Dust Particles: RMS Speed Calculation (3*10^13 mm/s)

[doggieslover]

calculate the rms (root-mean-square) speed v_rms of these particles, assuming them to be spheres of diameter 5 \; \mu {\rm m} and density 2 \; {\rm g}/{\rm cm}^3 = 2000 \; {\rm kg}/{\rm m^3}. The mass of such a dust particle is 1.31 \times 10^{-13} \; \rm kg.
Express your answer in millimeters per second to one decimal place only.

So I found out from the previous problem that the equation to use is:

v_rms =\sqrt{\left(\frac{\left(3k_{B}T\right)}{\left({\rho}\left(\frac{4}{3}\right){\pi}\left(\frac{d}{2}\right)^{3}\right)}\right)}


And after I plugged everything into the equation, I got 3*10^13 mm/s as the answer.

I am pretty sure that I converted everything to the right units, I double checked my work a few times already, I don't know what I 've done wrong.

Please help?

 

[tiny-tim]

Hi doggieslover!

Since nobody's replied in over 36 hours, maybe they can't read it … so here's a translation …

calculate the rms (root-mean-square) speed v_rms of these particles, assuming them to be spheres of diameter 5µm and density 2g/cm³ = 2000kg/m³. The mass of such a dust particle is 1.31 10^-13 kg.
Express your answer in millimeters per second to one decimal place only.

So I found out from the previous problem that the equation to use is:

$$v_{rms} \ =\ \sqrt{\left(\frac{\left(3k_{B}T\right)}{\left({\rho}\left(\frac{4}{3}\right){\pi}\left(\frac{d}{2}\right)^{3}\right)}\right)}$$And after I plugged everything into the equation, I got 3*10^13 mm/s as the answer.

I am pretty sure that I converted everything to the right units, I double checked my work a few times already, I don't know what I 've done wrong.

Please help?

 

[turin]

What did you use for T, and why? Do you understand the meaning of this formula?

 

[doggieslover]

T = 290K, it was given in a previous problem, I forgot to incorporate it in here.

I solved it already, the answer is .3mm/s, I probably just converted something wrong.