what is the area and uncertainty in area of one side of a rectangular plastic brick that has a length of (21.4 ± 0.3) cm and a width of (9.8 ± 0.1) cm? (a 210 cm2 (b) What If? If the thickness of the brick is (1.2 t0.3) cm, what is the volume of the brick and the uncertainty in this volume? (Give your answers in cm.) 021.73 x 421.30 x )cm3

Respuesta :

Answer:

a) A = (209.7 ± 5.0) cm^2

b) V = (251.6 ± 75.5) cm^3

Explanation:

The area will be simply

21.4 * 9.8 = 209.7 cm^2

For multiplication of uncertainties you have to first convert them to relative uncertainties

[tex]\frac{100 * 0.3}{21.4} = 1.4%[/tex]

[tex]\frac{100 * 0.1}{9.8} = 1.0%[/tex]

Then we add those together and this is the relative uncertainty of the product

(209.7 ± 2.4%) cm^2

And we can convert it to absolute uncertainty:

[tex]=\frac{2.4 * 209.7}{100} = 5.0[/tex]

Then the area is expressed as:

A = (209.7 ± 5.0) cm^2

For volume

209.7 * 1.2 = 251.6 cm^3

And the relative uncertainty is

[tex]\frac{100*0.3}{1.2} = 25%[/tex]

So the volume with relative uncertainty is:

(251.6 ± 30%) cm^3

And converting it to absolute uncertainty:

[tex]=\frac{30 * 251.6}{100} = 75.5[/tex]

V = (251.6 ± 75.5) cm^3

The area of the brick is [tex](209.7 \pm 5.1)\ cm^2[/tex], while the volume of the brick is [tex](251.7 \pm 69) cm^3[/tex]

(a) The area

The dimension of the plastic brick is given as:

Length = (21.4 ± 0.3) cm

Width = (9.8 ± 0.1) cm

The area of a rectangle is:

Area = Length * Width

So, we have:

[tex]Area = 21.4 * 9.8[/tex]

[tex]Area = 209.7[/tex]

The smallest dimension of the brick is:

Length = 21.1 i.e. 21.4 - 0.3

Width = 9.7 i.e. 9.8 - 0.1

So, the smallest area is:

[tex]Area = 21.1 * 9.7[/tex]

[tex]Area = 204.7[/tex]

The largest dimension of the brick is:

Length = 21.7 i.e. 21.4 + 0.3

Width = 9.9 i.e. 9.8 + 0.1

So, the largest area is:

[tex]Area = 21.7 * 9.9[/tex]

[tex]Area = 214.8[/tex]

Next, calculate the midpoint of the uncertain areas:

[tex]d = \frac{214.8 - 204.7}{2}[/tex]

[tex]d = 5.1[/tex]

Hence, the area of the brick is:

[tex]Area= (209.7 \pm 5.1)[/tex]

(b) The volume

The thickness is given as: (1.2 ± 0.3)

So, the volume of the brick is:

[tex]Volume = 21.4 * 9.8 * 1.2[/tex]

[tex]Volume = 251.7[/tex]

The smallest thickness of the brick is 0.9 i.e. 1.2 - 0.3

So, the smallest volume is:

[tex]Volume = (209.7 - 5.1) * 0.9[/tex]

[tex]Volume = 184.2[/tex]

The largest thickness of the brick is 1.5 i.e. 1.2 + 0.3

So, the largest volume is:

[tex]Volume = (209.7 + 5.1) * 1.5[/tex]

[tex]Volume = 322.2[/tex]

Next, calculate the midpoint of the uncertain volumes:

[tex]d = \frac{322.2 - 184.2}{2}[/tex]

[tex]d = 69[/tex]

Hence, the volume of the brick is:

[tex]Volume = (251.7 \pm 69) cm^3[/tex]

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