A cylindrical tank has a radius of 2 m and a height of 9 m. The tank is filled with water. Find the work needed to pump the top 3 m of water out the top of the tank. (Use 9.8 m/s2 for g and the fact that the density of water is 1000 kg/m3.)

3,325,140 Joules

Step-by-step explanation:

Work done by the pump = Force applied to pump * distance covered by the water.

Since Force = mass * acceleration due to gravity

Force = (density of water * volume of the tank) * acceleration due to gravity

F =ρVg

Workdone = (ρVg )* d

Given ρ = 1000kg/m³, g = 9.8m/s², d = 3m

Workdone by the pump = 1000 * 113.10 * 9.8 * 3

Workdone by the pump  = 3,325,140Joules

37/3

step-by-step explanation:

step 1: simplify both sides of the equation.

7x+8(x+

1

4

)=3(6x−9)−8

7x+(8)(x)+(8)(

1

4

)=(3)(6x)+(3)(−9)+−8(distribute)

7x+8x+2=18x+−27+−8

(7x+8x)+(2)=(18x)+(−27+−8)(combine like terms)

15x+2=18x+−35

15x+2=18x−35

tep 2: subtract 18x from both sides.

15x+2−18x=18x−35−18x

−3x+2=−35

step 3: subtract 2 from both sides.

−3x+2−2=−35−2

−3x=−37

step 4: divide both sides by -3.

−3x

−3

=

−37

−3

x=

37

3