Conversions

[Violagirl]

I'm having trouble on these two problems:

1) A runner wants to run 10.0 km. She knows that her running pace iss 7.5 miles per hour. How many minutes must she run?

Here is what I did:

10 km/1=.6214 mi/1 km. The km cancel out and I get 6.214 mi. Then I took 6.214 mi/1=1 hr/7.5 mi, the mi cancel out and I got .83. My next step was .83 hour/1=60 min/1 hour. The hours cancel out and I got 49.8 min as an answer. My book had an answer of 5.0 X 10^1 min and I'm not sure how they got that.


2) The density of titanium is 4.51 g/cm^3. What is the volume (in cubic inches) of 3.5 lb of titanium?

On this one, I know that Density=Mass/Volume. But here, I'm not sure where to start.

 

[mmm4444bot]

1) A runner wants to run 10.0 km. She knows that her running pace iss 7.5 miles per hour. How many minutes must she run?

... I got 49.8 min as an answer. My book had an answer of 5.0 X 10^1 min and I'm not sure how they got that.


2) The density of titanium is 4.51 g/cm^3. What is the volume (in cubic inches) of 3.5 lb of titanium?

... I know that Density=Mass/Volume ... I'm not sure where to start.

Hi Viola Girl:

On exercise 1, the book's author rounded the answer to the nearest minute. (Your answer is 12 seconds shorter than the book's answer.)

On exercise 2, start by substituting the given values for density and mass into the formula you posted. Then solve for volume.

Cheers,

~ Mark

 

[Deleted member 4993]

I'm having trouble on these two problems:

2) The density of titanium is 4.51 g/cm^3. What is the volume (in cubic inches) of 3.5 lb of titanium?

On this one, I know that Density=Mass/Volume. But here, I'm not sure where to start.

So

\(\displaystyle Volume = \frac{Mass}{Density}\)

first convert density to lb/cu.in units.

1 gm = ??? lbs

1 cm = ??? in

1 cm^3 = ??? cu.in

As an alternate approach, you could convert the given mass to gms and find the volume in cm^3 - then convert the volume to cu.in.

 

[Loren]

On number one you did it in separate steps, which is quite good. Nice going. As you move on to bigger stuff you might want to consider what some people call "dimensional analysis". Your problem would be worked like this, all in one statement, then let the computer do the arithmetic.

\(\displaystyle \frac{10km}{1}\times \frac{1 hr}{7.5 mi}\times \frac{60 min}{1 hr}\times \frac{0.6214 mi}{1 km}=49.712 min\)

It is easy to "cancel" the units. That is the "km" in the first fraction cancels with the "km" in the denominator of the 4th fraction. The "hr" in the second and third fractions cancel, etc., leaving only the desired units, namely, minutes. In fact this cancelation helps in setting up the fractions correctly because if they don't cancel, something is wrong somewhere.
The answer is more accurate in that no rounding was done until the final answer was obtained. The rounding off to the nearest unit in this problem negates the accuracy issue, but sometimes accuracy is important.