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soldi70 [24.7K]
2 years ago
6

Please answer the following question.

Physics
1 answer:
Dima020 [189]2 years ago
6 0

The oceanic lithosphere is a layer on the Earth's crust that consists of rocks of the mantle.

<h3>What is the lithosphere?</h3>

The lithosphere can be defined as the outer rock layer present on the Earth's surface.

The oceanic lithosphere is composed of different rocks of the mantle such as, among others, gabbro and/or basalt.

The structure of the oceanic lithosphere is similar around the world because it is slender compared to the continental lithosphere and always determined by the supply of magma.

In conclusion, the oceanic lithosphere is a layer on the Earth's crust that consists of rocks of the mantle.

Learn more about the lithosphere here:

brainly.com/question/10508547

#SPJ1

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The mean free path of a helium atom in helium gas at standard temperature and pressure is 0.2 um.What is the radius of the heliu
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The mean free path \lambda   of an atom is given by the following formula:

\lambda=\frac{RT}{\sqrt{2} \pi d^{2}N_{A}P}    (1)

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\lambda=0.2\mu m=0.2(10)^{-6}m

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d is the diameter of the helium atom

N_{A}=6.0221(10)^{23}/mol is the Avogadro's number

P=1atm=101.3kPa=101.3(10)^{3}Pa=101.3(10)^{3}J/m^{3} absolute standard pressure

Knowing this, let's find d from (1), in order to find the radius r of the helium atom:

d=\sqrt{\frac{RT}{\sqrt{2}\pi\lambda N_{A}P}}    (2)

d=\sqrt{\frac{(8.3145J/mol.K)(273.115K)}{\sqrt{2}\pi(0.2(10)^{-6}m)(6.0221(10)^{23}/mol)(101.3(10)^{3}J/m^{3})}}    (3)

d=2.0467(10)^{-10}m    (4)

If the radius is half the diameter:

r=\frac{d}{2}  (5)

Then:

r=\frac{2.0467(10)^{-10}m}{2}  (6)

r=1.0233(10)^{-10}m  (7)

However, we were asked to find this radius in nanometers. Knowing 1nm=(10)^{-9}m:

r=1.0233(10)^{-10}m.\frac{1nm}{(10)^{-9}m}=0.10233nm  (8)

Finally:

r=0.10233nm This is the radius of the helium atom in nanometers.

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