Question #306521

The following data are the measurements of the dimensions of a box.


š‘™ = 3.0 ± 0.3 cm

ā„Ž = 51.1 ± 0.5 mm

š‘¤ = 2.03 ± 0.01 m


How should the volume of the box be reported (central value ± uncertainty)? Write your calculations and box your final answer. 


Expert's answer

The volume

V=lhw=3.0āˆ—0.511āˆ—203=310 cm3V=lhw=3.0*0.511*203=310\:\rm cm^3

The relative uncertainty

ΔVV=Δll+Δhh+Δww\frac{\Delta V}{V}=\frac{\Delta l}{l}+\frac{\Delta h}{h}+\frac{\Delta w}{w}

ΔVV=0.33.0+0.551.1+0.012.03=0.11\frac{\Delta V}{V}=\frac{0.3}{3.0}+\frac{0.5}{51.1}+\frac{0.01}{2.03}=0.11

Ī”V=310āˆ—0.11=34 cm3\Delta V=310*0.11=34\rm\: cm^3

Finally

V=(310±34) cm3V=(310\pm34)\:\rm cm^3


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