How Many Trucks Can US Oil Production Fuel in a Year?

AI Thread Summary
The discussion revolves around calculating how many trucks can be fueled by US oil production, given that one truck consumes 6,000 barrels of oil annually. By dividing the total US oil production of 1.9 x 10^10 barrels by the annual consumption per truck, the result is 3,000,000 trucks. Participants clarify that while the initial calculation yields trucks per year, it can also be interpreted as the total number of trucks that can be fueled for a year. The conversation emphasizes understanding unit conversions and interpretations in the context of the problem. Ultimately, the calculation demonstrates that the US produces enough oil to fuel 3 million trucks annually.
cytochrome
Messages
163
Reaction score
3

Homework Statement


If one truck runs on 6000 barrels of oil per year, how many trucks could run on the oil produced in the US in one year (1.9*10^10 barrels)?

Homework Equations


(6000 barrels/year) used by one truck
(1.9*10^10 barrels/year) for US production

The Attempt at a Solution


I simply divided 1.9*10^10 by 6000 to get 3,000,000.

I'm confused about the fact that 3,000,000 is not in units of "trucks". How can I set up this problem properly to include trucks as my units?
 
Physics news on Phys.org
Hello cytochrome,

\left(\frac{1.9 \times 10^{10}~\textrm{barrels}}{1~\textrm{year}}\right) \cdot \left(\frac{1~\textrm{truck}}{6000~\textrm{barrels}}\right)

The unit of barrels cancels, leaving trucks and years, so your result is in trucks/year ("trucks per year").

EDIT: ACTUALLY, you can consider the quantity of 6000 to have units of "barrels per truck, per year" which would be barrels/(trucks*years). So, in that case, there would be 1 year in the numerator of the righthand factor in parentheses, and the year unit would cancel as well, leaving only trucks.

However, the way I did it is fine too. In this case, the way to think about it is that a "truck" is a unit of volume equal to the amount of oil needed to run one truck for a year. So, what your unit conversion is doing is saying that the US produces 3,000,000 "trucks" worth of oil per year. It's just a matter of interpretation.
 
Last edited:
Another version:

6000 bbl per year per truck

1.9 x 1010 bbl per year
 
Chestermiller said:
Another version:

6000 bbl per year per truck

1.9 x 1010 bbl per year

Yup. I had edited my post to include this version (just not LaTeXed).
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top