Any Chemistry or Math geniuses out there?

[Emit]

My teenager has surpassed me in my knowledge of such matters. She's stuck on these two questions from her AP Chem class that she's taking at our local University. Anyone know how to do these? Or can you point us in the right direction to get help to solve these? This was part of a summer assignment and she has not met with her professor yet, and she feels too shy to contact him yet. Thanks!!

The radius of an atom of krypton (Kr) is about 1.9 angstroms.

If the atom is assumed to be a sphere, what is the volume in centimeters cubed of a single Kr atom?

and

Gold can be hammered into extremely thin sheets called gold leaf. If a 200-mg piece of gold (denisty = 19.32 g/cm cubed) is hammered into a sheet measuring 2.4 X 1.0 ft, what is the average thickness in meters?

[pitohui]

1. i'm not seeing where you have to be a genius to answer these

maybe i'm stoopid but these are (appear to be?) standard not trick questions and it would be cheating to give the answers

[Emit]

2. Emit's daughter here.

I have the answers because they're in the back of my book, so I don't need them.

I need to know how to do them, I've been trying to do them for days now.

In the first one, I'm having trouble converting distance to volume, and in the second I just don't understand what the problem is asking.

[BlooInBloo]

4. Do you know the formula for volume? It probably starts with something like "V=....".

[Emit]

11. yes, I know the formula for volume.

there's 4/3(pi)(r)cubed.

and of course, density = mass/volume.

but those both still leave me needing to know how to convert distance to volume.

[Xipe Totec]

19. The r is a distance

when you cube it, it becomes a volume.

The distance you have is in angstrom, so you have to convert it to centimeters first. Then, when you cube it, it will become cubic centimeters.

(PS: density does not play any role in this problem)

[karlrschneider]

27. It does in a way, the value is needed to find the mass...

;-) But of course it's given in the problem.

[BlooInBloo]

31. Your utter mastery of this problem is hereby acknowledged.

[Xipe Totec]

33. you mean for the second problem?

Yea, I realized that after I posted. I was only thinking about the first problem. :hi:

[karlrschneider]

48. Yes. They added that to make it a little more complex, evidently.

:D

[Xipe Totec]

30. For the second problem,

use the density formula to convert from weight to volume.

volume is area times thickness, so plug in the volume and the area into the formula, and you get thickness.

[EstimatedProphet]

10. The first one:

Volume of a sphere is 4/3xPIx(r-cubed). You have the radius.

[Canuckistanian]

13. Sphere volume

sphere volume = 4/3 * PI * (radius * radius * radius)

[Xipe Totec]

15. Do you know the formula for the volume of a sphere?

and for the second problem, do you know the formula for the volume of a cube?

I'm just trying to figure out where you are at this moment; what you know already and what you're still trying to figure out.

[Emit]

68. yeah.

it's length x width x height.

do I need to find another measurement to go with 2.4 x 1.0 ft?


edit for typo.

[Xipe Totec]

69. Post #30 - Solve for height (thickness) n/t

[poverlay]

5. It's not cheating to help someone learn how to solve a problem. I'd help, but I haven't the

foggiest...

[Canuckistanian]

3. The questions really have nothing to do with chemistry

It's just standard geometry with a bit of algebra.

[BlooInBloo]

6. Let's keep in mind we're talking to a kid here, folks.

[pitohui]

20. the consequences of getting caught cheating are more serious for a kid EOM

.

[BlooInBloo]

28. Oh - I just meant go easy on the "this is so easy" criticism.

[pitohui]

35. ok point taken EOM

.

[Emit]

50. thank you

for sticking up for me.

[BlooInBloo]

56. :) Nobody needs the added "oh god - I don't want to say the wrong thing...

... and look like an idiot" pressure.

:)

[karlrschneider]

16. No algebra. A little geometry...

:D

[pitohui]

18. yeah that's what i'm thinking, this is very basic math

if the poster can't do this, we are not helping her by doing her homework, there is a bigger issue that is going to have to be addressed at some point -- and not by DU

[karlrschneider]

29. I was thinking that sounded like a 9th grade high school problem...

apparently things have changed.

