To solve this problem, let us recall that the formula for
gases assuming ideal behaviour is given as:
rms = sqrt (3 R T / M)
where
R = gas constant = 8.314 Pa m^3 / mol K
T = temperature
M = molar mass
Now we get the ratios of rms of Argon (1) to hydrogen (2):
rms1 / rms2 = sqrt (3 R T1 / M1) / sqrt (3 R T2 / M2)
or
rms1 / rms2 = sqrt ((T1 / M1) / (T2 / M2))
rms1 / rms2 = sqrt (T1 M2 / T2 M1)
Since T1 = 4 T2
rms1 / rms2 = sqrt (4 T2 M2 / T2 M1)
rms1 / rms2 = sqrt (4 M2 / M1)
and M2 = 2 while M1 = 40
rms1 / rms2 = sqrt (4 * 2 / 40)
rms1 / rms2 = 0.447
Therefore the ratio of rms is:
<span>rms_Argon / rms_Hydrogen = 0.45</span>
Answer:
It conserves both energy and momentum in the collision at the same time. By design, when the balls collide the strings that hold them up are vertical (assuming balls are only swung from one side).
Explanation:
Hope This Helps!!
I think yes because you won’t be able to smell
1 m/s
Explanation:
To solve this question we use the following formula:
momentum = mass × velocity
momentum of the first car = 1000 kg × 2.5 m/s
momentum of the second car = 2500 kg × X m/s
To bring the cars at rest the momentum of the first car have to be equal to the momentul of the second car.
momentum of the first car = momentum of the second car
1000 kg × 25 m/s = 2500 kg × X m/s
X (velocity of the second car) = (1000 × 25) / 2500 = 1 m/s
Learn more about:
momentum
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Answer:
The work done on the Frisbee is 1.36 J.
Explanation:
Given that,
Mass of Frisbee, m = 115 g = 0.115 kg
Initial speed of Frisbee, u = 12 m/s at a point 1 m above the ground
Final speed of Frisbee , v = 10.9674 m/s when it has reached a height of 2.00 m. Let W is the work done on the Frisbee by its weight. According to work energy theorem, the work done is equal to the change in its kinetic energy. So,
So, the work done on the Frisbee is 1.36 J. Hence, this is the required solution.
