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siniylev [52]
3 years ago
9

6) A person with mass 50.0 kg, standing still, throws an object with mass 12.0

Physics
1 answer:
matrenka [14]3 years ago
6 0

Explanation:

Momentum before = momentum after

m₁ u₁ + m₂ u₂ = m₁ v₁ + m₂ v₂

(50.0 kg) (0 m/s) + (12.0 kg) (0 m/s) = (50.0 kg) (2.0 m/s) + (12.0 kg) v

v = -8.33 m/s

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What happens to the energy of a photon if the wavelength is doubled?.
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What must a system owner do, once the threshold leak rate has been exceeded on any low-pressure system using an ozone-depleting
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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
ivolga24 [154]

Answer: 0.10233nm

Explanation:

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)

Where:

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

R=8.3145J/mol.K is the Universal gas constant

T=0\°C=273.115K is the absolute standard temperature

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.

5 0
3 years ago
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