The viewing portion of the rectangular glass window in a fish tank is 63 inches wide and runs from 1.5 inches below the water's surface to 40.5 inches below the surface. find the fluid force against this portion of the window. the weight-density of seawater is 64 lb/ft what is the fluid force against the window? lb te

fluid force = 1309 lb

Explanation:

Given data

wide = 63 inches

distance below water surface = 1.5 inches

distance below surface = 40.5 inches

weight-density of seawater W =  64 lb/ft

to find out

fluid force against the window

solution

we consider here A(y) is the depth of fluid and B(y) length of portion at y

so we can say y1 = 0 so y2 will be 33.5 - 0.5 = 33 inches = 2.75 feet

so  A(y) = 33.5/12 - y ) feet

and B(y) = 63 inches = 5.25 feet

fluid force = W A(y) B(y) dy

so take limit y = 0 to y = 2.77

fluid force = 64 ((33.5/12) -y ) (5.25) dy

fluid force = 64×5.25/12 ((33.5  -12y ) dy

fluid force = 28

fluid force = 28 × 46.75 - 0

so fluid force = 1309 lb

work: w = fd (given)

work done by chet   =50*1 = 50 j

work done by mika =40*2 = 80 j

work done by sara =30*1.5 = 45 j

work done by bill =60*0.5 = 30 j

students in order from the greatest amount of work done to the least is mika> chet> sara> bill

hence option (b) is correct.

the answer is a. at the equator