Answer
$P=1900Pa$
Work Step by Step
Given:
$\theta=55^{\circ}$
$v=1000\frac{m}{s}$
$m=3.3*10^{-27}kg$
$A=2.0cm^2=0.0002m^2$
$\Delta{t}=10^{23}$ (the time it takes for a single molecule to hit the wall)
Use the equation $F=\frac{\Delta{P}}{\Delta{t}}$ to find the force of the beam:
$F=\frac{mv_x-(-mv_x)}{\Delta{t}}=\frac{2mv_x}{\Delta{t}}=\frac{2(3.3*10^{-27}kg)(1000\frac{m}{s})cos(55^{\circ})}{10^{-23}s}\approx0.37856$
Now find the pressure that the force exerts on the area with $P=\frac{F}{A}$
$P=\frac{0.37856}{2*10^{-4}}\approx1900Pa$
I use $\approx$ as I rounded to the appropriate significant figures.
