Poll: Another Math Question

[Vim-Hogar]

I have a better Mathematics question:

If the volume of Earth is 1,083,210,000,000 km cubed and the Mass of Earth is 5.9742 × 10 to the power of 24 kilograms then what is the average mass of one cubed meter of Earth? And no I'm not converting anything into the archaic imperial system for you.

Show your working.

I put [(5.9742 × 10^24 kg) / 1,083,210,000,000 km cubed] into Google. The answer is 5 515.27405 kg / m^3. That's 46.0271926 pounds per US gallon, which seems reasonable (within an order of magnitude) since water is about 8 lb/gallon.

Also, when I did the first calculation, Google returned this page as the first result (under the calculator thing) as soon as I'd entered [1,083,210,000,000 km cubed], and this page was the only result (other than the answer) for the full query. Amazing.

EDIT: I should mention that I actually do know how to do the conversion myself (yes, km^3 -> m^3 is easy to get wrong), I'm just lazy and/or I think I'm clever.

Also, I just noticed that your × isn't a normal x, good job.

 

[Saelune]

One, Im a girl, and two you're being a jerk. An obnoxious one at that.

Thinking about it, I guess it would, since if I put it into a calculator, I would likely have done it in order. But in anything with more stuff, Id follow PEMDAS, and I dont think I should be put down for using what I was TAUGHT.

Well, sure you can't be blamed for doing what you were taught, but that doesn't mean you should parade around the fact that you're wrong as if the teacher's words invalidate hard fact. This isn't a court trial. I don't care about supporting my team, playing by the rules, or "winning" anything. I care about being right, whether or not I have to correct myself. Maybe that's a little beyond you.

Dont presume to know me whiel you sit on your high horse of snobbery. If you want to prove you are right, then prove you are right. So far you have just belittled me with personal attacks.

 

[gim73]

Instead of doing a poorly written problem you should do something like

d(cos(x^2))/dx
vs
d(cos(x)^2)/dx

 

[TiefBlau]

I have a better Mathematics question:

If the volume of Earth is 1,083,210,000,000 km cubed and the Mass of Earth is 5.9742 × 10 to the power of 24 kilograms then what is the average mass of one cubed meter of Earth? And no I'm not converting anything into the archaic imperial system for you.

Show your working.

Well, the formula for finding density is...

Density = Mass / Volume

Plugging all those nice little numbers into the formula,

Density[sub]Earth[/sub] = Mass[sub]Earth[/sub] / Volume[sub]Earth[/sub]

Density[sub]Earth[/sub] = 5.9742 × 10[sup]24[/sup] kg / 1,083,210,000,000 km[sup]3[/sup]

Density[sub]Earth[/sub] = 5.51527405 × 10[sup]12[/sup] kg/km[sup]3[/sup]

Now we convert it into meters cubed.

1 km[sup]3[/sup] = 10[sup]9[/sup] m[sup]3[/sup]

Using conversion factors,

Density[sub]Earth[/sub] = 5.51527405 × 10[sup]12[/sup] kg/km[sup]3[/sup] x (1 km / 10[sup]9[/sup] m[sup]3[/sup])

Density[sub]Earth[/sub] = 5.51527405 × 10[sup]21[/sup] kg/m[sup]3[/sup]

One cubic meter of Earth holds 5,515,274,050,000,000,000,000 kg

 

[TiefBlau]

Dont presume to know me whiel you sit on your high horse of snobbery. If you want to prove you are right, then prove you are right. So far you have just belittled me with personal attacks.

Hahaha, "personal attacks".

I don't have to prove anything. Multiplication is interchangeable with division, and addition is interchangeable with subtraction. It's up to you whether or not to believe me, but I must admit, if you don't, you're just gonna keep being wrong.