[msmcghee]

59. Yeah, but you guys have . .

. . minutes with 100 seconds in them . . so that will throw the whole thing off. B-)

[karlrschneider]

64. And circles with 400 degrees!

:rofl:

[mikelgb]

7. well im not a genius and i don't know the answers... but

but by my logic here is how I would go about each:

1> I would find the conversion of angstroms to cubic centimeters then find the formula for volume (i think it is 2 pi r², r = radius, but look it up) of a sphere and plug the results in

2>covert feet to cm, get the formula for density and reverse it to find height

[Toucano]

8. For volume of a sphere,

use 4/3*Pi*radius cubed

convert angstroms to centimeters

[mikelgb]

26. right

I was wrong

4/3¶r3

[karlrschneider]

9. It's pretty straightforward. Here's a hint

the volume of a sphere is 4/3 * Pi * R^3

(there's no chemistry involved in these)

0.2 grams of Au (200 mg) divided by 19.32 gives the volume of the gold. 2.4 ft is {2.4 x 12 x 25.4) cm
that should get her started
:D

edit: if you still can't get it, pm me and I will show you step by step but try it yourself first. :-)

[Emit]

24. okay thank you, but

now I am running into difficulty converting cm to cubic cm.
I can get the volume into angstroms and cm, but then what?

I can't just cube the number, because I tried that.

[karlrschneider]

34. You can't cube the number? It's just itself multiplied by itself 2 more times!

You are in college?

2 cubed is 8 (2 x 2 x 2)
3 cubed is 27 (3 x 3 x 3)
4 cubed is 64


etc.

[BlooInBloo]

45. Easy, tiger.

[Emit]

46. well of course I know how to cube a number.

but you can't just cube a the number.
you have to convert it to cubic cm.
there's a difference.

and no, I'm not in college.
I'm in high school, taking ap chemistry.

[hfojvt]

54. you have two ways to do that

one, convert your original distance into centimeters, then cube it, even if it is a small number like .000019 and cubing it makes it even ridiculously smaller.

The harder way is to cube the number you have and then convert cubic Angstroms into cubic centimeters. I has been so long I do not remember what that is, but it would be Angstroms cubed times (cm per angstrom) cubed. It's like multiplying fractions or algebra A* (C/A) = C where C is centimeters.

[karlrschneider]

61. One cm is 10 ^ 10 A (can't recall the HTML for angstrom) so

one cc. is 10^32 angstrom. Big numbers but not too difficult using sci notation...
:-)

[karlrschneider]

55. Sorry, your (mom?) originally mentioned a University class.

I didn't realize the situation. Was my first reply any help to you? What you have to do is get everything into the same units, whether angstroms, centimeters or whatever. I think you're just confused about the difference between length and volume, it can be daunting. Remember this, it may be helpful:

length is 1 dimension
area is 2, and
volume is 3.

Consider a cube 5 cm by 5 cm by 5 cm. Each edge is 5 cm. long (length)
each face is 5 x 5 (5 squared) or 25 cm^2 that's the area of the face and
the volume is 5 x 5 x 5 (5 cubed) or 125 cm^3

Does that help?



edit for typo

[Emit]

63. where did you get 12?

[karlrschneider]

66. Damn! I messed up SORRY. I wrote

2.4 ft is {2.4 x 12 x 25.4) cm

and was wrong. There are 12 inches in a foot and 2.54 NOT 25.4 cm per inch! I'm more used to thinking in millimeters.


:dunce:

[FogerRox]

12. try this

http://natsci.edgewood.edu/genchem/lab01310202.htm

[The Vinyl Ripper]

14. An Angstrom is 10^-8 cm

So the radius of a Kr atom = 1.9 * 10^-8 cm

Volume of a sphere is 4/3 pi * r^3

4/3 * pi * (1.9 * 10^-8)^3 = 9.14533333 × 10^-24 cm^3

If I got my algebra straight :D

[The Vinyl Ripper]

21. Oh, BTW..

Google is a scientific calculator, just enter your numbers into it and click go..

[Canuckistanian]

32. That's strange, 10^-8?

I'd have thought that an Angstrom would be some sort of Engineering notation standard.

You know, like 10^-6 or 10^-9.

[karlrschneider]

39. Okay so it's 10^-10 METERS

Happy?
:eyes:

[Canuckistanian]

52. It's still not Engineering notation

But then, Engineers don't usually work on these scales.

[AngryAmish]

57. Very small engineers do

Ever hear of quantum tunneling? Who do you think designs those tunnels?

