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Dovator [93]
3 years ago
11

The water molecules now in your body were once part of a molecular cloud. Only about onemillionth of the mass of a molecular clo

ud is in the form of water molecules, and the mass density of such a cloud is roughly 4.0×10−21 g/cm3 Estimate the volume of a piece of molecular cloud that has the same amount of water as your body.
Physics
1 answer:
Rudik [331]3 years ago
5 0

To solve the problem we first need to define the nomenclature by which there is an X amount of water in our body. Let's call this a variable m, corresponding to the total mass in us and whose units are Kilograms.

Amount_{water} = m (Kg)

We understand that one millionth of this mass, a small fraction of it, is the equivalent of that of a Water Molecule.

Assuming this to have the same body of water in our body that amount should be approximately

m_{water}=10^6

Density of water molecule

\rho = 10^{-18}kg/m^3

V = \frac{m*10^6}{10^{-18}}

V = m*10^{24}m^3

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A block of 250-mm length and 54 × 40-mm cross section is to support a centric compressive load P. The material to be used is a b
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P = 17.28*10⁶ N

Explanation:

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L = 250 mm = 0.25 m

a = 0.54 m

b = 0.40 m

E = 95 GPa = 95*10⁹ Pa

σmax = 80 MPa = 80*10⁶ Pa

ΔL = 0.12%*L = 0.0012*0.25 m = 3*10⁻⁴ m

We get A as follows:

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then, we apply the formula

ΔL = P*L/(A*E)  ⇒ P = ΔL*A*E/L

⇒  P = (3*10⁻⁴ m)*(0.216 m²)*(95*10⁹ Pa)/(0.25 m)

⇒  P = 24624000  N = 24.624*10⁶ N

Now we can use the equation

σ = P/A

⇒  σ = (24624000  N)/(0.216 m²) = 114000000 Pa = 114 MPa > 80 MPa

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⇒  P = σmax*A = (80*10⁶ Pa)*(0.216 m²) = 17280000 N = 17.28*10⁶ N

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