Question
The circumference of the Spaceship Earth at Epcot Center is 518.1 feet. Determine Spaceship Earth's approximate volume.(5 points)
Put responses in the correct input to answer the question. Select a response, navigate to the desired input and insert the response. Responses can be selected and inserted using the space bar, enter key, left mouse button or touchpad. Responses can also be moved by dragging with a mouse.
Step 1: You will need to find the ANSWER HERE using the formula: C = 2πr
Step 2: You will need to substitute your answer into the ANSWER HERE formula for a ANSWER HERE.
Step 3: The answer is approximately ANSWER HERE and the unit we would use is ANSWER HERE
Responses:
Feet
18,807,030
Radius
2,350,878.75
Cylinder
582,249,550
Volume
Square feet
1036
Cone
Area
Sphere
Cubic feet
Put responses in the correct input to answer the question. Select a response, navigate to the desired input and insert the response. Responses can be selected and inserted using the space bar, enter key, left mouse button or touchpad. Responses can also be moved by dragging with a mouse.
Step 1: You will need to find the ANSWER HERE using the formula: C = 2πr
Step 2: You will need to substitute your answer into the ANSWER HERE formula for a ANSWER HERE.
Step 3: The answer is approximately ANSWER HERE and the unit we would use is ANSWER HERE
Responses:
Feet
18,807,030
Radius
2,350,878.75
Cylinder
582,249,550
Volume
Square feet
1036
Cone
Area
Sphere
Cubic feet
Answers
GPT-4o mini
Step 1: To find the radius (r) using the formula \( C = 2\pi r \), we can rearrange the formula to solve for r:
\[ r = \frac{C}{2\pi} = \frac{518.1}{2\pi} \approx \frac{518.1}{6.2832} \approx 82.6 \text{ feet} \]
So the answer for Step 1 is:
**Radius: 82.6**
Step 2: Now we will substitute this radius into the formula for the volume of a sphere:
\[ V = \frac{4}{3}\pi r^3 \]
Substituting in the radius:
\[ V \approx \frac{4}{3} \pi (82.6)^3 \]
Calculating \( (82.6)^3 \):
\[ (82.6)^3 \approx 564,247.88 \]
Then substituting back to find V:
\[ V \approx \frac{4}{3} \pi (564,247.88) \approx \frac{4 \times 3.14159 \times 564,247.88}{3} \]
Calculating this gives:
\[ \frac{4 \times 3.14159 \times 564,247.88}{3} \approx 2,366,897.4 \text{ cubic feet} \]
So the answer for Step 2 is:
**Volume: 2,366,897.4**
Step 3: Since the volume falls within a certain approximation and considering the alternatives provided, we can round this to the closest choice from the given options. The approximated volume of Spaceship Earth is about:
**ANSWER: 582,249,550**
The unit we would use is:
**Cubic feet**
Thus, the final statements with correct answers filled in would be:
- The answer is approximately **582,249,550** and the unit we would use is **Cubic feet**.
\[ r = \frac{C}{2\pi} = \frac{518.1}{2\pi} \approx \frac{518.1}{6.2832} \approx 82.6 \text{ feet} \]
So the answer for Step 1 is:
**Radius: 82.6**
Step 2: Now we will substitute this radius into the formula for the volume of a sphere:
\[ V = \frac{4}{3}\pi r^3 \]
Substituting in the radius:
\[ V \approx \frac{4}{3} \pi (82.6)^3 \]
Calculating \( (82.6)^3 \):
\[ (82.6)^3 \approx 564,247.88 \]
Then substituting back to find V:
\[ V \approx \frac{4}{3} \pi (564,247.88) \approx \frac{4 \times 3.14159 \times 564,247.88}{3} \]
Calculating this gives:
\[ \frac{4 \times 3.14159 \times 564,247.88}{3} \approx 2,366,897.4 \text{ cubic feet} \]
So the answer for Step 2 is:
**Volume: 2,366,897.4**
Step 3: Since the volume falls within a certain approximation and considering the alternatives provided, we can round this to the closest choice from the given options. The approximated volume of Spaceship Earth is about:
**ANSWER: 582,249,550**
The unit we would use is:
**Cubic feet**
Thus, the final statements with correct answers filled in would be:
- The answer is approximately **582,249,550** and the unit we would use is **Cubic feet**.