# Thread: Finding Feet Given Density

1. ## Finding Feet Given Density

I am stuck!

qbkr21

2. Originally Posted by qbkr21
I am stuck!

qbkr21

I thought the radial value was 8.35 mm = .835 cm

So $\displaystyle V=\pi r^2h\implies 7.14\times10^3~cm^3=\pi(.835~cm)^2h$

Solving for h, we see that $\displaystyle h=\frac{7.14\times10^3~cm^3}{\pi(.835~cm)^2}\appro x\color{red}\boxed{3.26\times10^3 ~cm}$

Does this make sense?

--Chris

EDIT: I see now that .835 cm was the diameter...my mistake...

The equation should be $\displaystyle V=\pi r^2h\implies 7.14\times10^3~cm^3=\pi(.4175~cm)^2h$

Solving for h, we see that $\displaystyle h=\frac{7.14\times10^3~cm^3}{\pi(.4175~cm)^2}\appr ox\color{red}\boxed{1.3\times10^4 ~cm}$

3. Originally Posted by Chris L T521
I thought the radial value was 8.35 mm = .835 cm

So $\displaystyle V=\pi r^2h\implies 7.14\times10^3~cm^3=\pi(.835~cm)^2h$

Solving for h, we see that $\displaystyle h=\frac{7.14\times10^3~cm^3}{\pi(.835~cm)^2}\appro x\color{red}\boxed{1.02\times10^4 ~cm}$

Does this make sense?

--Chris

EDIT: I see now that .835 cm was the diameter...my mistake...

The equation should be $\displaystyle V=\pi r^2h\implies 7.14\times10^3~cm^3=\pi(.4175~cm)^2h$

Solving for h, we see that $\displaystyle h=\frac{7.14\times10^3~cm^3}{\pi(.4175~cm)^2}\appr ox\color{red}\boxed{3.26\times10^3 ~cm}$
Thanks for the response, but how do you figure the radial value = 8.35 when the problem states that it is the diameter. I divided by 2 (a half) to get the radius. From there I would still be stuck in millimeters. I was thinking about then going over to volume and converting it into millimeters cubed, then dividing it by pi and (4.175 mm)^2. I'd then be left will millimeters of which I could covert back into feet. Does this work or am I making it way to complicated?

4. Originally Posted by qbkr21
Thanks for the response, but how do you figure the radial value = 8.35 when the problem states that it is the diameter. I divided by 2 (a half) to get the radius. From there I would still be stuck in millimeters. I was thinking about then going over to volume and converting it into millimeters cubed, then dividing it by pi and (4.175 mm)^2. I'd then be left will millimeters of which I could covert back into feet. Does this work or am I making it way to complicated?
I noticed those mistakes once I posted...

I edited them as you can probably tell now. Sorry about that.

5. Originally Posted by qbkr21
Thanks for the response, but how do you figure the radial value = 8.35 when the problem states that it is the diameter. I divided by 2 (a half) to get the radius. From there I would still be stuck in millimeters. I was thinking about then going over to volume and converting it into millimeters cubed, then dividing it by pi and (4.175 mm)^2. I'd then be left will millimeters of which I could covert back into feet. Does this work or am I making it way to complicated?
Yes, this would work. I don't see how your way would be complicated. However, do you know the conversion from mm to ft?

--Chris

6. ## RE:

Originally Posted by Chris L T521
Yes, this would work. I don't see how your way would be complicated. However, do you know the conversion from mm to ft?

--Chris
Is this correct?

7. Originally Posted by qbkr21
Is this correct?

I get $\displaystyle \frac{7,140,000~mm^3}{\pi(4.175~mm)^2}\approx1.3\t imes10^5~mm$

Converting to feet, I got ~428 ft.

--Chris

8. Originally Posted by Chris L T521
I get $\displaystyle \frac{7,140,000~mm^3}{\pi(4.175~mm)^2}\approx1.3\t imes10^5~mm$

Converting to feet, I got ~428 ft.

--Chris
Woops I forgot to square the radius

How did you only get two sig figs for the 1.3* 10^5?

9. Originally Posted by qbkr21
Woops I forgot to square the radius

How did you only get two sig figs for the 1.3* 10^5?
Ahaha...I forgot about sig figs [and the rules]...

but I saw that $\displaystyle 130387.326\approx 1.3\times10^5~mm$

I didn't think we could have 3 sig figs...

--Chris

10. Originally Posted by Chris L T521
Ahaha...I forgot about sig figs [and the rules]...

but I saw that $\displaystyle 130387.326\approx 1.3\times10^5~mm$

I didn't think we could have 3 sig figs...

--Chris
Hey Chris just one other thing. In using Sig Figs do conversions count as Sig Figs such as multiplication, and also when I divide the diameter by 2 does the 2 count as a Sig Fig?

Thanks,

qbkr21

11. Originally Posted by qbkr21
Hey Chris just one other thing. In using Sig Figs do conversions count as Sig Figs such as multiplication, and also when I divide the diameter by 2 does the 2 count as a Sig Fig?

Thanks,

qbkr21
erm...I not sure at the moment