Answer
$\mathrm{a}.$
$145\mathrm{c}\mathrm{m}^{2}$
$\mathrm{b}.$
decreases
$\mathrm{c}.$
$43.7\ \mathrm{l}\mathrm{b}/\mathrm{i}\mathrm{n}^{2}$
Work Step by Step
$P=F/A \qquad $15-2
$($Units: area is in $\mathrm{m}^{2}$, pressure in $\mathrm{P}\mathrm{a},$ force in $ \mathrm{N}.)$
The pressure exerted by the amosphere is $P_{\mathrm{a}\mathrm{t}}=1.01\times 10^{5}\mathrm{N}/\mathrm{m}^{2}\approx 14.7$ lb $/\mathrm{i}\mathrm{n}^{2}$.
(use for unit conversion)
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Let $\mathrm{A}_{1}$ = area of contact of one tire.
Then,$\displaystyle \quad \mathrm{P}=\frac{\mathrm{m}\mathrm{g}}{4\mathrm{A}_{1}}$
$A_{1}=\displaystyle \frac{mg}{4P}=\frac{(1420\mathrm{k}\mathrm{g})(9.81\mathrm{m}/\mathrm{s}^{2})}{4(35.0\ \mathrm{l}\mathrm{b}/\mathrm{i}\mathrm{n}^{2})(\frac{1.01\times 10^{5}\mathrm{P}\mathrm{a}}{147\ \mathrm{l}\mathrm{b}/\mathrm{i}\mathrm{n}^{2}})}=0.0145\mathrm{m}^{2}$
$=0.0145\displaystyle \mathrm{m}^{2}\times\frac{100^{2}\mathrm{c}\mathrm{m}^{2}}{1\mathrm{m}^{2}}=145\mathrm{c}\mathrm{m}^{2}$
$\mathrm{b}.$
Reading 15-2, area and pressure are inversely proportional.
When the pressure increases, the area of contact decreases.
$\mathrm{c}.$
With $\mathrm{A}_{1}=116\mathrm{c}\mathrm{m}^{2}$
$P=\displaystyle \frac{mg}{4A_{1}}=\frac{(1420\mathrm{k}\mathrm{g})(9.81\mathrm{m}/\mathrm{s}^{2})}{4(116\mathrm{c}\mathrm{m}^{2}\times\frac{1\mathrm{m}^{2}}{100^{2}\mathrm{c}\mathrm{m}^{2}})}= 3.00\times 10^{5} \mathrm{P}\mathrm{a}$
$=(3.00\times 10^{5} \mathrm{P}\mathrm{a} )(\displaystyle \frac{14.7\ \mathrm{l}\mathrm{b}/\mathrm{i}\mathrm{n}^{2}}{1.01\times 10^{5}\mathrm{P}\mathrm{a}})=43.7\ \mathrm{l}\mathrm{b}/\mathrm{i}\mathrm{n}^{2}$
