Answer:
2 is the numerical answer.
Explanation:
Hello there!
In this case, according to the given information and formula, it is possible for us to remember that equation for the calculation of the average kinetic energy of a gas is:

Whereas R is the universal gas constant, NA the Avogadro's number and T the temperature.
Which means that for the given ratio, we can obtain the value as follows:

Regards!
Answer:
The observed wavelength on Earth from that hydrogen atom is
.
Explanation:
Given that,
The actual wavelength of the hydrogen atom, 
A hydrogen atom in a galaxy moving with a speed of, 
We need to find the observed wavelength on Earth from that hydrogen atom. The speed of galaxy is given by :

is the observed wavelength

So, the observed wavelength on Earth from that hydrogen atom is
. Hence, this is the required solution.
Answer : The average speed of the sprinter is, 34.95 Km/hr
Solution :
Average velocity : It is defined as the distance traveled by the time taken.
Formula used for average velocity :

where,
= average velocity
d = distance traveled = 200 m
t = time taken = 20.6 s
Now put all the given values in the above formula, we get the average velocity of the sprinter.

conversion :
(1 Km = 1000m)
(1 hr = 3600 s)
Therefore, the average speed of the sprinter is, 34.95 Km/hr
Answer:
option C
Explanation:
given,
diameter of circular room = 8 m
rotational velocity of the rider = 45 rev/min
= 
=4.712 rad/s
here in this case normal force is equal to centripetal force
N = m r ω²
N = m x 4 x 4.712²
N = 88.83m
frictional force = μ N
= 88.83m x μ
now, for the body to not to slide
gravity force is equal to frictional force
m g = 88.83 m x μ
g = 88.83 x μ
9.8 = 88.83 x μ
μ = 0.11
hence, the correct answer is option C
Answer:
The vapor pressure at 60.6°C is 330.89 mmHg
Explanation:
Applying Clausius Clapeyron Equation
![ln(\frac{P_2}{P_1}) = \frac{\delta H}{R}[\frac{1}{T_1}- \frac{1}{T_2}]](https://tex.z-dn.net/?f=ln%28%5Cfrac%7BP_2%7D%7BP_1%7D%29%20%3D%20%5Cfrac%7B%5Cdelta%20H%7D%7BR%7D%5B%5Cfrac%7B1%7D%7BT_1%7D-%20%5Cfrac%7B1%7D%7BT_2%7D%5D)
Where;
P₂ is the final vapor pressure of benzene = ?
P₁ is the initial vapor pressure of benzene = 40.1 mmHg
T₂ is the final temperature of benzene = 60.6°C = 333.6 K
T₁ is the initial temperature of benzene = 7.6°C = 280.6 K
ΔH is the molar heat of vaporization of benzene = 31.0 kJ/mol
R is gas rate = 8.314 J/mol.k
![ln(\frac{P_2}{40.1}) = \frac{31,000}{8.314}[\frac{1}{280.6}- \frac{1}{333.6}]\\\\ln(\frac{P_2}{40.1}) = 3728.65 (0.003564 - 0.002998)\\\\ln(\frac{P_2}{40.1}) = 3728.65 (0.000566)\\\\ln(\frac{P_2}{40.1}) = 2.1104\\\\\frac{P_2}{40.1} = e^{2.1104}\\\\\frac{P_2}{40.1} = 8.2515\\\\P_2 = (40.1*8.2515)mmHg = 330.89 mmHg](https://tex.z-dn.net/?f=ln%28%5Cfrac%7BP_2%7D%7B40.1%7D%29%20%3D%20%5Cfrac%7B31%2C000%7D%7B8.314%7D%5B%5Cfrac%7B1%7D%7B280.6%7D-%20%5Cfrac%7B1%7D%7B333.6%7D%5D%5C%5C%5C%5Cln%28%5Cfrac%7BP_2%7D%7B40.1%7D%29%20%3D%203728.65%20%280.003564%20-%200.002998%29%5C%5C%5C%5Cln%28%5Cfrac%7BP_2%7D%7B40.1%7D%29%20%3D%203728.65%20%20%280.000566%29%5C%5C%5C%5Cln%28%5Cfrac%7BP_2%7D%7B40.1%7D%29%20%3D%202.1104%5C%5C%5C%5C%5Cfrac%7BP_2%7D%7B40.1%7D%20%3D%20e%5E%7B2.1104%7D%5C%5C%5C%5C%5Cfrac%7BP_2%7D%7B40.1%7D%20%3D%208.2515%5C%5C%5C%5CP_2%20%3D%20%2840.1%2A8.2515%29mmHg%20%3D%20330.89%20mmHg)
Therefore, the vapor pressure at 60.6°C is 330.89 mmHg