 

[Jewrean]

I have a better Mathematics question:

If the volume of Earth is 1,083,210,000,000 km cubed and the Mass of Earth is 5.9742 × 10 to the power of 24 kilograms then what is the average mass of one cubed meter of Earth? And no I'm not converting anything into the archaic imperial system for you.

Show your working.

I put [(5.9742 × 10^24 kg) / 1,083,210,000,000 km cubed] into Google. The answer is 5 515.27405 kg / m^3. That's 46.0271926 pounds per US gallon, which seems reasonable (within an order of magnitude) since water is about 8 lb/gallon.

Also, when I did the first calculation, Google returned this page as the first result (under the calculator thing) as soon as I'd entered [1,083,210,000,000 km cubed], and this page was the only result (other than the answer) for the full query. Amazing.

Well done. Also keep in mind that Google Calculator has automatically converted the volume from KM to M for you so instead of the 10^12 it becomes 10^21. You have done well young Padawan learner. Your answer shows that on average every Cubed Metre of Earth is 5 and a half Tonnes.

NEXT QUESTION!
Assuming the population of Earth is now 7,000,000,000 to answer the following. If every person on Earth was standing in a massive army in the shape of a square and each person on average was allowed to occupy 1 metre squared, what are the dimensions of this square army?

FYI: The calculation is simpler than you think and I will tell you that the size of this square is comparable in size to the nearest city to me (Melbourne).
Area of Melbourne = 8806 km² according to Wikipedia and various other sources.

 

[Jewrean]

I have a better Mathematics question:

If the volume of Earth is 1,083,210,000,000 km cubed and the Mass of Earth is 5.9742 × 10 to the power of 24 kilograms then what is the average mass of one cubed meter of Earth? And no I'm not converting anything into the archaic imperial system for you.

Show your working.

Well, the formula for finding density is...

Density = Mass / Volume

Plugging all those nice little numbers into the formula,

Density[sub]Earth[/sub] = Mass[sub]Earth[/sub] / Volume[sub]Earth[/sub]

Density[sub]Earth[/sub] = 5.9742 × 10[sup]24[/sup] kg / 1,083,210,000,000 km[sup]3[/sup]

Density[sub]Earth[/sub] = 5.51527405 × 10[sup]12[/sup] kg/km[sup]3[/sup]

Now we convert it into meters cubed.

1 km[sup]3[/sup] = 10[sup]9[/sup] m[sup]3[/sup]

Using conversion factors,

Density[sub]Earth[/sub] = 5.51527405 × 10[sup]12[/sup] kg/km[sup]3[/sup] x (1 km / 10[sup]9[/sup] m[sup]3[/sup])

Density[sub]Earth[/sub] = 5.51527405 × 10[sup]21[/sup] kg/m[sup]3[/sup]

One cubic meter of Earth holds 5,515,274,050,000,000,000,000 kg

Good try. The problem you have made is that we needed to convert the volume of Earth from KM^3 to M^3 by multiplying 1.083x10^12 by 1000^3. It then becomes 1.083x10^21. So to solve the problem simply take the mass of Earth which is 5.9742x10^24 and divide it by 1.083x10^21. The answer then becomes approximately 5,515 kg rather than the massive number you have listed above. 5,515 kg is the same as saying 5 and a half Tonnes.

Try the next question I have listed in my previous post too.

 

[Vim-Hogar]

1 km[sup]3[/sup] = 10[sup]9[/sup] m[sup]3[/sup]

Good job! Not only are you not lazy like me, you got the tricky bit right!

Using conversion factors,
Density[sub]Earth[/sub] = 5.51527405 × 10[sup]12[/sup] kg/km[sup]3[/sup] x (1 km / 10[sup]9[/sup] m[sup]3[/sup])

Good, good.

Density[sub]Earth[/sub] = 5.51527405 × 10[sup]21[/sup] kg/m[sup]3[/sup]

And then it all goes wrong at the last step -- you went the wrong direction! D:

One cubic meter of Earth holds 5,515,274,050,000,000,000,000 kg

Look at that, does that even make sense? I mean, seriously.

