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dalvyx [7]
3 years ago
14

Gas A has molecules with small mass. Gas B has molecules with larger mass. They are at the same temperature.

Physics
1 answer:
Julli [10]3 years ago
6 0

Answer:

c)The gases have the same average kinetic energy.

Explanation:

As we know that the kinetic energy of gas is given as

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

here we know that

v = \sqrt{\frac{3RT}{M}}

so we have

K = \frac{1}{2}m (\frac{3RT}{M})

now we have

K = \frac{3}{2}n RT

now mean kinetic energy per molecule is given as

K_{avg} = \frac{3}{2}KT

so this is independent of the mass of the gas

so average kinetic energy will remain same for both the gas molecules

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Answer:

All these statement are true

Explanation:

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4 years ago
A truck moves 60 km West, and then 80 km North, and then
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▪▪▪▪▪▪▪▪▪▪▪▪▪  {\huge\mathfrak{Answer}}▪▪▪▪▪▪▪▪▪▪▪▪▪▪

Distance travelled by the truck is ~

  • \boxed{240 \:  \: km}

And it's displacement is ~

  • \boxed{0 \:  \: km}

\large \boxed{ \mathfrak{Step\:\: By\:\:Step\:\:Explanation}}

See the diagram in attachment for reference ~

Let O be the initial point, It travels 60 km towards west till point B and then 80 km towards north till point P and returns to initial point O in a straight line, now as we can observe here, it forms a right angled Triangle.

The measure of two legs is 60 km and 80 km, let's find the hypotenuse ~

According to Pythagoras theorem ~

hypotenuse² = sum of squares of other two legs

that is ~

  • h {}^{2}  = 60 {}^{2}  + 80 {}^{2}

  • {h}^{2}  = 3600 + 640 0

  • h {}^{2}  = 10000

  • h =  \sqrt{10000}

  • h =  \sqrt{100 \times 100}{}

  • h = 100 \:  \: km

So, the distance between the point A and O is 100 km

Now, The total distance is equal to the distance covered through actual path that is ~

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And displacement is the distance between the final point and initial point, but since the truck returns to the point from where it started the journey, so the final and initial point is same therefore displacement is equal to 0.

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Answer: Bohr's model

Explanation:

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The Bohr's model are introduced for overcoming the drawback in the Rutherford model as he proposed an idea that at different energy levels the electron can revolve in certain orbits around nucleus.  

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