To find the height of a cylindrical tank that can hold [tex]\( V \)[/tex] cubic meters of fuel and has a diameter [tex]\( d \)[/tex] meters across, we need to use the formula for the volume of a cylinder. The volume [tex]\( V \)[/tex] of a cylinder is given by:
[tex]\[ V = \pi r^2 h \][/tex]
where:
- [tex]\( r \)[/tex] is the radius of the cylinder (which is half of the diameter [tex]\( d \)[/tex]).
- [tex]\( h \)[/tex] is the height of the cylinder.
Given:
- [tex]\( V \)[/tex] (volume of the tank).
- [tex]\( d \)[/tex] (diameter of the tank).
First, we convert the diameter [tex]\( d \)[/tex] to radius [tex]\( r \)[/tex]:
[tex]\[ r = \frac{d}{2} \][/tex]
Next, we solve for the height [tex]\( h \)[/tex]:
[tex]\[ V = \pi \left(\frac{d}{2}\right)^2 h \][/tex]
[tex]\[ h = \frac{V}{\pi \left(\frac{d}{2}\right)^2} \][/tex]
Now simplifying the expression inside the denominator:
[tex]\[ h = \frac{V}{\pi \frac{d^2}{4}} \][/tex]
[tex]\[ h = \frac{V}{\frac{\pi d^2}{4}} \][/tex]
[tex]\[ h = \frac{4V}{\pi d^2} \][/tex]
Thus, the correct height [tex]\( h \)[/tex] of the cylindrical tank in meters is:
[tex]\[ h = \frac{4V}{\pi d^2} \][/tex]
Evaluating the choices given:
A. [tex]\(\frac{2V}{x n^2}\)[/tex]
- This does not match our simplified formula.
B. [tex]\(\frac{4V}{d}\)[/tex]
- This does not match our simplified formula.
C. [tex]\(\frac{V}{x^2}\)[/tex]
- This does not match our simplified formula.
D. [tex]\(\frac{4V}{m^2}\)[/tex]
- This does not match our simplified formula.
E. [tex]\(8 V\)[/tex]
- This does not match our simplified formula.
Therefore, none of the provided options seem to exactly match the expression [tex]\( \frac{4V}{\pi d^2} \)[/tex]. Based on the calculations, there might be an oversight or misstatement within the problem's choices.
However, based on our derived calculations, the height [tex]\( h \)[/tex] of the cylindrical tank in meters is:
[tex]\[ h \approx 0.637 \][/tex]
(derived from the calculation) when substituting in actual values like [tex]\( V = 50 \)[/tex] and [tex]\( d = 10 \)[/tex].
