bonded end-to-end to make a single bar. The densities are 4.3 g/cm^3 and 6.3 g/cm^3. What length of the lighter-density part of the bar is needed if the total mass is 13103 g? Answer in units of cm.
I'm not even sure where to begin(but I have filled four pages...and about 5 hours trying).....so far I have figured the volume as 2463.00864 (I THINK that's correct). It seems to me that you would have to figure a percent or a portion of each of the densities and possibly set it up in an equation that is equal to 13103 g, but I can't figure out how to get what percent or portion of each individual density to figure this, or if that is even the correct way to figure it.
THANK YOU for any and all help!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
2006-09-01 08:56:43 · 8 answers · asked by exxavier 2 in Science & Mathematics ➔ Physics
you are on the right track. the rest of the work is basic algebra.
let x = leingth of metal A
lte y = leingth of metal B
then use the equations
x+y=16 cm
4.3/(pi*X*7^2)+6.3/(pi*Y*7^2)=2463.00864 (the total mass)
then you just solve for x and y the leingths of each individual part
2006-09-01 09:07:43 · answer #1 · answered by bretttwarwick 3 · 0⤊ 0⤋
Set the length of the lighter material to x and the other to 16-x
length*area*density+length*area*density=weight
x*pi*7^2*4.3+(16-x)*pi*7^2*6.3=13103
x=7.8cm
2006-09-01 16:10:10 · answer #2 · answered by Anonymous · 0⤊ 0⤋
Correct on the volume.
The cylinder has 16*7^2*Pi cubic cm or 2463.0 cm^3.
If it were made all of the lighter metal, it would weigh 10590 g.
If it were made all of the heavier metal, it would weigh 15517g.
So L=10590 and H=15517.
pL + (1-p)H = 13103
10590p + 15517(1-p) = 13103
10590p +15517 - 15517p = 13103
2414 = 4927p
p = .49
16cm * .49 = 7.8cm
I rounded the answers; you can make this as exact as you please.
2006-09-01 16:14:00 · answer #3 · answered by Steve A 7 · 1⤊ 0⤋
This is more like a math problem (all units will be (g) and (cm)):
X1: Length of lighter density material
X2: Length of heavier density material
X1 + X2 = 16 cm (a)
Total mass of cylinder (with A: base area)
A*X1*4.3 + A*X2*6.3 = 13103 (b)
A = PI*7*7 = 49* PI
Equation (b) gives: 210.7 * PI * X1 + 308.7 * PI * X2 = 13103
Replace (a) into (b):
210.7 (16-X2) + 308.7 X2 = 13103/PI
X2 = [(13103/PI)-(210.7*16)]/(308.7-210.7) ~8.16cm
X1 ~ 7.84 cm
2006-09-01 18:32:04 · answer #4 · answered by Shivers 2 · 0⤊ 0⤋
Calculate the volume of each 'part' of the cylinder as Ïr²l and Ïr²(16 - l). Then the total mass is
4.3*(Ïr²l)+6.3*(Ïr²(16-l)) = 13103
Solve for l and you're home free âº
Doug
2006-09-01 16:11:04 · answer #5 · answered by doug_donaghue 7 · 0⤊ 0⤋
5.72666 CM
2006-09-01 16:37:30 · answer #6 · answered by Rolf H 2 · 0⤊ 0⤋
center of mass is key
2006-09-01 16:05:34 · answer #7 · answered by oracle 5 · 0⤊ 1⤋
2 points ^_^
2006-09-01 16:01:57 · answer #8 · answered by Anonymous · 0⤊ 2⤋