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Natalka [10]
3 years ago
7

The CEO, ellen misk, left her martian office but accidentally left a cylindricall can of coke (3.1 inches in diameter, 5.42 inch

es in height) on her desk. If the can exerts a pressure of 510 Pascals, what is the specific gravity of the can?
Physics
1 answer:
iren2701 [21]3 years ago
8 0

Answer:

Specific Gravity = 0.378

Explanation:

First, we will find the force exerted by the can on the table. This force will be equal to the weight of the can:

Pressure = Force/Area = Weight/Area

Weight = Pressure*Area

where,

Area = πdiameter²/4 = π[(3.1 in)(0.0254 m/1 in)]²/4 = 4.8 x 10⁻³ m²

Weight = (510 N/m²)(4.8 x 10⁻³ m²)

Weight = 2.48 N

Now, the weight is given as:

Weight = mg

2.48 N = m(9.8 m/s²)

m = (2.48 N)/(9.8 m/s²)

m = 0.25 kg

Now, we calculate volume of can:

Volume = (Area)(Height) = (4.8 x 10⁻³ m²)(5.42 in)(0.0254 m/1 in)

Volume = 6.6 x 10⁻⁴ m³

Hence, the density of can will be:

Density of Can = m/Volume = 0.25 kg/6.6 x 10⁻⁴ m³

Density of Can = 378.32 kg/m³

So, the specific gravity of Can will be:

Specific Gravity = Density of Can/Density of Water

Specific Gravity = (378.32 kg/m³)/(1000 kg/m³)

<u>Specific Gravity = 0.378</u>

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ElenaW [278]

Answer:

The convection coefficient is 15456.48\ W/m^{2}K

Solution:

Mass flow rate, \dot{m} = 0.4\ kg

Inner diameter of the tube, d = 0.014 m

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Thermal Conductivity, K = 0.74

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Now,

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N_{u} = 0.023\times 46036.253^{0.8}\times 8.6^{0.4} = 292.42

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292.42 = \frac{0.014\times h}{0.74}

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Answer:

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