min, meaning that about 1.85 ✕ 103 cubic feet
densk [106]

Answer:

0.8726  m^3/s

Explanation:

We are to convert 1.85 x 10^3 ft^3/min to m^3/s

First, let us convert the numerator from ft3 to m3

1 ft3 = 0.0283 m3

Hence,

1.85 x 10^3 ft3 = 1.85 x 10^3 x 0.0283 m3

     = 52.355 m3

Now, let us convert the denominator from minutes to seconds

1 min = 60 sec

Therefore;

1.85 x 10^3 ft^3/min = 52.355/60  m^3/s

        = 0.8726  m^3/s

7 0
3 years ago
38.4 mol of krypton is in a rigid box of volume 64 cm^3 and is initially at temperature 512.88°C. The gas then undergoes isobari
kolbaska11 [484]

Answer:

Final volumen first process V_{2} = 98,44 cm^{3}

Final Pressure second process P_{3} = 1,317 * 10^{10} Pa

Explanation:

Using the Ideal Gases Law yoy have for pressure:

P_{1} = \frac{n_{1} R T_{1} }{V_{1} }

where:

P is the pressure, in Pa

n is the nuber of moles of gas

R is the universal gas constant: 8,314 J/mol K

T is the temperature in Kelvin

V is the volumen in cubic meters

Given that the amount of material is constant in the process:

n_{1} = n_{2} = n

In an isobaric process the pressure is constant so:

P_{1} = P_{2}

\frac{n R T_{1} }{V_{1} } = \frac{n R T_{2} }{V_{2} }

\frac{T_{1} }{V_{1} } = \frac{T_{2} }{V_{2} }

V_{2} = \frac{T_{2} V_{1} }{T_{1} }

Replacing : T_{1} =786 K, T_{2} =1209 K, V_{1} = 64 cm^{3}

V_{2} = 98,44 cm^{3}

Replacing on the ideal gases formula the pressure at this piont is:

P_{2} = 3,92 * 10^{9} Pa

For Temperature the ideal gases formula is:

T = \frac{P V }{n R }

For the second process you have that T_{2} = T_{3}  So:

\frac{P_{2} V_{2} }{n R } = \frac{P_{3} V_{3} }{n R }

P_{2} V_{2}  = P_{3} V_{3}

P_{3} = \frac{P_{2} V_{2}}{V_{3}}

P_{3} = 1,317 * 10^{10} Pa

7 0
3 years ago
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