Oxygen and hydrogen gas are at the same temperature T.What is the ratio of kinetic energies of oxygen molecule and hydrogen mole
1 answer:
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
...............(1)
Here k is Boltzmann's constant ().
From above equation it is clear that average kinetic energy of a gas molecule only depends on temperature.
Therefore kinetic energy of oxygen molecule ....2
and kinetic energy of hydrogen molecule ....3
the ratio of kinetic energies of oxygen molecule and hydrogen molecule if oxygen is 16 times heavier than hydrogen is given as
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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Answer:
The shell hit at a distance of 1.9 x 10² km
The time of flight of the shell was 5.3 x 10² s
Explanation:
The position of the shell is given by the vector "r":
r = (x0 + v0 * t * cos α ; y0 + v0 * t * sin α + 1/2 g t²)
where:
x0 = initial horizontal position
v0 = magnitude of the initial velocity
t = time
α = launching angle
y0 = initial vertical position
g = acceleration of gravity
When the shell hit, the vertical component (ry) of the vector position r is 0. See figure.
Then:
ry = 0 = y0 + v0 * t * sin α + 1/2 g t²
Since the gun is at the center of our system of reference, y0 and x0 = 0
0 = t (v0 sin α + 1/2 g t)
t= 0 is discarded as solution
v0 sin α + 1/2 g t = 0
t = -2v0 sin α / g
t = (-2 * 2610 m/s * sin 81.9°)/ (-9.8 m/s²) = 5.3 x 10² s. This is the time of flight of the shell until it hit.
Then, the distance at which the shell hit is:
Distance = Module of r = ( x0 + v0 * t * cos α; 0) = x0 + v0 * t * cos α
Distance = 2.61 km/s * 5.3 x 10² s * cos 81.9 = 1.9 x 10² km
