Physics, 17.09.2019 01:00 tylerallen0918102

The length and width of a rectangular room are measured to be 3.92 ± 0.0035 m and 3.15 ± 0.0055 m. in this problem you can approximate the error on a product or quotient of quantities by the sum of the percent error on each quantity. a.) calculate the floor area of the room in square meters b.) calculate the uncertainty in this measurement in square meters.

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Answer from: allendraven89pds755

A) $A=12.2480\ m^2$

B) $12.2480\pm 0.1029\ m^2$

Explanation:

Given:

Length of the room l= 3.92 ± 0.0035

Width of the room w= 3.15 ± 0.0055

A) Let A be the area of the room

$A=l\times w\\A=3.92\times3.15\\A=12.2480\ \rm m^2$

B)We will calculate uncertainty in each dimension

%uncertainty in length $=\dfrac{0.0035}{3.92}\times 100=0.0892\ %$

%uncertainty in width = $\dfrac{0.0055}{3.15}\times 100=0.0174%$

The uncertainty in area will be sum of uncertainty in length and width

%uncertainty in Area= %uncertainty in length + %uncertainty in width

%uncertainty in Area $=0.0892\ % + 0.0174\ %$

%uncertainty in Area=0.0106

Uncertainty in Area $=0.0106\times 12.2480=0.1029\ \rm m^2$

There Area is $12.2480 ± 0.1029\ \rm m^2$

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