Question

In a pipeline transporting crude oil at 1200 L/min, what is the volume transported in 3.2 hours?

a) 2.88 x 10⁵ L
b) 2.40 x 10⁵ L
c) 3.84 x 10⁵ L
d) 4.32 x 10⁵ L

Answer 1

To find the volume transported by the pipeline in 3.2 hours, multiply the flow rate (1200 L/min) by the total time in minutes (192), resulting in 230,400 liters or 2.304 x 10⁵ L, which does not match the provided options.

Step-by-step explanation:

To calculate the volume of crude oil transported in a pipeline over a period of time, we multiply the flow rate by the time during which the oil flows. Given a flow rate of 1200 L/min and a time period of 3.2 hours, we can find the total volume transported as follows:

• First convert the time from hours to minutes:
  3.2 hours × 60 minutes/hour = 192 minutes
• Then multiply the time in minutes by the flow rate:
  1200 L/min × 192 minutes = 230,400 L

The volume transported in 3.2 hours is therefore 230,400 liters, which can be written in scientific notation as 2.304 × 10⁵ L, matching none of the options provided in the question. It appears there may be a mistake in the options.