(I secretly hope you were trying to be sarcastic, because that would be super-duper-win. In that case, though, you should know that sarcasm is even more impossible to convey in equations than it is in normal writing.)

Also,

 

[TiefBlau]

I have a better Mathematics question:

If the volume of Earth is 1,083,210,000,000 km cubed and the Mass of Earth is 5.9742 × 10 to the power of 24 kilograms then what is the average mass of one cubed meter of Earth? And no I'm not converting anything into the archaic imperial system for you.

Show your working.

Well, the formula for finding density is...

Density = Mass / Volume

Plugging all those nice little numbers into the formula,

Density[sub]Earth[/sub] = Mass[sub]Earth[/sub] / Volume[sub]Earth[/sub]

Density[sub]Earth[/sub] = 5.9742 × 10[sup]24[/sup] kg / 1,083,210,000,000 km[sup]3[/sup]

Density[sub]Earth[/sub] = 5.51527405 × 10[sup]12[/sup] kg/km[sup]3[/sup]

Now we convert it into meters cubed.

1 km[sup]3[/sup] = 10[sup]9[/sup] m[sup]3[/sup]

Using conversion factors,

Density[sub]Earth[/sub] = 5.51527405 × 10[sup]12[/sup] kg/km[sup]3[/sup] x (1 km / 10[sup]9[/sup] m[sup]3[/sup])

Density[sub]Earth[/sub] = 5.51527405 × 10[sup]21[/sup] kg/m[sup]3[/sup]

One cubic meter of Earth holds 5,515,274,050,000,000,000,000 kg

Good try. The problem you have made is that we needed to convert the volume of Earth from KM^3 to M^3 by multiplying 1.083x10^12 by 1000^3. It then becomes 1.083x10^21. So to solve the problem simply take the mass of Earth which is 5.9742x10^24 and divide it by 1.083x10^21. The answer then becomes approximately 5,515 kg rather than the massive number you have listed above. 5,515 kg is the same as saying 5 and a half Tonnes.

Try the next question I have listed in my previous post too.

1 km[sup]3[/sup] = 10[sup]9[/sup] m[sup]3[/sup]

Good job! Not only are you not lazy like me, you got the tricky bit right!

Using conversion factors,
Density[sub]Earth[/sub] = 5.51527405 × 10[sup]12[/sup] kg/km[sup]3[/sup] x (1 km / 10[sup]9[/sup] m[sup]3[/sup])

Good, good.

Density[sub]Earth[/sub] = 5.51527405 × 10[sup]21[/sup] kg/m[sup]3[/sup]

And then it all goes wrong at the last step -- you went the wrong direction! D:

One cubic meter of Earth holds 5,515,274,050,000,000,000,000 kg

Look at that, does that even make sense? I mean, seriously.

(I secretly hope you were trying to be sarcastic, because that would be super-duper-win. In that case, though, you should know that sarcasm is even more impossible to convey in equations than it is in normal writing.)

Also,

Lol, whoops, I multiplied instead of divided xD

Writing down equations on a single line does have its disadvantages...

 

[Vim-Hogar]

NEXT QUESTION!
Assuming the population of Earth is now 7,000,000,000 to answer the following. If every person on Earth was standing in a massive army in the shape of a square and each person on average was allowed to occupy 1 metre squared, what are the dimensions of this square army?

FYI: The calculation is simpler than you think and I will tell you that the size of this square is comparable in size to the nearest city to me (Melbourne).

7 billion people, 1 m^2 each, that's 7e9 m^2. Taking "square root" rather literally, that gives us a square about 84 km wide/tall.

 

[Jewrean]

NEXT QUESTION!
Assuming the population of Earth is now 7,000,000,000 to answer the following. If every person on Earth was standing in a massive army in the shape of a square and each person on average was allowed to occupy 1 metre squared, what are the dimensions of this square army?

