Answer
$\mathrm{a}.$ greater
$\mathrm{b}. 52 \mathrm{P}\mathrm{a}$
Work Step by Step
Pressure, P, is force F per area A:
$\mathrm{P}=\mathrm{F}/\mathrm{A} \qquad $15-2
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$\mathrm{a}.$
The water is accelerating upwards. An upward net force is acting on it. This net force is the result of the bottom of the glass pushing the water up with force greater than the water's weight. The water, in reaction (Newton's third law) pushes down on the glass with the same magnitude force.
Observe 15-2. F has increased. Then, so has P.
$\mathrm{b}.$
Given $\Delta \mathrm{v}$ and t, we find the acceleration.
$\displaystyle \mathrm{a}=\frac{2.4\mathrm{m}/\mathrm{s}-0}{3.2\mathrm{s}}=0.75\mathrm{m}/\mathrm{s}^{2}$
The mass of the water is $\mathrm{m}=\rho \mathrm{V}=\rho(\mathrm{A}\mathrm{h})=(1000\mathrm{k}\mathrm{g}/\mathrm{m}^{3})\cdot(0.069\mathrm{m})\cdot \mathrm{A}$
$\displaystyle \Delta P=\frac{\mathrm{F}_{\mathrm{n}\mathrm{e}\mathrm{t}}}{\mathrm{A}}=\frac{\mathrm{m}\mathrm{a}}{\mathrm{A}}=\frac{(1000\mathrm{k}\mathrm{g}/\mathrm{m}^{3})\cdot(0.069\mathrm{m})\cdot \mathrm{A}\cdot(0.75\mathrm{m}/\mathrm{s}^{2})}{\mathrm{A}}$
$=(1000\mathrm{k}\mathrm{g}/\mathrm{m}^{3})(0.069\mathrm{m})(0.75\mathrm{m}/\mathrm{s}^{2})=52 \mathrm{P}\mathrm{a}$
