Answer: [tex]35 L[/tex] in [tex]5 s[/tex] and [tex]5.25 L[/tex] in [tex]3/4 s[/tex]
Explanation:
We can use the Rule of Three to solve this problem:
If:
[tex]7 L[/tex]-----[tex]1 s[/tex]
[tex]?[/tex]-----[tex]5 s[/tex]
Then:
[tex]?=\frac{(7 L)(5 s)}{1 s}[/tex]
[tex]?=35 L[/tex] This is the number of liter that will escape in [tex]5 s[/tex]
Now, in 3/4 s we use the Rule of Three again:
If:
[tex]7 L[/tex]-----[tex]1 s[/tex]
[tex]?[/tex]-----[tex]\frac{3}{4} s[/tex]
Then:
[tex]?=\frac{(7 L)(\frac{3}{4} s)}{1 s}[/tex]
[tex]?=5.25 L[/tex] This is the number of liter that will escape in [tex]3/4 s[/tex]
