Answer
a) .00369 Ns
b) 20 meters
Work Step by Step
a) We use the equation for impulse:
$J=F\Delta t = (.041)(.09)=\fbox{.00369 Ns}$
b) We first find acceleration:
$a = \frac{.041}{8.3\times 10^{-6}}=-4939.759 \ m/s^2$
This means:
$v_0=(4939.759 \ m/s^2)(.09s)=444.57 \ m/s$
This means that the distance traveled is:
$\Delta x = (444.57 \ m/s)(.09)+\frac{1}{2}(-4939.759 \ m/s^2)(.09s)^2=\fbox{20 meters}$
