A little Physics help from a newbie...

Printable View

June 21st 2007, 12:28 PM
John 5

A little Physics help from a newbie...

I am stuck on the following problems:

1.) One student, using unit analysis, says that the equation V=√2ax is dimensionally correct. Another says it isn't. With whom do you agree, and why?

2.) A student has a car that gets, on average, 25.0 mi/gal of gasoline. She plans to spend a year in Europe and take the car with her. a- What should she expect the car's average gas mileage to be in kilometers per liter? b- During the year there, she drove 6,000 km. Assuming that gas costs $5.00/gal in Europe, how much did she spend on fuel? (Approximate to the nearest dollar.)

3.) Approximately 118 miles wide and 307 miles long and averaging 279 feet in depth, Lake Michigan is the second-largest Great Lake by volume. Estimate its volume of water in m³.
June 21st 2007, 12:35 PM
topsquark

Quote:

Originally Posted by John 5

1.) One student, using unit analysis, says that the equation V=√2ax is dimensionally correct. Another says it isn't. With whom do you agree, and why?

I am assuming this equation is
$v = \sqrt{2ax}$

The unit for v is $m/s$ .
The unit for a is $m/s^2$ .
The unit for x is $m$ .
There is no unit for "2."

So the RHS reads:
$\sqrt{\frac{m}{s^2}m} = \sqrt{\frac{m^2}{s^2}} = \frac{m}{s}$
which is the same as the LHS.

So yes, the units on this equation are correct.

-Dan

PS I'll get to the other two later, unless someone else gets there first.
June 21st 2007, 12:50 PM
John 5

^^^^ Thanks for the quick reply!!! :)

Excuse my ignorance, but what do "RHS" and "LHS" stand for?
June 21st 2007, 01:06 PM
topsquark

Quote:

Originally Posted by John 5

^^^^ Thanks for the quick reply!!! :)

Excuse my ignorance, but what do "RHS" and "LHS" stand for?

Sorry! :o
LHS = Left Hand Side (of the equation)
RHS = Right Hand Side

-Dan
June 21st 2007, 11:27 PM
Jhevon

Quote:

Originally Posted by John 5

2.) A student has a car that gets, on average, 25.0 mi/gal of gasoline. She plans to spend a year in Europe and take the car with her. a- What should she expect the car's average gas mileage to be in kilometers per liter? b- During the year there, she drove 6,000 km. Assuming that gas costs $5.00/gal in Europe, how much did she spend on fuel? (Approximate to the nearest dollar.)

This is a unit conversion problem. First we need to know our conversion factors.

1 mile = 1.609344 km ------------> 1 km = 0.621371192 miles

1 gallon = 3.7854118 liters -------> 1 liter = 0.264172051 gallons

Now on to question (a), there are two main methods to go about this:

Method 1: Start with the ratio given and multiply by the ratios of the conversion factors in such a way that the units you don't want cancel out, and the units you want stay. Part (a) is essentially asking 25 mi/gal = ? km/L. So here goes:

$\frac {25 \mbox { miles}}{ \mbox {gallon}} \times \frac {1.609344 \mbox { km}}{1 \mbox { mile}} \times \frac {1 \mbox { gallon}}{3.7854118 \mbox { liters}} = \frac {10.62859264 \mbox { km}}{\mbox { liter }}$

So her average mileage in km/L is $\approx 10.63 \mbox { km/L}$

Notice that the gallons and liters units canceled out.

Method 2: Simply replace a unit you don't want with its corresponding conversion factor in a unit you do want.

25 miles / gal = 25 (1.609344 km) / (3.7854118 L) $\approx$ 10.63 km / L

(b)

She drove 6000 km, how many liters of gas did she use?

$6000 \mbox { km } \times \frac {1 \mbox { liter}}{10.62859264 \mbox { km}} = 564.5150024 \mbox { liters}$

The price is per gallon, so how many gallons is that?

$564.5150024 \mbox { liters} = 564.5150024 (0.264172051 \mbox { gallons}) = 149.129086 \mbox { gallons}$

Now it's $5 per gallon, so the amount of money she spent is:

5(149.129086) = $745.65
June 21st 2007, 11:32 PM
Jhevon

Quote:

Originally Posted by John 5

3.) Approximately 118 miles wide and 307 miles long and averaging 279 feet in depth, Lake Michigan is the second-largest Great Lake by volume. Estimate its volume of water in m³.

This is also a unit conversion question, so seeing that I did question 2 for you completely, I will leave this one to you. Here are the conversion factors you need:

1 mile = 1609.344 meters

1 foot = 0.3048 meters

Change everything to meters, then just use the formula for the volume (which is volume = length * width * height if you didn't know)
June 22nd 2007, 05:59 AM
John 5

MANY MANY thanks to you Ghevon and Topsquark. Thanks for giving me a sense of direction for the third problem.

Have a great weekend! :)