A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

anonymous

  • 5 years ago

A rectangular plate has a length of (22.4 ± 0.2) cm and a width of (8.4 ± 0.1) cm. Calculate the area of the plate, including its uncertainty.

  • This Question is Closed
  1. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    i understand how to get the area of the plate i just don't understand how to get the uncertainty.

  2. radar
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    probably has something to do with the tolerance in the dimensions the \[\pm.2cm \pm.1cm\]

  3. radar
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    What is maximum error in the area of the plate?

  4. radar
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    The area could be 22.6 X 8,5=192.1 or it could be 22.2 X 8.3 = 184.26

  5. radar
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    looks like there could be a discrepancy of 7.84 sq cm

  6. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    that's what i got and that's not the answer they want

  7. radar
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Lets see what gogind has to offer.

  8. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    For this kind of uncertainty (with 2 variables) you need to find the percentage uncertainties of each of the variable (sides of the rectangle). When you find them, just add them together to get the percentage uncertainty of the total area. From that you should be able to get the absolute uncertainty. I'll explain in latex if needed :D

  9. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    oh ok thanks that helps a lot

  10. radar
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Like one dimension if 22.4 with tolerance of .2 cm(+/2) To convert this to a percentage do we use 2 or 4 Would it be 2 or 4 divided by 22.4 to get the percentage?

  11. radar
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    * .2 or .4

  12. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    If you have an uncertainty of +/- 2cm of a bar with the length lets say 200cm, the percentage uncertainty of this bar would be 200cm +/- (2/200)*100 %. Typo corrected :D

  13. radar
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Thanks I am sure ncrosby can take it from there how about it ncrosby?

  14. radar
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    One more gogind, when you get it for both dimensions, how do you combine the +/- percentages, or will it be left as a +/- percentage?

  15. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    You just sum them. In this example (question of this thread) you get 0.89% error for 1 side of the rectangle and 1.19% for the other. So the total percentage uncertainty for the area will be the sum of those two. Since it is a percentage, it already contains the information that the actual value of the area can be 2.08% smaller than 188.16 or 2.08% grater then 188.16(that is. a*b= 22.4*8.4= 188.16)

  16. radar
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Thanks again gogind,I got around .9% a little less than you, but I understand the procedure. Thankfully I am not a student and this is not my problem. I am an ole timer who likes to play lol.

  17. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    We could also use the Kline-McClintock uncertainty formula. \[A = bh\]\[U = \sqrt{ \left( {\partial A \over \partial b} w_b \right)^2 + \left( {\partial A \over \partial h} w_h \right)^2 }\]where \(w_b\) is the given uncertainty of the base dimension, and \(w_h\) is the given uncertainty of the height dimension.

  18. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    thanks everyone I got it and it was right

  19. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Indeed we could. :D I would also like to note that what I said earlier : "For this kind of uncertainty (with 2 variables) you need to find the percentage uncertainties of each of the variable (sides of the rectangle)." I meant, two variables that are multiplied together, like eashmore wrote(A=b*h) .

  20. radar
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Point taken. Good day to all.

  21. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.