Unit analysis help (possible significant figure error?)

[mbisCool]

Okay I have been doing unit analysis homework and it is generally easy however a few questions I use the same logic but cannot come up with the correct answer. Keep in mind the homework is turned in electronically so the answer must be exact but without the aid of expressions or ratios etc to express them. simply decimal notation not even scientific notation or i get a syntax error. My errors might simply be in rounding; i believe the answer must be to correct significant figures

(1) A "gry" is an old English measure for length, defined as 1/10 of a line, where "line" is another old English measure for length, defined as 1/12 inch. A common measure for length in the publishing business is a "point", defined as 1/72 inch. What is an area of 0.75 gry2 in terms of points squared (points2)?

I do (0.75 gry) (1line/10gry) (6pts/line)= 0.45 points (left out squares for simplicity) But this answer is wrong. Even working backwords from .45points i get .75 gyr

(2)Hydraulic engineers often use, as a unit of volume of water, the "acre-foot", defined as the volume of water that will cover 1 acre of land to a depth of 1 ft. A severe thunderstorm dumps 1.7 in. of rain in 30 min on a town of area 36 km2. What volume of water, in acre-feet, fell on the town?|

I calculated the volume of water to be 16732.28346ft^3. After i divided that number by 43560ft^3 and my answer was 0.3841203733 acre-ft

(3)Gold, which has a mass of 19.32 g for each cubic centimeter of volume, is the most ductile metal and can be pressed into a thin leaf or drawn out into a long fiber.
(a) If a sample of gold, with a mass of 77.36 g, is pressed into a leaf of 2.800 µm thickness, what is the area of the leaf?b) If, instead, the gold is drawn out into a cylindrical fiber of radius 2.000 µm, what is the length of the fiber?

The volume in cm^3 would be (77.36/19.32)/100= V I calculated the 2.8 micron thickness to be 0.00028m. After that i divded the volume by the thickness to get the area(pretty sure they don't want surface area) to get A=143.005m^2

For B, using the same volume i divided the area (pi(2x10^-6)^2) to get the length of 3186393995m

A cubic centimeter in a typical cumulus cloud contains 50 to 500 water drops, which have a typical radius of 10 µm. For that range, give the lower value and the higher value, respectively, for the following.
(a) How many cubic meters of water are in a cylindrical cumulus cloud of height 2.0 km and radius 1.0 km?
to m3
(b) How many 1 liter pop bottles would that water fill?
to bottles
(c) Water has a mass per unit volume (or density) of 1000 kg/m3. How much mass does the water in the cloud have?


I had to assume they wanted me to use the radius of the water drop to caluclate the volume of one drop as a sphere then multiply it by the number of drops (50 and/or 500) and then by the number of cubic centimeters in the cloud. Again my answer was wrong

Any insight into what i am doing wrong on these would be greatly appreciated.

 

[tiny-tim]

(1) A "gry" is an old English measure for length, defined as 1/10 of a line, where "line" is another old English measure for length, defined as 1/12 inch. A common measure for length in the publishing business is a "point", defined as 1/72 inch. What is an area of 0.75 gry2 in terms of points squared (points2)?

I do (0.75 gry) (1line/10gry) (6pts/line)= 0.45 points (left out squares for simplicity) But this answer is wrong. Even working backwords from .45points i get .75 gyr

Hi mbisCool!

Nooo … you left out squares for the wrong answer!

this is an area, in gry² … 0.45 is the right answer for length, in gry.

 

[baba89]

It is possible that your errors are due to rounding and significant figures. In unit analysis, it is important to keep track of units and to use the correct conversion factors. However, it is also important to use the correct number of significant figures in your calculations and final answer.

In the first problem, the calculation for the area of 0.75 gry2 in terms of points squared should be (0.75 gry) (1 line/10 gry) (6 pts/line)^2 = 0.2025 points^2. The answer should then be rounded to the appropriate number of significant figures.

In the second problem, the volume of water should be rounded to 0.38 acre-ft, as the original given values have only two significant figures.

In the third problem, for part a, the area should be calculated as (77.36 g/19.32 g/cm^3) / (0.00028 cm) = 110,000 cm^2. When converting to meters, the answer should be 1.1 m^2. For part b, the length should be calculated as (77.36 g/19.32 g/cm^3) / (3.14 x (2 µm)^2) = 1.3 x 10^8 cm. When converting to meters, the answer should be 1.3 x 10^-2 m.

In the last problem, you are correct in using the radius of the water drop to calculate the volume of one drop. However, when multiplying by the number of drops and the number of cubic centimeters in the cloud, make sure to use the appropriate number of significant figures. The final answer for part a should be 3.1 x 10^10 m^3. For part b, the number of 1 liter pop bottles would be 3.1 x 10^13 bottles. And for part c, the mass of the water in the cloud would be 3.1 x 10^13 kg.