[dropkickpa]

93. The monks, of course!

:hi:

[karlrschneider]

62. It's called scientific notation and engineers most surely do use them.

Try calculating Reynolds numbers without it.

[Canuckistanian]

71. I know engineers use them

But it's just that it doesn't follow the milli/micro/nano/pico model of using exponents in multiples of 3.

[karlrschneider]

76. Maybe things have changed but it has always been standard practice to

use one significant digit before the decimal followed by the appropriate exponent. We would never express a number like 794.3245 as 0.7943245E3 (your suggestion). Or even 79.43245E1

[Canuckistanian]

82. In science, yes. engineering, no.

F'r instance, we would say 900mW, not 9.0 X 10^-1 Watts.

Engineering notation uses 1 to 999, followed by the exponent.

And we would never use 0.794, as you say in your example. To us, that would be 794 milli<whatever unit>.

[karlrschneider]

84. Well, scientists have ideas and engineers turn them into reality.

But I still disagree - sure if you're describing some physical quantity in general terms, you will most certainly say
something like "the distance from the earth to the sun is 93 million miles" or "the speed of light is 300,000,000 meters per second" but if you're doing actual calculations those figures will always be expressed in proper mathematical terms.

Here is my opinion as both an engineer by education and a scientist by avocation: Most engineers are scientists but not many scientists are engineers.

:-)

[Canuckistanian]

90. I agree (with your last statement)

:D

[sailor65]

92. Very well put

....from another engineer
:)

[The Vinyl Ripper]

49. It's actually 10^-10 meter..

But 10^-8 cm was the first definition that popped up when I Googled it.

I thought I was right but wanted to be sure..

Trust but verify. -R Reagan

[Emit]

60. that's the right answer.

but what is 10^-8 a conversion for?

[The Vinyl Ripper]

65. 10^-8 centimeters/angstrom..

Which in this case actually makes the problem much easier since the answer is in cubic centimeters.

[Bill McBlueState]

83. The angstrom isn't an SI unit

That's why it seems odd -- SI units are some base unit times 10^X, where X is a multiple of 3. An angstrom is 10^-10 meters. It's an older unit used in spectroscopy, chosen as a way to conveniently express visible wavelengths. Eyes can detect light from about 4000 to 7000 angstroms.

[Canuckistanian]

98. Thanks, that's a good explanation

I never did get the point of an angstrom.

[The Vinyl Ripper]

51. Yep, I did screw up...

The answer is actually.

2.8730912 × 10^-23

I left the pi out when doing the problem.. :D

[dsc]

17. These really aren't all that bad

For a sphere V = 4/3 pi r^3. Using that formula find the volume in angstroms and then convert. (Use google to find the conversion factor if need be).

For the second one, convert feet to meters in the given dimensions. 1 meter = 39/12 feet. Use the first part of the question to figure out the cubic meters of gold you have. Then volume = width X length X height. You have v, w, and l.

[Emit]

37. I tried to find the conversion on google.

from angstrom to cubic cm.
actually, pretty much every distance to volume ratio possibly.

I can't find it anywhere, and that's what is making this problem hard.

[karlrschneider]

42. You have to get all the values in the same form. You can't convert

angstroms to cubic centimeters because one is a length and the other is a volume! Think of it this way: How many inches are there in a gallon? Obviously it's a meaningless question!

[The Vinyl Ripper]

43. Angstrom is a unit of length, not volume..

It's actually an archaic term now and not really used much.

Nanometers (nm) are the accepted measuring unit for those sort of distances now.

One nm = 10^-9 meter.

[dsc]

47. I should have been clearer

My formula would give a volume in cubic centimeters but you would have to convert once for each dimension. What ever the factor is (apparently 10^-8) would have to be cubed

[bluerum]

22. Hints: convert angstroms to centimeters and use the standard formula for

calculating the volume of a sphere (see any physics/chem text).

The second problem is similar. Convert g/cm^3 to g/m^3. Convert feet to meters. Then set the volume of the piece of gold equal to the volume of the sheet and solve for the thickness.

These are classic stoichiometry conversion problems.

[karlrschneider]

23. Uh, stoichiometry has to do with oxidation, not arithmetic...

:D

[bluerum]

25. uh - ratios and balancing equations.

on edit: although in an extremely narrow and literal sense you are absolutley correct.

