Author Topic: physics help needed........plz  (Read 6113 times)

Offline ashwinkandel

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physics help needed........plz
« on: October 02, 2010, 02:18:31 pm »
The mass of a cube of aluminium is found to be 580 g with an uncertainty in the measurement of 10 g. Each side of the cube has a length of (6.0 ± 0.1) cm.
Calculate the density of aluminium with its uncertainty. Express your answer to an appropriate number of significant figures.

density= ......................±................... g cm-3

elemis

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Re: physics help needed........plz
« Reply #1 on: October 02, 2010, 03:59:04 pm »
Whats the answer ?

elemis

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Re: physics help needed........plz
« Reply #2 on: October 02, 2010, 04:06:17 pm »
d= \frac{m}{v} \rightarrow\frac{580}{6^3} equation 1

adding the fractional uncertainties :\frac{10}{580}+(3*\frac{0.1}{6}) equation 2

Hence, multiplying 1 and 2 gives an absolute uncertainty of 0.2

Therefore, ans = 2.7\pm0.2 g/dm^3

Offline Deadly_king

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Re: physics help needed........plz
« Reply #3 on: October 02, 2010, 04:10:07 pm »
The mass of a cube of aluminium is found to be 580 g with an uncertainty in the measurement of 10 g. Each side of the cube has a length of (6.0 ± 0.1) cm.
Calculate the density of aluminium with its uncertainty. Express your answer to an appropriate number of significant figures.

density= ......................±................... g cm-3

hey that's fro, June 09 p2, right??

Anyway it's not that difficult. Let me get this clear for you.

Density = Mass / Volume ---> density = 580/63 = 2.685

Volume of cube = l3

Therefore density = m/l3

Delta D/density * 100 = delta m/mass *100 + 3deltaL/Length * 100

Delta D = [(10/580 * 100) + (3*0.1/6 *100) ] * 2.685/100 =0.18

Since all the values given were correct to 2 sf, your answer should be correct to 2sf.

Hence answer is 2.7  +- 0.2 gcm-3

elemis

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Re: physics help needed........plz
« Reply #4 on: October 02, 2010, 04:12:16 pm »
I got there before you :P

Good work man ;)

Offline Deadly_king

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Re: physics help needed........plz
« Reply #5 on: October 02, 2010, 04:12:45 pm »
d= \frac{m}{v} \rightarrow\frac{580}{6^3} equation 1

adding the fractional uncertainties :\frac{10}{580}+(3*\frac{0.1}{6}) equation 2

Hence, multiplying 1 and 2 gives an absolute uncertainty of 0.2

Therefore, ans = 2.7\pm0.2 g/dm^3
Well done Ari.....except that the units are g/cm3 :P

elemis

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Re: physics help needed........plz
« Reply #6 on: October 02, 2010, 04:18:05 pm »
Well done Ari.....except that the units are g/cm3 :P

I did that purposely because I wanted to see of you were paying attention :P

I was doing chem all this time..... concentrations and all :D

Offline Deadly_king

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Re: physics help needed........plz
« Reply #7 on: October 02, 2010, 04:20:23 pm »
I did that purposely because I wanted to see of you were paying attention :P

I was doing chem all this time..... concentrations and all :D

Haha........take it as if I believed ya :P

Yeah......i know you have loads on your head.....just chill and relax a bit dude :)

Offline ashwinkandel

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Re: physics help needed........plz
« Reply #8 on: October 02, 2010, 04:38:20 pm »
i looked up the marking scheme. it was given like this:
(answer 2.69 ± 0.09 g cm-3 scores 4 marks).
Can anyone teach me about the use of significant figures while solving the problems like the above one.

Offline Deadly_king

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Re: physics help needed........plz
« Reply #9 on: October 02, 2010, 05:25:35 pm »
i looked up the marking scheme. it was given like this:
(answer 2.69 ± 0.09 g cm-3 scores 4 marks).
Can anyone teach me about the use of significant figures while solving the problems like the above one.

Hmm.....this question carries 5 marks.

Ok........the values given to you and which you used in your calculations were 580g and 6.0cm.

In both cases the values were correct to 2 significant figures.

NOTE : The 0 in 580 is not considered as a significant figure but the one in 6.0 is considered as one. In decimals a zero in front of the other numbers is not considers as a significant figure (Example 2) but a zero after the decimal point which has a number in front is a significant figure.(Example 3)

Examples :
1. Express 5.89 correct to 2 sf ---> 5.9
2. Express 0.00875 correct to 2sf ---> 0.0086
3. Express 1.99 correct to 2 sf ---> 2.0

Hope it helps :)




Offline ashwinkandel

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Re: physics help needed........plz
« Reply #10 on: October 02, 2010, 05:40:38 pm »
thanks for the info. but i have one doubt. how many significant digits are there in the digit 1.001010. and how do we know how many significant figures should we use in the question i  asked.

Offline iluvme

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Re: physics help needed........plz
« Reply #11 on: October 02, 2010, 05:42:38 pm »
thanks for the info. but i have one doubt. how many significant digits are there in the digit 1.001010. and how do we know how many significant figures should we use in the question i  asked.

6 significant figures. You ignore the last 0.
Usually its to 3.
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Offline Deadly_king

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Re: physics help needed........plz
« Reply #12 on: October 02, 2010, 05:44:34 pm »
6 significant figures. You ignore the last 0.
Usually its to 3.
Indeed it is 6 :)

Good job girl :)
+rep

Offline iluvme

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Re: physics help needed........plz
« Reply #13 on: October 02, 2010, 05:49:42 pm »
Indeed it is 6 :)

Good job girl :)
+rep

Thanks.
Isn't this like grade 5 math?
No offense meant to ashwinkandel.
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Offline Deadly_king

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Re: physics help needed........plz
« Reply #14 on: October 02, 2010, 05:51:07 pm »
Thanks.
Isn't this like grade 5 math?
No offense meant to ashwinkandel.
Yeah......it's somehow similar though in physics it's much more complex  ;)