Step-by-Step Solution
Step 1
We are given the values $58342\;{\rm{m}}$, $68.534\;{\rm{s}}$, $2553\;{\rm{N}}$ and$7555\;{\rm{kg}}$.
We are asked to round off the number to three significant figures.
Step 2
(a)
To round off the number to three significant figures we will use the following relation.
Since ${10^3}\;{\rm{m}} = 1\,{\rm{km}}$, then\[\begin{array}{c} 58342\;{\rm{m}} = \left( {58.342 \times {{10}^3}\;{\rm{m}}} \right)\left( {\frac{{1\;{\rm{km}}}}{{1000\;{\rm{m}}}}} \right)\\ = 58.3\;{\rm{km}} \end{array}\]
Step 3
(b)
To round off the number to three significant figures we will use the following relation.
As a general rule, any numerical figure ending in a number more than five is rounded up and a number less than five is not rounded up.\[68.534\;{\rm{s}} = 68.5\;{\rm{s}}\]
Step 4
(c)
To round off the number to three significant figures we will use the following relation.
Since $1000\;{\rm{N}} = 1\;{\rm{kN}}$, then\[\begin{array}{c} {\rm{2553 N}} = \left( {2.553 \times {{10}^3}\,{\rm{N}}} \right)\left( {\frac{{1\;{\rm{kN}}}}{{1000\;{\rm{N}}}}} \right)\\ = 2.55\;{\rm{kN}} \end{array}\]
Step 5
(d)
To round off the number to three significant figures we will use the following relation.
Since $1\;{\rm{kg}} = {10^3}\;{\rm{Mg}}$, then\[\begin{array}{c} {\rm{7555 kg}} = \left( {7.555 \times {{10}^3}\,{\rm{kg}}} \right)\left( {\frac{{1\;{\rm{Mg}}}}{{1000\;{\rm{kg}}}}} \right)\\ = 7.56\;{\rm{Mg}} \end{array}\]