FYI: The calculation is simpler than you think and I will tell you that the size of this square is comparable in size to the nearest city to me (Melbourne).

7 billion people, 1 m^2 each, that's 7e9 m^2. Taking "square root" rather literally, that gives us a square about 84 km wide/tall.

Very good! See? My questions are more interesting yet simple to solve!

Want another one?

 

[Jewrean]

For a moment let's forget about the laws of gravity and mass for a fun problem. If every human on Earth decided to stand on top of one an-others shoulders and we made a gigantic tower that reached into space how many times would this tower go past the Moon (if at all)?

Information required:
- Earth Population = 7,000,000,000
- Average distance the Moon is from Earth = 384,403 kilometers
- We will say average human height is 160cm.
- We will also assume that the average height from the tip of your head to your shoulders is 25cm.

 

[Wuffykins]

Your calculator is high then.
[pic]

And you're adding clarity to an equation that has none. Let's do this phoenetically then. The equation above can be spoken out two ways:

1) Six Halves by bracket One plus Two [or you could say "Six over Two"]
2) Six divided by Two bracket One plus Two [in the style of a(x+b)]

As written out it can be taken as either a two or three term expression, where even as a two term expression it's not quite clear what the two terms are (6/2 & (1+2) or 6 & 2(1+2)). Also not helping is that the "/" can be taken either as a divide sign or as part of a rational number because not many people bother to figure out how to put the ÷ sign. Then there's the fact that 2(1+2) is written in the style of a polynomial in it's factored form (hell, it could be the first step in 2(x+2) if x=1) where multiplying by juxtaposition can* qualify as 'solving the brackets' and is therefore ahead of multiplication/division.

Yes, it's unclear and badly written in that form.

That said, as written on your calculator you did clarify it by separating it into terms of 6/2 & (1+2), just as Google does by adding brackets around 6/2 as well as a * separator, and yes, 6 divided by 2 times 3 does equal nine. But then again it's much like the well placed comma, where you can change the expression into something different (or more clear, as is this case).

Note: Oh, yes that was an asterisk next to can, and that's because it would seem that you can get a different ruling on that depending on what you read, as there seems to be no one clear cut answer to it. *shrug* What can you do? Oh, yes, I didn't forget that you wanted a pic of my stoner calculator. Unfortunately it doesn't display the input along with the end product, so I had to borrow someone else's if you don't mind ...

Spoiler: Here's what 2 calculators say:

Yup, one saying it's '1', and a second (newly installed) calc confirming that the 6/2*(1+2) is in fact 9, but also asking me what the hell is he supposed to do with the OP's equation. Take that as you will.

 

[Ladette]

It's 9 I believe, also fuck math.

Fuck college algebra too. Worthless information as far as my major was concerned, such a waste of my time. Yeah, i'm bitter about it 2 years later.

Then why did you choose the subject for the degree? Or are you talking about High School and therefore still just whinging like every other high-schooler?

Maybe it's different elsewhere, but where I went to school (UNLV) you need to take up to Math 120 to meet the University requirments for graduation, regardless of major. As a History major is doesn't help me in the slightest though.

 

[Jewrean]

It's 9 I believe, also fuck math.

Fuck college algebra too. Worthless information as far as my major was concerned, such a waste of my time. Yeah, i'm bitter about it 2 years later.

Then why did you choose the subject for the degree? Or are you talking about High School and therefore still just whinging like every other high-schooler?

Maybe it's different elsewhere, but where I went to school (UNLV) you need to take up to Math 120 to meet the University requirments for graduation, regardless of major. As a History major is doesn't help me in the slightest though.

That's strange. Very strange. I went to an Australian University and took a IT / Math degree. I was forced to do a 'English' based subject which I despised but it mainly focused on how to make a resume and apply for jobs and crap like that. I would have thought High-school would have covered the Mathematics required for everyday life in America.