[karlrschneider]

44. I pick nits for a living.

:rofl:
Well, sort of...;-)

[The Vinyl Ripper]

36. I've noticed that a lot of people who can do the math have real difficulty with word problems..

It's converting the problem outlined in the words into a sequence of mathematical manipulations that stumps one hell of a lot of people that could do the actual math.

If you are mathematically inclined such things are pretty easy, if you are not, they can be hell.

For instance, I like music but have not a clue as to the theory behind it. I can usually tell when a note is the wrong one, but telling you why it's wrong is totally beyond me.

I have mathematical talent but lack musical talent.

[karlrschneider]

77. I'm pretty good at math, calculus even, but I HATE these kinds of problems:

If 6 people can paint 5 rooms in 8 hours, how many rooms can 9 people paint in 20 hours?

AAAAAAAAARRRGGGHHHHHHHHHH

:silly:

[BlooInBloo]

79. yah - d=rt problems kill a lot of people.

[karlrschneider]

80. Yeah, solving the equations is trivial...formulating them is what's a PITA

...

[reichstag911]

88. What if one...

...of the additional people works only half as hard as evryone else, and one works three times as hard as everyone else? }(

[karlrschneider]

96. Then it becomes a question about illegal immigrants and I have absolutely no idea.

:rofl:

[reichstag911]

99. The illegal immigrant...

...being the one working three times as hard, and a Republican being the one working half as hard? :eyes:

[karlrschneider]

101. Of course!

:D
(Uh, do Republicans actually work?)
:eyes:

[reichstag911]

103. Only the ones...

...totin' a bible, watchin' NASCAR and/or professional wrasslin', listenin' to Rushbo talkin' 'bout them durn immigrants ruinin' OUR country, and voting against their own economic interests.

[The Vinyl Ripper]

97. 48 man hours for 5 rooms is 9.6 man hours/room

180 man hours divided by 9.6 man hours/room = 18.75 rooms.

My mind just happens to work that way.

Funny thing you should mention painting. I have a family member with a small painting business and when my daughter was moaning about learning math I used him as an example of why nearly everyone needs math.

Most people wouldn't think that it takes any math to paint, they would be wrong. It actually takes quite a bit of arithmetic to bid a job. On a lot of jobs all you have to work from is a blueprint and missing one tiny detail out of a myriad of details can cost you thousands very easily.

[karlrschneider]

100. So true! I remember from years ago when I was flying a plane for a construction company,

they were sweating out the deadline on a several million dollar bid for some project, I forget the details, but they got it together about 3 hours before it had to be submitted...the boss collared me and said "this has to be in St. Louis before 11:00 a.m., TAKE IT THERE NOW! :D

I grabbed it and drove like a madman to the airport, hopped in the plane and took off. On the way, I opened the folder up and took a look at the paperwork - and noticed someone had neglected to multiply something by something , or maybe it was a decimal point wrong, can't recall, anyway I wasn't sure what to do, no way to contact them (this was before cell phones)...so I marked out the wrong numbers and wrote in the correct amount. Borrowed a car at the airport and got it delivered in time, barely and waited...we didn't get the bid. I thought "oh shit, I screwed up big time"

Flew back and went to the office figuring I'd be in trouble but actually they gave me a bonus when they saw what
happened! :D

[The Vinyl Ripper]

106. Figuring bids is very stressful..

My family member does all the math in his head when figuring prints. I'm a lot better at anything beyond arithmetic but he is far better at mental arithmetic than I.

You may well have saved your company tens or even hundreds of thousands of dollars.

Getting a contract you have bid wrong can hurt you way worse than not getting one you bid right.

As my family member says, if you get everything you bid, you're bidding too low.

[karlrschneider]

108. Reminds me of a toon from years ago: Astronauts sitting atop a rocket...

one says "You ever think about the fact this thing was built by the lowest bidder?"

:P

[reichstag911]

102. Or you could skip the intermediate step...

...of calculating the 9.6mh/r (man X hour / room) and set up two equal fractions:

48mh 180 mh
------ = --------
5 r x r

then cross-multiply:

48x mhr = 900mhr, then divide both sides by 48mhr to get

x = 18.75

Big fan of math shortcuts here, and cross-mult's one of my faves.

OH, WELL, I CON'T GET THE SECOND NUMERATOR AND DENOMINATOR TO LINE UP PROPERLY; THE SECOND NUMERATOR IS (180mh), AND THE SECOND DENOMINATOR IS (x r).

