All Solutions
Section 1.2: Measurement
To clarify more, the useق for example wants to measure the length of a pen cap which is 3 cm long $textit{but underline{textbf{he can’t}} measure the pen cap length textbf{accurately}}$ because the end of his ruler is worn out, thus $textit{measuring the pen cap length textbf{with a value different from the actual value}}$.
Thus a tiny space is left before zero count to make it less probable that the accuracy of the ruler gets affected due to the worn out or faded zero mark, thus enhancing the accuracy of measurement as overall.
A micrometer can measure lengths up to the nearest 0.01 mm. Since the micrometer can measure values in smaller steps, we can say that the micrometer is **more precise** as compared to the meter stick.
However, since the micrometer is bent, it means that it cannot measure the actual lengths properly. Hence, we can say that the micrometer is **less accurate** as compared to the meter stick.
However, this micrometer will *not be more accurate* because it is bent and it will be off in the precise measurement it provides compared to the meter stick.
[ V = l times wtimes h tag{1}]
enumerate[bfseries (a)]
item Knowing that the length of the box is 18.1 cm, and the width of the box is 19.2 cm, and the height of the box is 20.3 cm, thus using equation (1) the volume of the box is
begin{align*}
V&= 18.1 times 19.2 times 20.3 \
&= 7054.66 ~ rm{cm}^3
intertext{But, since the measured lengths have only 3 significant number, hence the product of these 3 numbers must have only 3 significant number, therefor the volume of the box rounded to 3 significant figures is}
&= fbox{$7050 ~ rm{cm}^3$}
end{align*}
item The precision of the length is to the nearest 0.1 cm or 1 mm, while the precision of the volume is to the nearest 10 cm$^3$ where the volume from part (b) is rounded to 3 significant figures thus it is rounded to the nearest 10 cm$^3$.
item The height of 12 stack of boxes knowing that one box is 20.3 cm height is
begin{align*}
text{Tall} &= 12 times 20.3 \
&= fbox{$243.6 ~ rm{cm}$}
intertext{textbf{underline{textit{note:}}}: It is like adding 20.3 cm 12 times, thus the final answer must have at least one digit after the decimal thus the final answer is not rounded as it only have 1 significant figures after the digit.}
end{align*}
item The precision of the measurement of one box is length is to the nearest 1 mm, the precision of the height of the 12 box is the same as 1 mm, where in addition the final answer must have the same number of significant figures after the digit as before therefor in both cases they have 1 significant figures after the digit, therefor both of measurements are to the nearest mm.
item 7050 cm$^3$.
item To the nearest 1 mm, and to the nearest 10 cm$^3$.
item 243.6 cm.
item Both to the nearest 1 mm.
If the average time was 65.4s, the result would be more acceptable.
