Uche pumps gasoline at a rate of 18\,\dfrac{\text{L}}{\text{min}}18 min L 18, start fraction, start text, L, end text, divided by, start text, m, i, n, end text, end fraction. What is Uche's pumping rate in \dfrac{\text{mL}}{\text{s}} s mL start fraction, start text, m, L, end text, divided by, start text, s, end text, end fraction? \dfrac{\text{mL}}{\text{s}} s mL

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MrRoyal MrRoyal · Answer 1 · 2020-11-08T09:49:25+01:00

Question:

Uche pumps gasoline at a rate of 18 L/min. What is Uche's pumping rate in mL/s?

Answer:

[tex]Rate = 300mL/s}[/tex]

Step-by-step explanation:

Given

[tex]Rate = \frac{18L}{min}[/tex]

Required

Express as mL/s

First, we need to convert L to mL

[tex]1L = 1000mL[/tex]

So:

[tex]Rate = \frac{18 * 1000mL}{min}[/tex]

[tex]Rate = \frac{18000mL}{min}[/tex]

Next, convert min to seconds

[tex]1\ min = 60\ s[/tex]

So:

[tex]Rate = \frac{18000mL}{min}[/tex]

[tex]Rate = \frac{18000mL}{60s}[/tex]

[tex]Rate = 300mL/s}[/tex]