To determine the unused space in the container, we first need to calculate the total volume occupied by the marbles.
- Calculate the Volume of One Marble:
The volume \( V \) of a sphere is given by the formula: \[ V = \frac{4}{3} \pi r^3 \] where \( r \) is the radius.
Given that the radius of each marble is 2 mm, we first convert this to meters: \[ 2 \text{ mm} = 0.002 \text{ m} \]
Now, we can calculate the volume of one marble: \[ V = \frac{4}{3} \times 3.14 \times (0.002)^3 \]
Calculating \( (0.002)^3 \): \[ (0.002)^3 = 0.000000008 \text{ m}^3 \]
Now substituting this back into the volume formula: \[ V \approx \frac{4}{3} \times 3.14 \times 0.000000008 \approx \frac{4 \times 3.14 \times 0.000000008}{3} \approx 0.00000003351 \text{ m}^3 \]
- Volume of 50 Marbles:
To find the total volume occupied by 50 marbles, we multiply the volume of one marble by 50: \[ \text{Total Volume of Marbles} = 50 \times 0.00000003351 \approx 0.0000016755 \text{ m}^3 \]
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Volume of the Container: The volume of the container is given as 4000 m³.
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Calculate Unused Space: Now we can find the unused space in the container by subtracting the total volume occupied by the marbles from the total volume of the container: \[ \text{Unused Space} = 4000 - 0.0000016755 \approx 3999.9999983245 \text{ m}^3 \]
Since we're looking for the unused space rounded to the nearest tenth, we manage a rounding: \[ \text{Unused Space} \approx 4000.0 \text{ m}^3 \]
In this case, none of the given options are directly applicable because they all refer to much smaller values. Instead, based on our calculations, there is virtually no unused space as calculated.
Given this situation, it seems necessary to verify the volume calculations due to discrepancies in implications because the numbers provided in the options aren't aligned with the computed values based on the parameters. Please double-check your container volume and marble volume if seeking a match with given choices.
However, the closest sensible interpretation considering volume rounding would still yield "4000.0 m³" - thus, this would go happily unreported against any structural confines as none of the provided figures contextually align.