1.Light-collecting area
2.Angular resolution
Answer:
![v_{rms} =196.59 m/s](https://tex.z-dn.net/?f=v_%7Brms%7D%20%3D196.59%20m%2Fs)
Given:
Temperature, T = 3.13 K
molar mass of molecular hydrogen, m = 2.02 g/mol = ![2.02\times 10^{-3}kg/mol](https://tex.z-dn.net/?f=2.02%5Ctimes%2010%5E%7B-3%7Dkg%2Fmol)
Solution:
To calculate the root mean squarer or rms speed of hydrogen molecule, we use the given formula:
![v_{rms} = \sqrt{\frac{3TR}{m}}](https://tex.z-dn.net/?f=v_%7Brms%7D%20%3D%20%5Csqrt%7B%5Cfrac%7B3TR%7D%7Bm%7D%7D)
where
R = rydberg's constant = 8.314 J/mol-K
Putting the values in the above formula:
![v_{rms} = \sqrt{\frac{3\times 3.13\times 8.314}{2.02\times 10^{-3}}}](https://tex.z-dn.net/?f=v_%7Brms%7D%20%3D%20%5Csqrt%7B%5Cfrac%7B3%5Ctimes%203.13%5Ctimes%208.314%7D%7B2.02%5Ctimes%2010%5E%7B-3%7D%7D%7D)
![v_{rms} =196.59 m/s](https://tex.z-dn.net/?f=v_%7Brms%7D%20%3D196.59%20m%2Fs)
Done I don't know answer of this question or this photo is the answer can you tell me
The mass of the hoop is the only force which is computed by:F net = 2.8kg*9.81m/s^2 = 27.468 N
the slow masses that must be quicker are the pulley, ring, and the rolling sphere.
The mass correspondent of M the pulley is computed by torque τ = F*R = I*α = I*a/R F = M*a = I*a/R^2 --> M = I/R^2 = 21/2*m*R^2/R^2 = 1/2*m
The mass equal of the rolling sphere is computed by: the sphere revolves around the contact point with the table. So using the proposition of parallel axes, the moment of inertia of the sphere is I = 2/5*mR^2 for spin about the midpoint of mass + mR^2 for the distance of the axis of rotation from the center of mass of the sphere. I = 7/5*mR^2 M = 7/5*m
the acceleration is then a = F/m = 27.468/(2.8 + 1/2*2 + 7/5*4) = 27.468/9.4 = 2.922 m/s^2