# I thought conversions were simple

• indepth
In summary, the conversation revolves around a problem involving finding the volume of a cloud with 50 water drops and a radius of 10 um. The participants discuss using the formula for a sphere to find the volume and then applying this to a cylinder with a height of 2.9 km and a radius of 1.4 km. They also discuss converting units and finding the total volume of water in the entire cloud. The conversation then shifts to a different problem involving finding the volume of a gold fiber and using the density equation to solve for the length of the fiber. The participants remind each other to pay attention to units and convert them when necessary.
indepth
I thought conversions were simple, but this one is giving me some trouble. If anyone could help out, it would be greatly appreciated. Thanks in advance.

http://img225.imageshack.us/img225/6229/pjhyscb0.png

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If there are 50 water drops of radius 10 um, what total volume is that?

im confused to find volume with only radius..:(

What's the volume of a sphere? (The radius is the only thing you need to know)

4.18879x10^-10 in meters

It's r^3, I think you may have done r^2. The units would be m^3 since it's volume, right?

$$V = \frac{4 \pi r^3}{3}$$

sorry, yeah ^3 i did squared. so.. 4.18879x10^-15

OK, so what's the volume of a cylinder of height 2.9 km and radius 1.4 km?

The way to approach this problem is to find out how many cubic centimeters there are in the whole cloud. Once you have that, you can figure out how much water there is in the entire cloud because you now know how much there is in one cubic centimeter.

indepth said:
sorry, yeah ^3 i did squared. so.. 4.18879x10^-15

Remember to write down your units. That's the whole point of this problem and you will make mistakes if you lose track of them.

So this should be in m^3.

Don't forget to multiply this by 50 to get the total volume of the 50 drops in the cubic centimeter of cloud.

in the cylinder, volume is 1.7856x10^10 m^3, right?

indepth said:
in the cylinder, volume is 1.7856x10^10 m^3, right?

Yes, that's right.

hage567 said:
Remember to write down your units. That's the whole point of this problem and you will make mistakes if you lose track of them.

So this should be in m^3.

Don't forget to multiply this by 50 to get the total volume of the 50 drops in the cubic centimeter of cloud.

after x by 50 the answer is 2.0949x10^-15. what do i do with the cylinder now?

So if there is 2.09x10^-15 m^3 of water for every cm^3 of cloud, how much water is there is 1 m^3 of cloud? (Just convert cm^3 to m^3)

Then, you can find out how much water is in the entire cloud by multiplying by the entire volume of the cloud you just found.

2.09x10^-12 per m^3. so now this times 1.7856x10^10 m^3?

indepth said:
2.09x10^-12 per m^3. so now this times 1.7856x10^10 m^3?

1 m^3 = 100^3 cm^3. You must cube the 100.

So it should look like $$\frac{2.09x10^-15 m^3 water}{1 cm^3 cloud} * \frac{100^3 cm^3}{1 m^3}$$

Does that make sense?

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yeah that looks good, i forgot about the cubing of the 100. and now same for the 500 correct? but also. that's not the final is it

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Yes. What answer did you get?

2.09x10^-9 is what i got for the 50. now i will do it for the 500. any idea for part b and or c?

indepth said:
2.09x10^-9 is what i got for the 50. now i will do it for the 500. any idea for part b and or c?

Don't forget to multiply by the entire volume of the cloud.

For (b) How many m^3 in a Liter?

For (c), what's density?

Give these a try and see what you can do.

3.7319x10^1 correct for 50?

B) .00 m^3
c) p= m/v

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indepth said:
2.09x10^-9 i thought that was the asnwer after multiplying the vol of cloud.
No, the 2.09x10^-9 m^3 is the volume of water per m^3 of cloud. To find the total volume of water you must multiply by the total volume of the cloud.

B) .00 m^3

c) p= m/v
Yes. So how might you use that to figure out the mass of water in the cloud?

indepth said:
3.7319x10^1 correct for 50?

OK, that looks better (watch your sig figs). Don't forget to state your units! It's a good habit to get into.

hage567 said:
OK, that looks better (watch your sig figs). Don't forget to state your units! It's a good habit to get into.

but is correct answer right? only 1 try on this homework problem :(.

.001 i meant to say sorry.

use mass of cloud divided by the volume we just found?

A) is correct thanks

B) .001 m^ = 1 L?

C) just use density formula?

indepth said:
B) .001 m^ = 1 L?

Yes. (m^3)

C) just use density formula?

Yes, you know the volume now and you were given the density.

B) so i have .001 m^3. divide the #'s from a by this #?

C) i tried 1000/(answer from a) and it is not correct :(

indepth said:
B) so i have .001 m^3. divide the #'s from a by this #?
Yes.

C) i tried 1000/(answer from a) and it is not correct :(
But mass = density * volume. You have to arrange the equation for mass.

If you check your units to see if they come out right you can catch these kinds of errors!

Thanks alot, i got all parts right. Yeah i get confused a lot with my units, I am sure that's what messes me up the most. If you still have time i have one last problem.

http://img225.imageshack.us/img225/1765/pjhys2po4.png

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Actually i figured it out. Thanks for your help, i am sure i will be in this forum board through the semester so i look forward to seeing you around. :) thanks

Start with the density equation again. Can you find the volume of 44.21 g of gold? How do you think you might find the area of the leaf using that information?

For (b), if you know the volume the gold fiber needs to be, what length will give you that?

Be careful with your units and remember to convert them when necessary.

## 1. How do you convert between units?

Converting between units involves multiplying the original value by a conversion factor. This conversion factor is derived from the relationship between the two units. For example, to convert from meters to feet, you would multiply by 3.28084, since 1 meter is equal to 3.28084 feet.

## 2. What is the difference between metric and imperial units?

Metric units are based on the International System of Units (SI) and are used in most countries around the world. Imperial units, also known as British units, are primarily used in the United States and are based on the old English system of measurement. The main difference between the two is the base unit of length - meters for metric and feet for imperial.

## 3. Why is it important to convert between units?

Converting between units is important because it allows for easier communication and comparison of measurements. It also allows for more accurate calculations and ensures that measurements are consistent and standardized.

## 4. Can you convert between any units?

In theory, yes, you can convert between any units as long as you have the correct conversion factor. However, some units may not have a direct conversion factor and may require multiple steps or conversions to get to the desired unit.

## 5. What are some common conversion mistakes to avoid?

Some common conversion mistakes to avoid include using the wrong conversion factor, forgetting to convert the unit symbol, and rounding too early in the conversion process. It is also important to pay attention to the direction of the conversion - for example, converting from meters to feet is different than converting from feet to meters.

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