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A rectangular tank with a square base of 2 meters and a height of 4 meters is half filled with water. Letting y -correspond to the bottom of the tank. Find the work required to pump the water to the top of the tank. The density of water is 1000 and the force of gravity is 9. 8. Round your answer to one decimal place. 2m 4m

User Rennat
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Answer: To find the work required to pump the water to the top of the tank, we need to calculate the change in potential energy of the water as it is moved from the halfway point (y = 2 meters) to the top of the tank (y = 4 meters).

Given data:

Square base side length (s) = 2 meters

Height of the tank (h) = 4 meters

Density of water (ρ) = 1000 kg/m³

Acceleration due to gravity (g) = 9.8 m/s²

Let's proceed with the calculation:

Volume of water in the tank (V_water):

The tank is half-filled with water, so the volume of water is half the total volume of the tank.

Volume of water = (1/2) * Base area * Height

V_water = (1/2) * (s²) * h

V_water = (1/2) * (2 m)² * 4 m

V_water = 4 m³

Mass of the water (m_water):

Mass = Volume * Density

m_water = V_water * ρ

m_water = 4 m³ * 1000 kg/m³

m_water = 4000 kg

Change in height (Δy):

The change in height of the water is from the halfway point (y = 2 meters) to the top of the tank (y = 4 meters).

Δy = 4 m - 2 m

Δy = 2 m

Work required (W) to pump the water to the top:

The work done is equal to the change in potential energy.

Potential energy (PE) = Mass * Gravitational acceleration * Change in height

PE = m_water * g * Δy

PE = 4000 kg * 9.8 m/s² * 2 m

PE = 78,400 kg·m²/s²

Since 1 Joule (J) is equal to 1 kg·m²/s², the work required (W) is 78,400 Joules.

Rounded to one decimal place, the work required to pump the water to the top of the tank is approximately 78,400.0 Joules.

User Akourt
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