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Arisa [49]
3 years ago
10

Oxygen and hydrogen gas are at the same temperature T.What is the ratio of kinetic energies of oxygen molecule and hydrogen mole

cule if oxygen is 16 times heavier than hydrogen.
Physics
1 answer:
VashaNatasha [74]3 years ago
6 0

Answer

the ratio of kinetic energies of oxygen molecule and hydrogen molecule if oxygen is 16 times heavier than hydrogen is 1

Explanation

The average kinetic energy of a gas molecule at temperature T is given by following formula

E =\frac{3kT}{2}...............(1)

Here k is Boltzmann's constant (k=1.38\times 10^{-23} JK^{-1}).

From above equation it is clear that average kinetic energy of a gas molecule only depends on temperature.

Therefore kinetic energy of oxygen molecule E_{1}=\frac{3kT}{2}....2

and kinetic energy of hydrogen molecule E_{2}=\frac{3kT}{2}....3

the ratio of kinetic energies of oxygen molecule and hydrogen molecule if oxygen is 16 times heavier than hydrogen is given as

\frac{E_{1}}{E_{2}} =\frac{\frac{3kT}{2}}{\frac{3kT}{2}}

\frac{E_{1}}{E_{2}}=1

Therefore, the ratio of kinetic energies of oxygen molecule and hydrogen molecule if oxygen is 16 times heavier than hydrogen is 1

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Explanation:

i hope you understand me and this helps, sorry if it doesn't TT

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What kind of habitat do we live in
Aleksandr-060686 [28]

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3 years ago
Two vectors are given by a = 8.6i + 5.1 j and b = 931 + 9.5.
skelet666 [1.2K]

Answer:

<u>Assuming b = 9.3i + 9.5j</u> <em>(b = 931 + 9.5 is wrong):</em>

a) a×b = 34.27k

b) a·b = 128.43

c) (a + b)·b = 305.17

d) The component of a along the direction of b = 9.66

Explanation:

<u>Assuming b = 9.3i + 9.5j</u> <em>(b = 931 + 9.5 is wrong)</em> we can proceed as follows:  

a) The vectorial product, a×b  is:

a \times b = (8.6*9.5 - 5.1*9.3)k = 34.27k

b) The escalar product a·b is:

a\cdot b = (8.6*9.3) + (5.1*9.5) = 128.43

c) <u>Asumming (a</u><u> </u><u>+ b)·b</u> <em>instead a+b·b</em> we have:

(a + b)\cdot b = [(8.6 + 9.3)i + (5.1 + 9.5)j]\cdot (9.3i + 9.5j) = (17.9i + 14.6j)\cdot (9.3i + 9.5j) = 305.17

d) The component of a along the direction of b is:

a*cos(\theta) = \frac{a\cdot b}{|b|} = \frac{128.43}{\sqrt{9.3^{2} + 9.5^{2}}} = 9.66

I hope it helps you!                        

5 0
3 years ago
What will the stopping distance be for a 2,000-kg car if -2,000 N of force are applied when the car is traveling 20 m/s?
astraxan [27]

Answer is B- 200 m

Given:

m (mass of the car) = 2000 Kg

F = -2000 N

u(initial velocity)= 20 m/s.

v(final velocity)= 0.

Now we know that

<u>F= ma</u>

Where F is the force exerted on the object

m is the mass of the object

a is the acceleration of the object

Substituting the given values

-2000 = 2000 × a

a = -1 m/s∧2

Consider the equation

<u>v=u +at</u>

where v is the initial velocity

u is the initial velocity

a is the acceleration

t is the time

0= 20 -t

t=20 secs


s = ut +1/2(at∧2)

where s is the displacement of the object

u is the initial velocity

t is the time

v is the final velocity

a is the acceleration

s= 20 ×20 +(-1×20×20)/2

<u>s= 200 m</u>


3 0
3 years ago
Read 2 more answers
Two spheres are cut from a certain uniform rock. One has radius 4.50 cm. The mass of the other is five times greater. Find its r
Sonja [21]
Vol of sphere = 4/3 pi r^2.density of sphere = mass/volume.mass = densityxvolumesphere 1. mass = density x 4/3 pi 4.5^2sphere 2 5mass = density x 4/3 pi r^25=4/3 pi r^2 divided by 4/3 pi 4.5^25=r^2 divided by  4.5^25x4.5^2=r^2root(5x4.5^2)=r4.5 root 5 = r
4 0
3 years ago
Read 2 more answers
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