Calculate the mass of water in a bank of fog hovering over Beaverton on a moderately foggy day. Assume that the surface area of Beaverton is 53 km² and the fog reaches to an altitude of 100 m. On a moderately foggy day, the density of water vapor in air will be 5 x 10⁻⁸ g/cm³. A) 3 x 10³ g
B) 2 x 10⁴ g
C) 3 x 10⁸ g
D) 2 x 10⁹ g
E) 3 x 10⁶ g

a: 45.9 meters

b: 4.29 seconds

step-by-step explanation:

a: h(t)=1/2(-9.8)(3)^2+0(3)+90

h(t)=-4.9(3)^2+90

h(t)=-44.1+90

h(t)=45.9 meters

b: 0=1/2(-9.8)(t^2)+0t+90

0=-4.9t^2+90

-90=-4.9t^2

18.37=t^2

t=4.29 seconds

in the second question, the numbers aren't exact, but they're a close approximation.

i think it's c. if you look search electronic symbol on google and then click on wiki it shows you a

bcde is the incorrect answer it is a