Aswimming pool has the shape of a box with a base that measures 25 m by 11 m and a uniform depth of 3.1 m. how much work is required to pump the water out of the pool when it is full? use 1000 kg divided by m cubed for the density of water and 9.8 m divided by s squared for the acceleration due to gravity. draw a y-axis in the vertical direction (parallel to gravity) and choose one corner of the bottom of the pool as the origin. for 0 less than or equals y less than or equals 3.1 , find the cross-sectional area a(y).

Step-by-step explanation:

Volume of water in the swimming pool

= 25 x 11 x 3.1

= 852.5 m³

mass of water in the pool

= volume x density of water

m = 852.5 x 10³ kg

We shall consider the height H  by which the centre of mass of the water-body is increased to lift the water .

H = 3.1 / 2 = 1.55 m

Work done in pumping the water out of pool

mgH  ,

=  852.5 x 10³ x 9.8 x 1.55

= 13 x 10⁶ J

-3

step-by-step explanation:

a drawing really does here:

so i draw the horizontal line y=2 and the vertical line x=-2 at this intersection i put b (this is where my right angle is going to be).   so point b is at (-2,2).   point a is 5 units long and is also on y=2 so this point has to be (3,2) or (-7,2) -doesn't matter i will go with (3,2).

now we know c is at (-2,y) we don't know the y yet bet we know that c is on x=-2.

we have the area of the triangle is .5bh=12.5

we know the base length is 5 (ab)

we have the height is (2-y) if you look at your

plug in

.5(5)(2-y)=12.5

solve for y to get the y-coordinate for c

2.5(2-y)=12.5

divide both sides by 2.5 giving:

2-y=5

now subtract 2 on both sides

-y=3

y=-3