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The problem involves finding the work \( W \) required to pump liquid out of a spherical tank. The key details are:

- The tank is a sphere with radius \( r = 4 \) m.
- The liquid fills half of the tank.
- The liquid's density is \( 900 \, \text{kg/m}^3 \).
- The spout is located 1 meter above the top of the sphere.
- Gravitational acceleration \( g = 9.8 \, \text{m/s}^2 \).

### Work Calculation Strategy
To find the total work, we need to:
1. Compute the volume of liquid at a small height \( h \) above the tank's bottom.
2. Determine the force needed to lift the liquid at that height.
3. Integrate this force over the height of the liquid to find the total work.

### Step-by-step Breakdown:

1. **Tank Geometry:**
   - The volume element of the liquid at a height \( h \) is a thin disc.
   - The radius of this disc, at height \( h \), is given by the relationship \( r = \sqrt{16 - y^2} \) where \( y \) is the distance from the center of the sphere (and \( y = h - 4 \)).
   - The area of this disc is \( \pi r^2 = \pi(16 - y^2) \).
   
2. **Work of lifting an element:**
   - The volume of the disc at height \( h \) is \( \pi(16 - y^2) \cdot dh \).
   - The weight of the liquid in this disc is \( \rho \cdot g \cdot \text{volume} = 900 \cdot 9.8 \cdot \pi(16 - y^2) \cdot dh \).
   - The work to lift this volume to the spout (located at a height of \( h + 1 \)) is \( (h + 1) \cdot \text{weight} \).

3. **Integrate over the height of the liquid:**
   - The liquid fills half of the sphere, so the limits of integration will be from \( y = -4 \) to \( y = 0 \) (as the height is measured from the center of the sphere).
   - The total work \( W \) is the integral of the work to lift each disc element.

I'll now set up the integral expression and calculate the work required.The total work required to pump the liquid out of the spherical tank is approximately \( 7,684,587 \, \text{J} \) (joules).

Would you like more details or have any further questions? Here are some related questions you might explore:

1. How would the work change if the tank were fully filled with liquid?
2. What if the spout was positioned at a different height?
3. How does the liquid's density affect the required work?
4. What would be the work if the radius of the tank was larger, say 5 meters?
5. How does the gravitational constant \( g \) influence the work in different planets or environments?

**Tip:** Integrating volume or area-related functions often helps calculate physical quantities like work, as seen in this case with spherical tanks.

A spherical tank with radius 4 m is half full of a liquid that has a density of 900 kg/m^3. The tank has a 1 m spout at the top. Find the work W required to pump the liquid out of the spout. Use 9.8 m/s^2 for g.

This problem involves calculating the work required to pump a liquid from a half-full spherical tank with a radius of 4 meters, using the liquid's density of 900 kg/m^3 and gravitational acceleration of 9.8 m/s^2. Detailed calculus integration steps and formulas are used.

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