mass problem dealing with areas and lengths

missrikku

Hello again,

The problem states that:

Gold has a mass of 19.32 g for each cubic centimeter of volume.

a) If 1.000 oz of gold, with a mass of 27.63 g, is pressed into a leaf of 1.000 micrometers thickness, what is the area of the leaf?

b) If the gold is drawn out into a cylindrical fiber of radius 2.500 micrometers, what is the length of the fiber?

Well,

From the first part of the problem I got this ratio:

1 cubic centimeter of gold : 19.32 g of gold

For part A:

I did a proportion:
1 cubic centimeter / 19.32 g = x / 27.63 g
x = 1.43 cubic centimeters = 1.43 x 10^-2 cubic meters

but that is where i got stuck. i think that the x that i found is the volume for the 1 oz of gold, but how do i get the length of the leaf so I can get the area? also, if the problem says that the leaf has a certain thickness, then is the area they are looking for a surface area?

For part B:

I first changed the 2.500 micrometers into meters and got: 2.500 x 10^-6 meters

the volume of a cylinder is: V = pi * r^2 * h

then from part A, V = 0.143 cubic meters

0.143 = pi * ( 2.500 x 10^6 )^2 * h

solving for h, i got: h = 7.283 x 10^8 meters

I think I can only get the right answer if I have the volume from part A correct, so I am unsure if my answer is correct.

Tom Mattson

Originally posted by missrikku
From the first part of the problem I got this ratio:

1 cubic centimeter of gold : 19.32 g of gold

For part A:

I did a proportion:
1 cubic centimeter / 19.32 g = x / 27.63 g
x = 1.43 cubic centimeters = 1.43 x 10^-2 cubic meters

OK, good start.

but that is where i got stuck. i think that the x that i found is the volume for the 1 oz of gold, but how do i get the length of the leaf so I can get the area?

You don't need the length. If the leaf is of uniform thickness (T), then the volume (V) and area (A) are related by:

V=AT

You can solve for the area directly.

also, if the problem says that the leaf has a certain thickness, then is the area they are looking for a surface area?

Yes.

For part B:

I first changed the 2.500 micrometers into meters and got: 2.500 x 10^-6 meters

the volume of a cylinder is: V = pi * r^2 * h

then from part A, V = 0.143 cubic meters

0.143 = pi * ( 2.500 x 10^6 )^2 * h

solving for h, i got: h = 7.283 x 10^8 meters

I think I can only get the right answer if I have the volume from part A correct, so I am unsure if my answer is correct.

I do not have a calculator handy, so I can't check your number, but I can say that your method is correct. Assuming you did the calculator work correctly, this is A-OK.

missrikku

about converting the volume of 1.430 cm^3 into m^3, am i correct in doing the following:

(1.430 cm^3)(1m/10^2cm)^3 = 1.430 x10^-6 m^3

is that correct? or should i have just done this:

(1.430 cm^3)(1m/10^2cm) = 1.430 x10^-2 m^3

HallsofIvy

Your first calculation is correct. There are 100 cm in a m and so 100^3= 1000000 cubic cm in a cubic m. Although I wonder why yo are converting to cubic meters. Wouldn't it be easier to convert everything to cm?

Similar Threads for: mass problem dealing with areas and lengths
Thread	Forum	Replies
Calculus: volumes, areas, lengths and Work	Calculus & Beyond Homework	0
Dealing with mass, acceleration and velocity	Introductory Physics Homework	2
Mathmatics-problem on areas	General Math	2
Problem dealing with mAH, mA, and volts...	Introductory Physics Homework	6
Summation of rectangular areas (calculus) problem.	Introductory Physics Homework	4

missrikku	about converting the volume of 1.430 cm^3 into m^3, am i correct in doing the following: (1.430 cm^3)(1m/10^2cm)^3 = 1.430 x10^-6 m^3 is that correct? or should i have just done this: (1.430 cm^3)(1m/10^2cm) = 1.430 x10^-2 m^3

HallsofIvy Recognitions: Gold Member Science Advisor Staff Emeritus	mass problem dealing with areas and lengths Your first calculation is correct. There are 100 cm in a m and so 100^3= 1000000 cubic cm in a cubic m. Although I wonder why yo are converting to cubic meters. Wouldn't it be easier to convert everything to cm?