Chapter 10: Q3P (page 260)
(I) If you tried to smuggle gold bricks by filling your backpack, whose dimensions are 54cm× 31cm× 22cm what would its mass be?
Short Answer
The mass of the gold bricks is 710.78Kg.
Chapter 10: Q3P (page 260)
(I) If you tried to smuggle gold bricks by filling your backpack, whose dimensions are 54cm× 31cm× 22cm what would its mass be?
The mass of the gold bricks is 710.78Kg.
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Get started for freeThree containers are filled with water to the same height and have the same surface area at the base, but the total weight of water is different for each (Fig. 10–46). In which container does the water exert the greatest force on the bottom of the container?
(a) Container A.
(b) Container B.
(c) Container C.
(d) All three are equal
An airplane has a mass of \({\bf{1}}{\bf{.7 \times 1}}{{\bf{0}}^{\bf{6}}}\;{\bf{kg}}\) and the air flows past the lower surface of the wings at 95 m/s. If the wings have a surface area of \({\bf{1200}}\;{{\bf{m}}^{\bf{2}}}\), how fast must the air flow over the upper surface of the wing if the plane is to stay in the air?
Show that the power needed to drive a fluid through a pipe with uniform cross-section is equal to the volume rate of flow, \({\bf{Q}}\), times the pressure difference \({{\bf{P}}_{\bf{1}}}{\bf{ - }}{{\bf{P}}_{\bf{2}}}\), Ignore viscosity.
Water at a gauge pressure of \({\bf{3}}{\bf{.8}}\;{\bf{atm}}\) at street level flows into an office building at a speed of \({\bf{0}}{\bf{.78}}\;{\bf{m/s}}\) through a pipe \({\bf{5}}{\bf{.0}}\;{\bf{cm}}\)in diameter. The pipe tapers down to \({\bf{2}}{\bf{.8}}\;{\bf{cm}}\) in diameter by the top floor, \({\bf{16}}\;{\bf{m}}\) above (Fig. 10–53), where the faucet has been left open. Calculate the flow velocity and the gauge pressure in the pipe on the top floor. Assume no branch pipes and ignore viscosity.
Figure 10-53
In Fig. 10-54, take into account the speed of the top surface of the tank and show that the speed of fluid leaving an opening near the bottom is \({{\bf{v}}_{\bf{1}}}{\bf{ = }}\sqrt {\frac{{{\bf{2gh}}}}{{\left( {{\bf{1 - A}}_{\bf{1}}^{\bf{2}}{\bf{/A}}_{\bf{2}}^{\bf{2}}} \right)}}} \),
where \({\bf{h = }}{{\bf{y}}_{\bf{2}}} - {{\bf{y}}_{\bf{1}}}\), and \({{\bf{A}}_{\bf{1}}}\) and \({{\bf{A}}_{\bf{2}}}\) are the areas of the opening and of the top surface, respectively. Assume \({{\bf{A}}_{\bf{1}}}{\bf{ < < }}{{\bf{A}}_{\bf{2}}}\) so that the flow remains nearly steady and laminar.
Figure 10-54
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