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Sagot :
Step-by-step explanation:
To find the depth of water in the cone, we need to know how much water is in the cone or its volume. If we assume the cone is partially filled, we need additional information like the volume of water. However, if the cone is fully filled, the water depth equals the height of the cone.
Given:
- Height (\(h\)) of the cone: 16 cm
- Diameter (\(d\)) of the base: 12 cm
- Radius (\(r\)) of the base: \(d/2 = 12/2 = 6\) cm
The volume (\(V\)) of a cone is given by the formula:
\[ V = \frac{1}{3} \pi r^2 h \]
Let's calculate the volume:
\[ V = \frac{1}{3} \pi (6^2) (16) \]
\[ V = \frac{1}{3} \pi (36) (16) \]
\[ V = \frac{1}{3} \pi (576) \]
\[ V = 192 \pi \text{ cubic centimeters} \]
If we need to calculate how deep the water is in the cone given a specific volume of water, we would use the inverse process. Let's assume we know the volume of water, \(V_w\), and we need to find the corresponding height of water, \(h_w\).
The formula to relate the volume of a partial cone filled with height \(h_w\) is:
\[ V_w = \frac{1}{3} \pi r_w^2 h_w \]
Since the cone is similar in shape at any height, the ratio of the radius to the height of the cone will be the same as the ratio of the radius of the water surface (\(r_w\)) to the height of the water (\(h_w\)):
\[ \frac{r_w}{h_w} = \frac{r}{h} \]
\[ r_w = \frac{r}{h} \cdot h_w \]
\[ r_w = \frac{6}{16} \cdot h_w \]
\[ r_w = \frac{3}{8} \cdot h_w \]
Substituting \(r_w\) back into the volume formula:
\[ V_w = \frac{1}{3} \pi \left(\frac{3}{8} h_w\right)^2 h_w \]
\[ V_w = \frac{1}{3} \pi \left(\frac{9}{64} h_w^2\right) h_w \]
\[ V_w = \frac{1}{3} \pi \frac{9}{64} h_w^3 \]
\[ V_w = \frac{3 \pi}{64} h_w^3 \]
Solving for \(h_w\) in terms of \(V_w\):
\[ V_w = \frac{3 \pi}{64} h_w^3 \]
\[ h_w^3 = \frac{64 V_w}{3 \pi} \]
\[ h_w = \left(\frac{64 V_w}{3 \pi}\right)^{\frac{1}{3}} \]
If the cone is fully filled, \(V_w = 192 \pi\):
\[ h_w = \left(\frac{64 \cdot 192 \pi}{3 \pi}\right)^{\frac{1}{3}} \]
\[ h_w = \left(\frac{12288}{3}\right)^{\frac{1}{3}} \]
\[ h_w = \left(4096\right)^{\frac{1}{3}} \]
\[ h_w = 16 \text{ cm} \]
Thus, if the cone is fully filled with water, the water depth is 16 cm. If you have a specific volume of water \(V_w\) in mind, you can use the formula \( h_w = \left(\frac{64 V_w}{3 \pi}\right)^{\frac{1}{3}} \) to find the depth.
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