[The Vinyl Ripper]

109. Ever heard of Trachtenberg math?

http://www.speed-math.com/

What is Trachtenberg Speed Math?

* Trachtenberg Speed Math (TSM) is a software program based on Professor Jakow Trachtenberg's system of simple methods to perform high speed mathematics.
* TSM is an interactive, speed math learning tool that allows you to quickly learn Trachtenberg's system.
* The simple methods were invented by this brilliant Russian engineer while imprisoned in the Nazi death camps during World War II.
* Because Trachtenberg did not have much paper to write on in prison, he invented methods based on mental calculations.
* Children extremely weak in mathematics displayed genius-like mathematical abilities after learning the system.

My dad was a big fan of Trachtenberg and managed to calculate about as fast as a math savant by using Trachtenberg methods and a lot of other tricks he had figured out.

[reichstag911]

111. Actually, no, I never had,...

...but I'm going to take a look at it. There's probably some stuff in there that I've used as I've always run numbers/problems/etc. in my head for fun, but I wasn't really exposed to any formalized system of math simplification. I was taught the hard way; I used my own shortcuts to make it faster/easier. Believe me, I was forever being scolded for not showing calculations that I'd done in my head, or for doing math problems/tests in pen. Thanks for the link!

[The Vinyl Ripper]

113. My dad spent time in India before WWII

The Indian currency system is base sixteen, sixteen annas to the the rupee.

Dad could add, subtract, divide and multiply base sixteen in his head faster than you could do it on a hex calculator.

And this was a long time before computers made base sixteen familiar to non Indians.

I know what you mean about getting in trouble for not showing every step. :D

There's another trick I remember my dad using called "casting out the nines"..

http://www.jimloy.com/number/nines.htm

[mudesi]

38. Number 1:

1 angstrom = 1.0 x 10E-10 meters

Volume of a sphere = (4/3)*pi*(radius^3)

So the volume of the atom in angstroms is

(4/3)*(3.14)*(1.9^3) = 28.7163467 angstroms cubed.

To convert that to cm, we do this:

((1.0 x 10E-30 meter cubed) / (1.0 ang cubed)) * ((1000000 cm cubed) / (1.0 m cubed)) * 28.7163467 angstroms cubed = 2.87163467 x 10E-23 cm cubed

If the answer in your book is 2.87163467 x 10E-23 cm cubed, then I haven't made any mistakes.

[Emit]

41. that's the answer in my book.

and I was able to get the volume in angstroms.

but I'm confused about how you got "((1.0 x 10E-30 meter cubed) / (1.0 ang cubed)) * ((1000000 cm cubed) / (1.0 m cubed))" these numbers.

[mudesi]

58. This is an easy mechanical trick

From google you find out that 1 ang = 1.0 x 10E-10 m.

First, write that out as a fraction:

1.0 x 10E-10 m
--------------
    1.0 ang

Now, the question asks for your answer in cm, so do the same thing with meters and centimeters. You know that 100 cm = 1m, so write that out as a fraction:

100 cm
------
 1 m

Finally, you know your answer in angstroms cubed.
28.71 ang^3

Now, put these three things together as a chain of multiplications:

1.0 x 10E-10 m        100 cm      28.71 ang^3
--------------   *  ----------  *
   1.0 ang                1 m

Pretend that the units are algebraic variables. The trick is to cancel out the units in such a way that you are left with cm^3, because that is what you want your answer to be in. So, cube the first fraction and cube the second fraction:

(1.0 x 10E-10 m)^3     (100 cm)^3      28.71 ang^3
--------------   *  --------------  *
   (1.0 ang)^3          (1 m)^3

After cubing, you get this (remember to pretend that the units are algebraic variables!)

1.0 x 10E-30 m^3       1000000 cm^3         28.71 ang^3
--------------   *  -----------------  *
   1.0 ang^3             1 m^3

Notice now that after you cancel out all the "units" (because they are variables), you end up with cm^3 as your only units left. This means that you have laid out your fractions and multiplications correctly. Multiply all the numbers out and you end up with your answer in cm^3.

I don't know how clear I've been, but this is how a lot of people do this!

[BlooInBloo]

67. It's called "dimensional analysis" in math/physics/chemisty...

... and "type checking" in computer science. The Google will know all about it, for those who are interested.

