Are you not supposed consider sig digs for mass?

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The answers section in my book gives let statements that always have significant digits. However, it never seems to take mass into consideration. For example, if everything in the question has 3 sig digs and the mass is 2000kg, the final answer will have 3 sig digs. If everything in the question has 3 sig digs and the mass is 250kg, the final answer will still have 3 sig digs. I'm a little confused as to why there is an "exception'. I thought you were supposed to use the lowest number of significant digits given in the question for your final statements. Is my book wrong or is there a reason for this?

Also, do degrees have any affect on significant digits?
 

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  • #2
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Also, do degrees have any affect on significant digits?
3 sig digs and the mass is 2000kg
"2000" has four significant figures; 2.00 x 103 would be three; 2.0 two.
3 sig digs and the mass is 250kg
Same game.

Also, do degrees have any affect on significant digits?
Angular measure? Temperature? Trig. functions of angular measure?
 
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I dont understand that then. I thought "2000" would have one significant digit? Are you saying that because 2000 is the equivalent to 2.00x10^3 that it can be considered as having more than one sig dig?

I mean if the questions says something like: "Object is moving 14.0m/s at an angle 7 degrees above the horizontal. Would the final answer have 3 or 1 significant figures?
 
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And I remember a different example. The minimum number of sig digs given in the question was 3, and the object had a mass of 1.5kg. Final answer was given with 3 significant figures, which doesnt make sense to me.
 
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First sig. fig. is "2." Second is "0." Third is the second "0." Fourth is the third "0." If you're given 2.00 x 103, you have only the "2" and two zeroes.
I thought "2000"
14.0m/s at an angle 7 degrees
If the direction is in some way involved in calculations, yes. If it's so much window dressing, no.
 
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1.5kg. Final answer was given with 3 significant figures, which doesnt make sense to me.
That is not a correct statement of significant figures; only two are given, and only two can be reported at the end of the calculation.
 
  • #7
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Zeroes can be both significant and not significant. If they say the mass is 2000 kg with 3 significant digits, it can be 2001 kg or 2002.5 kg... it is rounded to the tens. But correctly it should be written as 2.00 x 103 kg.
1.5 kg is given with 2 significant digits. If you want to give it with 3 significant figures you should write 1.50 kg.
 
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  • Zeros at the end of a number but to the left of a decimal are significant if they have been measured or are the first estimated digit; otherwise, they are not significant. (example: 2000 has just 1 significant digit)
 
  • #9
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if they have been measured
It's all up to you what you want to call measured or unmeasured. Life is much easier if you get used to certain "conventions" surrounding use of the concept.
 
  • #10
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Zeros at the end of a number but to the left of a decimal are significant if they have been measured or are the first estimated digit; otherwise, they are not significant. (example: 2000 has just 1 significant digit)
That's a very poor example. The number of significant figures in 2000 is ambiguous since you don't know if the 0s are significant or merely placeholders. The use of scientific notation removes this ambiguity, and that's one reason for using it.

Keep in mind that using the lowest number of sig figs is only a rule of thumb. You can sometimes keep more digits sometimes. For example, 0.99 m + 0.01 m = 1.00 m — if the two measurements are to the nearest centimeter, adding them isn't suddenly going to cause the uncertainty to jump to the nearest decimeter. Similarly, (3.2 m)π = 10.1 m. To figure out the proper number of digits to keep, you'd really have to do error analysis.
 
  • #11
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I mean if the questions says something like: "Object is moving 14.0m/s at an angle 7 degrees above the horizontal. Would the final answer have 3 or 1 significant figures?
There are calculations where the concept of significant figures does not work any more, and a better error analysis becomes necessary.
For example, the horizontal velocity here is 14.0m/s * cos(7°) = 13.9m/s. You can give three significant figures as the answer does not depend much on the angle, 8° and 6° still lead to the same answer.
 

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