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s344n2d4d5 [400]
3 years ago
13

Boltzmann’s constant is 1.38066 × 10−23 J/K, and the universal gas constant is 8.31451 J/K · mol.

Physics
1 answer:
djyliett [7]3 years ago
4 0

Answer:

4.95\cdot 10^{-21} J

Explanation:

First of all, let's convert everything into SI units:

n = 2.4 mol (number of gas moles)

p=11 atm = 1.11\cdot 10^6 Pa (gas pressure)

V=4.3 L=4.3\cdot 10^{-3} m^3 (gas volume)

R = 8.31451 J/K · mol (gas constant)

The ideal gas equation states that

pV=nRT

Solving for T, we find the gas temperature

T=\frac{pV}{nR}=\frac{(1.11\cdot 10^6)(4.3\cdot 10^{-3})}{(2.4)(8.31451)}=239.2 K

And now we can find the average kinetic energy of the gas:

E_K = \frac{3}{2}kT

where

k = 1.38066 × 10−23 J/K is the Boltzmann's constant

Substituting,

E_K = \frac{3}{2}(1.38066\cdot 10^{-23} J/K)(239.2 K)=4.95\cdot 10^{-21} J

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A 4-kg object is moving with a speed of 5 m/s at a height of 2 m. The kinetic
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Hello!

\large\boxed{KE = 50J}

Use the formula for kinetic energy:

KE = \frac{1}{2}mv^{2}

Plug in the given mass and velocity:

KE = \frac{1}{2} (4)5^{2}

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KE = \frac{1}{2} (100)\\\\KE = 50 J

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In the ground state of hydrogen, according to the Bohr model, an electron orbits 5.3 x 10-11 m from the nucleus. It undergoes a
Readme [11.4K]

Answer:

Explanation:

Given

radius of electron(r)=5.3\times 10^{-11} m

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we know

a_c=\frac{v^2}{r}

v=\sqrt{r\times a_c}

v=\sqrt{5.3\times 10^{-11}\times 9\times 10^{22}}

v=\sqrt{47.7\times 10^11}

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(b)For n=10

r=100\times 5.3\times 10^{-11} m\approx 5.3\times 10^{-9} m

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Kipish [7]
Technically this is a Biology question;

The 'amount' we can see depends on how much light can get through our pupil to hit our retina.
When there is a lot of light the pupil is small; it doesn't need to be big to let a lot of light in.
When we move to a dark space there is much less light, so the pupil 'dilates' to let enough light so we can see properly.
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