[karlrschneider]

72. Oh I see you beat me to it...was out feeding the doggies.

;-)

[BlooInBloo]

73. As long as somebody says it.

[karlrschneider]

70. A nice little exemplar of dimensional analysis.

I wish they would teach that in middle school, it sure makes things easier!

[Speck Tater]

40. Here ya go...

Volume of a sphere is V = (4*pi*r^3)/3

Thus: V = (4*3.14159*1.9^3)/3 = 86.19266324/3 = 28.73 Angstroms cubed.

Converting from Angstroms to centimeters the conversion factor is 1 * 10^-8 thus 28.73 Angstroms cubed is 28.73 * 10^-8 Centimeters cubed.

--------------

Convert 200 mg to grams:

200mg * 0.001 = 0.2 grams

0.2 grams * 1cm / 19.32 grams = 0.01035 cubic cm.

Converting cubic cm to cubic inches use conversion factor 0.06102: 0.01035 * 0.06102 = 6.3168 * 10^-4

Converting feet to inches gives the sheet dimensions 28.8 X 12.0 inches.
The volume of a rectangular solid is L*W*T = V so 28.8 * 12.0 * T = 6.3168 * 10^-4

Solving for T gives (6.3168*10^-4)/(28.8*12.0) = 1.828 * 10-6 inches.

Converting inches to meters use the conversion factor 0.0254 giving 4.643 * 10^-4 meters.

[reichstag911]

86. You didn't cube the exponent in the Kr problem...

...you should have wound up, using pi extended to 3.14159, with 28.73 x 10^-24 cm³, or as lynyrd_skynyrd has it, ~2.873 x 10^-23 cm³.

And to Emit's daughter, the entire basis of lynyrd's conversions is the grade school-taught Identity Property of Multiplication. The ratios by which he multiplies to convert from cubic angstroms to cubic centimeters are all equal to 1, and multiplying any number by 1 equals that number.

His calculations can also be expressed as

28.73ų x 1 x 1... with the actual value of the original spherical volume calculation never changing (though the coefficient will change as it is multiplied by numbers in the numerators, or divided by numbers in the denominators, which fractions, remember, taken as a whole, equal 1), but with its units of measure -- because of the cancellations of the numerator/denominator units of measure -- changing to the form in which the problem demands, in this case cm³.

[MadMaddie]

53. Umm... I was a Communications Major...

And avoided all of these classes like the plague.

We have some really smart folks on DU!!

A kick for Emit for asking and a Kick for all the Brainy DU folks giving instruction on how to figure out the problem.:think:

[L. Coyote]

74. Calling all plagiarists, calling all plagiarists. Do my kids homework! n/t

[Emit]

75. That's clearly NOT the case -- Emit here. n/t

[BlooInBloo]

78. To avoid this kind of silliness, you may want to consider the lounge for this sort of thing...

.... next time. GD is full of critics, pedants, and looky-how-smart-I-am-boy-are-you-stupid types - none of which are helpful to kids.

[Emit]

81. Yes, I considered posting in the lounge, but

I rarely post there and know more DUers here in GD. Thought I'd give it a try -- two hours up late last night with a tearful teenager, re-reading on a subject I haven't cracked open a book for in 20 years or more had us both feeling a bit desperate for guidance. And, I considered other message boards, but, again, I thought I'd stick to what I know. She's a bright kid -- she's just stuck on these two problems out of, like, 50 or so problems that were assigned as summer work. She has appreciated all the responses, and especially the ones that make her think about it and work through those areas she's stuck on. She's got the answers in the back of the book -- she just needs help getting from point a to b on these 2, so to speak.

And least anyone think she's lazy, this is a 16 yo whose worked f/t as a lifeguard all summer long, and has been taking AP History, Chem and English -- which all required summer assignments. This is just the tailend of it all.

Thanks, BlooInBloo! for your kind advice.
:hi:

edit typo

[karlrschneider]

87. Some of us (I was, anyway) a little confused by your OP which at first glance

seemed to indicate she was in college already! I don't think those problems are really appropriate for a chemistry class but of course would be for one in math...they really are just mostly number-crunching questions. :-)

And I don't think you need to apologize for asking for a little help either.

