## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

$v = 1360~m/s$
We can find the kinetic energy of one helium atom. $KE = \frac{3700~J}{6.02\times 10^{23}~atoms}$ $KE = 6.146\times 10^{-21}~J$ We can find the speed of each helium atom. $\frac{1}{2}mv^2 = KE$ $v^2 = \frac{2~KE}{m}$ $v = \sqrt{\frac{2~KE}{m}}$ $v = \sqrt{\frac{(2)(6.146\times 10^{-21}~J)}{6.68\times 10^{-27}~kg}}$ $v = 1360~m/s$