A tank in the shape of an inverted right circular cone has height $ 10$ meters and radius $ 8$ meters. It is filled with $ 6$ meters of hot chocolate.

Find the work required to empty the tank by pumping the hot chocolate over the top of the tank. Note: the density of hot chocolate is $ \delta = 1470 kg/m^3$

What do the $ signs mean before and after the numbers? What does $ \ mean in front of "delta"?

Does <$ 6$ meters> mean 6 cubic meters or a liquid height of 6 meters?

To find the work required to empty the tank by pumping the hot chocolate over the top, we need to calculate the total volume of hot chocolate in the tank and then use the equation for the work done in lifting the liquid.

1. Calculate the volume of the inverted cone-shaped tank: The volume of a cone can be calculated using the formula $V = \frac{1}{3} \pi r^2 h$, where $r$ is the radius and $h$ is the height. Since we have an inverted cone, we need to subtract the volume of the smaller cone at the top. The volume of the inverted cone can be calculated as $V_{\text{cone}} = \frac{1}{3} \pi R^2 H$, where $R$ is the radius of the larger base and $H$ is the height.

In this case, the radius of the larger base $R$ is $8$ meters, and the height $H$ is $10$ meters. So the volume of the inverted cone is $V_{\text{inverted cone}} = \frac{1}{3} \pi (8^2) (10)$.

2. Calculate the volume of the hot chocolate: The volume of the hot chocolate is equal to the height of the hot chocolate. In this case, the volume of the hot chocolate is $V_{\text{hot chocolate}} = 6$ cubic meters.

3. Calculate the volume of air in the tank: The volume of air in the tank can be calculated by subtracting the volume of the hot chocolate from the volume of the inverted cone: $V_{\text{air}} = V_{\text{inverted cone}} - V_{\text{hot chocolate}}$.

4. Calculate the mass of the hot chocolate: The mass of the hot chocolate can be calculated using the formula $m = \delta V$, where $\delta$ is the density of the hot chocolate and $V$ is the volume of the hot chocolate. In this case, the density $\delta$ is given as $1470 \, \text{kg/m}^3$.

So the mass of the hot chocolate is $m = (1470 \, \text{kg/m}^3) \cdot (6 \, \text{m}^3)$.

5. Calculate the work required: The work required to lift the hot chocolate over the top of the tank is given by the formula $W = mgh$, where $m$ is the mass, $g$ is the acceleration due to gravity (approximately $9.8 \, \text{m/s}^2$), and $h$ is the height the liquid is lifted.

In this case, the work required is $W = (1470 \, \text{kg/m}^3) \cdot (6 \, \text{m}^3) \cdot (9.8 \, \text{m/s}^2) \cdot (10 \, \text{m})$.

Now we can calculate the work required by plugging in the values and simplify the expression.

Note: The final answer will be in joules (J).