[pitohui]

105. i know coyote

this is really NOT a good idea

if the teacher ever learns of this thread (small chance, i realize) this kid is toast and getting an F in the course just for starters

[mudesi]

85. Number 2:

1 g = 1000 mg

So 200 mg = 200 x 10E-3 g.

The volume of the sheet is therefore:

(200 x 10E-3 g) / (19.32 g / cm^3) = 0.010351966 cm^3

The surface area of the sheet is 2.4 ft * 1.0 ft = 2.4 ft^2

Converting the surface area from ft^2 to m^2:
1 ft = 30.48 cm

((30.48 cm)^2 / (1 ft^2)) * ((1 m^2) / (100 cm)^2) * (2.4 ft^2) = 0.222967296 m^2

The volume of the sheet (which we have) is equal to the surface area (which we also have) times the thickness. First convert the volume from cm^3 to m^3, then solve for it:

Volume = (0.010351966 cm^3) / (100 cm)^3 = 1.0 x 10E-8 m^3

Thickness = (1.0 x 10E-8 m^3) / (0.222967296 m^2) = 4.6 x 10E-8 m.

Answer in your book?

[Emit]

89. thank you sooo much.

that is the answer in my book.

one quick question though, "((1 m^2) / (100 cm)^2)" why is this included in the equation?

[mudesi]

94. If you lay it out as mutliplication of fractions...

...you'll see that you have to cancel out cm^2, because when I googled the conversion I googled feet to centimeters, and the question asks for meters.

(30.48 cm)^2       1 m^2         2.4 ft^2
------------  *  ----------  *
    ft^2            (100 cm)^2

[karlrschneider]

95. In case L_S doesn't see your question, that is to convert

square meters to square centimeters. It needs to be there to keep everything in the proper units.
:D

[karlrschneider]

91. Looks right...I didn't run the numbers but your methodology is right. Here's one for YOU

:evilgrin:

I get a long rope and wrap it around the world at the equator, pull it tight (assume an even 25000 miles and no mountains in the way)...now I untie the rope and splice in a ten foot piece, and arrange for it to be evenly supported
all the way round above the surface. How high off the ground/ocean is it?

[Canuckistanian]

104. About one foot, 7 inches?

Or am I missing a trick here?

BTW, you never mentioned what unit you wanted the answer in.

[karlrschneider]

107. 5 feet/ Pi to be precise. :D Units don't matter, I can convert

but you're right I should have specified. :D

Lots of people don't analyze the problem and will guess usually something like a hundredth of an inch or thereabouts.
:D
Nicely done

edit: By the way, I've been pretty much all over Canada, from PEI to the Yukon but I never made it to Ottawa...I regret that. :cry:

[reichstag911]

110. Pretty simple.

Since C=2πr, the additional circumference is 10, and we are seeking r (since the rope will be equidistant from the earth at all points), just solve for r in 10=2πr.

[fortyfeetunder]

112. Dimensional Analysis will set you free

I see you got a lot of good answers, though I offer to you this website for future questions on dimensional analysis -- converting length to volume...I use this even as a professional and it's a very handy tool!

http://www.chem.tamu.edu/class/fyp/mathrev/mr-da.html

Good luck with your studies! :hi:

[slutticus]

114. Are you serious?

Are these supposed to be hard?


Shame on you for trying to cheat anyway.

[Name removed]

115. Deleted message

Message removed by moderator. Click here to review the message board rules.

[MiniMe]

116. She is trying to learn, not cheat

She clearly stated that the answers are in the book. She is trying to understand how to get from point A to point B. That is not cheating, that is learning how to do the problem.

[slutticus]

117. The point of problems like these....

... is to teach someone how to extrapolate knowledge to a problem that is a little different from what they learned in class.

I assume they taught the students how to calculate the volume of spheres and cubes and are now extending the problem to incorporate things like density and unit conversions.

By asking someone to think through the problem for you, you are not learning. When I was a TA in grad school, my biggest pet peeve was when students came to me with a question like ("how do i do this problem?") without offering any ideas of their own about how they approached it and what they have already tried....even if it didn't work.

With the internet now...it's even easier to find the solutions to problems without thinking through it yourself. This is a huge problem (at least when i was a TA). By encouraging this behavior the OP is doing their child a disservice in my opinion.

And of course the answer is in the book. That's not the point. The point of these problems isn't to get the answer right....it's to think through the problem and find the solution. (and you need the final answer to know if you're on the right track!)