I use a thermometer… - QuestionCove
OpenStudy (diamondboy):

I use a thermometer graduated in 1/5 degree Celsius to measure outside air temperature. Measured to the nearest 1/5 degree, yesterday's temperature was 22.4 degrees Celsius and today's is 24.8 degrees Celsius. What is the relative uncertainty in the temperature difference between yesterday and today?

1 year ago
OpenStudy (michele_laino):

the uncertainty in the difference is $$2 \times (1/5)=0.4$$ degrees, so the relative uncertainty, is: $\huge \frac{{0.4}}{{\left( {24.8 - 22.4} \right)}} \times 100 = ...\%$

1 year ago
OpenStudy (diamondboy):

Thanks though mehn but that's not correct.

1 year ago
OpenStudy (diamondboy):

I tried that

1 year ago
OpenStudy (diamondboy):

1 year ago
OpenStudy (michele_laino):

the measured temperatures, are: $\Large \begin{gathered} \left( {22.4 \pm 0.2} \right) \hfill \\ \hfill \\ \left( {24.8 \pm 0.2} \right) \hfill \\ \end{gathered}$

1 year ago
OpenStudy (diamondboy):

yes....I don't know why the answer behind the book itself is different

1 year ago
OpenStudy (diamondboy):

Apart from me and my textbook, you are the second person solving this question this way so I guess I am just going to write the answer that way .Could you please help with another question?

1 year ago
OpenStudy (michele_laino):

so, the difference, is: $\Large \Delta T = \left( {2.4 \pm 0.4} \right)$

1 year ago
OpenStudy (diamondboy):

yes

1 year ago
OpenStudy (diamondboy):

My digital watch gives a time reading as 09:46. What is the absolute uncertainty of the measurement?

1 year ago
OpenStudy (michele_laino):

I tried to use the theory of propagation of uncertainties, namely I write this: $\Large \delta = \sqrt {{{0.2}^2} + {{0.2}^2}} \simeq 0.283$ as uncertainty on the temperature difference nevertheless I haven't got the result above

1 year ago
OpenStudy (michele_laino):

is your watch able to read the $$1/100$$ of second ?

1 year ago
OpenStudy (diamondboy):

the question doesn't state

1 year ago
OpenStudy (michele_laino):

what is the minimum quantity which can be measured by your watch, then?

1 year ago
OpenStudy (diamondboy):

the question doesn't state, but I just assumed it should be like 1 minute

1 year ago
OpenStudy (michele_laino):

therefore, we can write this: $\Large \left( {9.77 \pm 0.02} \right)\;hour$

1 year ago
OpenStudy (diamondboy):

but the answer in the book is 0.5 minutes

1 year ago
OpenStudy (diamondboy):

|dw:1457979535301:dw| This was what I did I said that the actual time should be between 9:45 to 9:46

1 year ago
OpenStudy (michele_laino):

ok! It is a possible choice usually, when I was at laboratory course, I took the minimum measured quantity as uncertainty

1 year ago
OpenStudy (diamondboy):

so it's correct then? Is that what you are saying?

1 year ago
OpenStudy (michele_laino):

if the measure instrument is new, then we can take $$1/2$$ of the minimum measured quantity as uncertainty, otherwise, it is better to take the same minimum measured quantity as uncertainty

1 year ago
OpenStudy (diamondboy):

ok thanks then...God bless you. Bye

1 year ago
OpenStudy (michele_laino):

:)

1 year ago