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Answer :
* First, the radius of the hemisphere is calculated by dividing the diameter by 2: $r = \frac{37.6}{2} = 18.8 \text{ m}$.
* Then, the volume of the hemisphere is calculated using the formula $V_{hemisphere} = \frac{2}{3} \pi r^3$.
* Substituting the radius into the formula: $V_{hemisphere} = \frac{2}{3} \pi (18.8)^3 \approx 13916.6 \text{ m}^3$.
* Finally, the volume of the hemisphere, rounded to the nearest tenth, is: $\boxed{13916.6 \text{ m}^3}$.
### Explanation
1. Problem Analysis
We are given a hemisphere with a diameter of 37.6 meters. We need to find its volume, rounded to the nearest tenth of a cubic meter.
2. Calculate the Radius
First, we need to find the radius of the hemisphere. The radius is half of the diameter, so we have:
$$r = \frac{d}{2} = \frac{37.6}{2} = 18.8 \text{ m}$$
3. Calculate the Volume of the Sphere
Next, we calculate the volume of the entire sphere using the formula:
$$V_{sphere} = \frac{4}{3} \pi r^3$$
Substituting the value of the radius, we get:
$$V_{sphere} = \frac{4}{3} \pi (18.8)^3$$
$$V_{sphere} = \frac{4}{3} \pi (6643.072)$$
4. Calculate the Volume of the Hemisphere
Since we need the volume of a hemisphere, which is half of the sphere, we divide the volume of the sphere by 2:
$$V_{hemisphere} = \frac{V_{sphere}}{2} = \frac{1}{2} \times \frac{4}{3} \pi (18.8)^3$$
$$V_{hemisphere} = \frac{2}{3} \pi (18.8)^3$$
$$V_{hemisphere} = \frac{2}{3} \pi (6643.072)$$
$$V_{hemisphere} \approx 13916.6 \text{ m}^3$$
5. Round to the Nearest Tenth
Finally, we round the volume to the nearest tenth of a cubic meter, which gives us 13916.6 m³.
6. State the Final Answer
Therefore, the volume of the hemisphere is approximately $13916.6 \text{ m}^3$.
### Examples
Understanding the volume of hemispheres is useful in various real-world applications. For example, when designing large dome-shaped structures like planetariums or observatories, engineers need to calculate the volume of the hemispherical dome to determine the amount of material required for construction and to estimate the interior space available. Similarly, in the manufacturing of storage tanks or containers with hemispherical ends, accurate volume calculations are essential for determining the capacity and optimizing the design.
* Then, the volume of the hemisphere is calculated using the formula $V_{hemisphere} = \frac{2}{3} \pi r^3$.
* Substituting the radius into the formula: $V_{hemisphere} = \frac{2}{3} \pi (18.8)^3 \approx 13916.6 \text{ m}^3$.
* Finally, the volume of the hemisphere, rounded to the nearest tenth, is: $\boxed{13916.6 \text{ m}^3}$.
### Explanation
1. Problem Analysis
We are given a hemisphere with a diameter of 37.6 meters. We need to find its volume, rounded to the nearest tenth of a cubic meter.
2. Calculate the Radius
First, we need to find the radius of the hemisphere. The radius is half of the diameter, so we have:
$$r = \frac{d}{2} = \frac{37.6}{2} = 18.8 \text{ m}$$
3. Calculate the Volume of the Sphere
Next, we calculate the volume of the entire sphere using the formula:
$$V_{sphere} = \frac{4}{3} \pi r^3$$
Substituting the value of the radius, we get:
$$V_{sphere} = \frac{4}{3} \pi (18.8)^3$$
$$V_{sphere} = \frac{4}{3} \pi (6643.072)$$
4. Calculate the Volume of the Hemisphere
Since we need the volume of a hemisphere, which is half of the sphere, we divide the volume of the sphere by 2:
$$V_{hemisphere} = \frac{V_{sphere}}{2} = \frac{1}{2} \times \frac{4}{3} \pi (18.8)^3$$
$$V_{hemisphere} = \frac{2}{3} \pi (18.8)^3$$
$$V_{hemisphere} = \frac{2}{3} \pi (6643.072)$$
$$V_{hemisphere} \approx 13916.6 \text{ m}^3$$
5. Round to the Nearest Tenth
Finally, we round the volume to the nearest tenth of a cubic meter, which gives us 13916.6 m³.
6. State the Final Answer
Therefore, the volume of the hemisphere is approximately $13916.6 \text{ m}^3$.
### Examples
Understanding the volume of hemispheres is useful in various real-world applications. For example, when designing large dome-shaped structures like planetariums or observatories, engineers need to calculate the volume of the hemispherical dome to determine the amount of material required for construction and to estimate the interior space available. Similarly, in the manufacturing of storage tanks or containers with hemispherical ends, accurate volume calculations are essential for determining the capacity and optimizing the design.
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