Natural Gas Consumption Calculation in Cubic Feet and Seconds

So the final equation in cubic feet and seconds would be:V = 3.888E6(t sec.) + 0.800(t sec.)^2In summary, the given empirical equation for the consumption of natural gas by a company can be expressed in units of cubic feet and seconds as V = 3.888E6(t sec.) + 0.800(t sec.)^2. The coefficients represent the rate of change in volume with respect to time and are properly assigned units in cubic feet per second and cubic feet per second squared. This conversion was done by first converting the months variable to seconds and then solving for t in terms of seconds.
  • #1
Mivz18
44
0
Problem: The consumption of natural gas by a company satisfies the empirical equation V=1.50t + 0.800 00(t squared), where V is the volume in millions of cubic feet and t the time in months. Express this equation in units of cubic feet and seconds. Assign proper units to the coefficients. Assume a month is equal to 30.0 days.

My work:

From the equation given, I took each part of the right and converted the months variable to seconds.

1.50(t months) * (t 30 days) * (t 24 hours) * (t 60 min) * (t 60 sec.)
-------------- ---------- ---------- --------- ----------
1 (t month) (t 1 day) (t 1 hr.) (t 1 min)


= 3.888E6(t sec.)

Then I repeated the conversion multiplication fractions with 0.008 00 and squaring the other numbers and reaching a value of 5.37477E9(t sec)^2

My new equation is then V= 3.888E6(t sec.) + 5.37477E9(t sec.)^2

dividing by a million to convert millions of cubic feet to cubic feet, I get:

V=3.889E6(t sec.) + 53747.7(t sec.)^2

However, the book gives an answer of 0.579t (ft cubed/sec.) + 1.19E-9t^2 (ft cubed/sec. squared)
Is not assigning units to the coefficients screwing me up or is there another problem with my calculations?
 
Physics news on Phys.org
  • #2
Mivz18 said:
Problem: The consumption of natural gas by a company satisfies the empirical equation V=1.50t + 0.800 00(t squared), where V is the volume in millions of cubic feet and t the time in months. Express this equation in units of cubic feet and seconds. Assign proper units to the coefficients. Assume a month is equal to 30.0 days.

My work:

From the equation given, I took each part of the right and converted the months variable to seconds.

1.50(t months) * (t 30 days) * (t 24 hours) * (t 60 min) * (t 60 sec.)
-------------- ---------- ---------- --------- ----------
1 (t month) (t 1 day) (t 1 hr.) (t 1 min)


= 3.888E6(t sec.)

Remember that the given formula requires t (the time in months). When dealing with t you have to be careful... I'd do it like this:

let a=time in seconds.

a = t months * (30days/1 month) * (24 hours/1day)*(60min/1hr)*(60s/1min)
so a= 2.592E6 * t seconds

So now solve for t, and then plug back into the original equation.

So t=a/2.592E6 . Plug a/2.592E6 into the original equation. Think of it like this. You've got the time in seconds a, then getting the time in months t, in terms of a...

So you'll then get an equation in terms of a, where a is in seconds. Volume is still in millions of cubic feet.

Mivz18 said:
dividing by a million to convert millions of cubic feet to cubic feet, I get:

To get from millions of cubic feet to cubic feet, you need to multiply by a million, not divide.
 
  • Like
Likes MarianC
  • #3


Your calculations seem to be correct, however, you did not assign units to the coefficients in your final equation. This can lead to confusion and difficulty in understanding the units of the equation. The units of the first coefficient should be (ft^3/sec) and the units of the second coefficient should be (ft^3/sec^2). It is important to always include units in your calculations to avoid any confusion or errors.
 

1. What are simple calculation problems?

Simple calculation problems are mathematical equations or operations that involve basic arithmetic operations such as addition, subtraction, multiplication, and division. These problems typically require the use of only one or two numbers and do not involve complex formulas or concepts.

2. Why are simple calculation problems important?

Simple calculation problems are important because they help develop and improve basic mathematical skills such as number sense, mental math, and problem-solving. These skills are essential for everyday tasks such as budgeting, cooking, and shopping.

3. How can I improve my skills in solving simple calculation problems?

Practice is key to improving your skills in solving simple calculation problems. You can start by using flashcards, online quizzes, or math worksheets to practice different types of calculations. It is also helpful to understand the underlying concepts and strategies for solving these problems.

4. Are there any tips for solving simple calculation problems more efficiently?

Yes, there are a few tips that can help you solve simple calculation problems more efficiently. One tip is to break down the problem into smaller steps and solve each step separately. Another tip is to use mental math tricks such as rounding or using known facts to solve the problem quickly.

5. How can I apply simple calculation problems in real-life situations?

Simple calculation problems are used in various real-life situations, such as calculating tips at a restaurant, determining discounts while shopping, or calculating ingredient measurements while cooking. These problems are also essential for more complex calculations in fields such as science, engineering, and finance.

Similar threads

  • Introductory Physics Homework Help
Replies
15
Views
1K
  • Introductory Physics Homework Help
Replies
10
Views
1K
  • Introductory Physics Homework Help
Replies
5
Views
2K
  • Introductory Physics Homework Help
Replies
8
Views
3K
  • Introductory Physics Homework Help
Replies
6
Views
1K
  • Introductory Physics Homework Help
Replies
7
Views
3K
  • Introductory Physics Homework Help
Replies
5
Views
2K
Replies
2
Views
1K
  • Introductory Physics Homework Help
Replies
1
Views
2K
  • Introductory Physics Homework Help
Replies
11
Views
1K
Back
Top