In a fluid column, pressure increases with depth by $\rho gh$, so blood pressure is greater at lower parts of the body. Atmospheric pressure is due to the weight of air above a point; because air is compressible, density is greatest near sea level and falls with height, making pressure drop rapidly in the first few kilometres. Hydrostatic pressure is scalar because it is specified by magnitude at a point; the force on a surface is normal to that surface.
(a) Feet are at greater depth below the heart/brain, so hydrostatic pressure is larger. (b) Air density decreases rapidly with height, so much of the atmosphere's mass lies near the earth's surface. (c) Pressure at a point in a fluid has the same value in all directions; the force direction comes from the surface normal, not from pressure itself.
The contact angle depends on relative cohesive and adhesive forces. Water-glass adhesion is strong, so water spreads and has an acute contact angle. Mercury atoms cohere strongly and adhere weakly to glass, so mercury has an obtuse contact angle and forms drops. Surface tension does not depend on total area; increasing area needs proportionate work. Detergents act as wetting agents and lower the angle of contact. With no external force, a drop minimises surface energy by taking the shape of least area for fixed volume, a sphere.
(a) Mercury has stronger cohesive attraction than adhesion to glass, while water adheres strongly to glass. (b) Water wets glass because adhesion dominates; mercury forms drops because cohesion dominates. (c) Surface tension is force per unit length or surface energy per unit area and is a property of the interface. (d) Detergents reduce surface tension and improve wetting, reducing the contact angle. (e) A sphere has minimum surface area for a given volume.
Surface tension of liquids usually falls with temperature. Gas viscosity rises with temperature, while liquid viscosity falls. Solids resist shear strain; fluids resist the rate of shear strain. Speed increase at a constriction follows continuity, $Av=\text{constant}$. For a smaller model, the Reynolds number reaches the turbulent value only at greater speed.
(a) decreases. (b) increases; decreases. (c) shear strain; rate of shear strain. (d) conservation of mass. (e) greater.
By Bernoulli's principle, faster air above paper has lower pressure, so the higher pressure below supports the paper. Continuity makes water speed increase through small gaps. For laminar flow through a narrow tube, Poiseuille's law gives $Q\propto r^4\Delta P$, so needle radius controls flow strongly. Momentum conservation gives a reaction thrust opposite the emerging fluid jet. A spinning ball drags air differently on its two sides, producing pressure difference and lift/side force, so its path is not a simple parabola.
(a) Blowing over the paper lowers pressure above it. (b) The small openings force water to emerge at high speed. (c) Flow rate is extremely sensitive to needle radius. (d) The outgoing fluid carries momentum, giving the vessel a recoil thrust. (e) Spin creates unequal air speeds and pressures around the ball, causing a sideways Magnus force.
Force is $mg=50\times9.8=490\,\text{N}$. Heel radius is $0.005\,\text{m}$, so area $A=\pi r^2=7.85\times10^{-5}\,\text{m}^2$. Pressure $P=F/A=490/(7.85\times10^{-5})=6.24\times10^6\,\text{Pa}$.
$6.2\times10^6\,\text{Pa}$.
For a barometer, $P=\rho gh$. Thus $h=P/(\rho g)=1.013\times10^5/(984\times9.8)=10.5\,\text{m}$.
$10.5\,\text{m}$.
The pressure due to seawater at depth $h$ is roughly $\rho gh\approx1000\times9.8\times3000=2.94\times10^7\,\text{Pa}$. This is much less than the maximum stress $10^9\,\text{Pa}$, so the structure is suitable if currents and other effects are ignored.
Yes. The hydrostatic pressure at 3 km depth is about $2.9\times10^7\,\text{Pa}$, well below $10^9\,\text{Pa}$.
Pressure transmitted through the fluid is $P=F/A=mg/A$. Here $F=3000\times9.8=2.94\times10^4\,\text{N}$ and $A=425\,\text{cm}^2=4.25\times10^{-2}\,\text{m}^2$. Thus $P=2.94\times10^4/(4.25\times10^{-2})=6.92\times10^5\,\text{Pa}$.
$6.9\times10^5\,\text{Pa}$.
Since mercury levels are the same, pressures at the two mercury interfaces are equal. Thus $\rho_w g(10.0\,\text{cm})=\rho_s g(12.5\,\text{cm})$. Therefore $\rho_s/\rho_w=10.0/12.5=0.80$.
$0.80$.
From Exercise 9.9, specific gravity of spirit is $0.80$. Let the mercury level difference be $h$ cm, with mercury lower on the water side. Accounting for the changed liquid-column heights in equal arms gives $25+h/2=0.80(27.5-h/2)+13.6h$. Hence $3=12.7h$ and $h=0.236\,\text{cm}$.
About $0.23\,\text{cm}$, with mercury lower in the water arm.
Bernoulli's equation assumes steady, incompressible, non-viscous flow along a streamline. Flow through a rapid is turbulent, dissipative and highly non-steady, so Bernoulli's equation is not directly applicable.
No, not reliably.
Bernoulli's equation involves pressure differences along a streamline. Gauge pressure differs from absolute pressure by the constant atmospheric pressure. Adding or subtracting the same constant from all pressure terms does not change the equation.
No, as long as the same reference pressure is used throughout.
Volume flow rate is $Q=\dot m/\rho=4.0\times10^{-3}/(1.3\times10^3)=3.08\times10^{-6}\,\text{m}^3\text{s}^{-1}$. Poiseuille's law gives $\Delta P=8\eta LQ/(\pi r^4)=8(0.83)(1.5)(3.08\times10^{-6})/[\pi(0.01)^4]=9.75\times10^2\,\text{Pa}$. The mean speed gives Reynolds number about $0.31$, far below the turbulent range, so laminar flow is consistent.
$9.8\times10^2\,\text{Pa}$ approximately; the flow is laminar.
By Bernoulli's principle, pressure is lower where speed is higher. Pressure difference is $\Delta P=\frac12\rho(v_{upper}^2-v_{lower}^2)=\frac12(1.3)(70^2-63^2)=605\,\text{Pa}$. Lift $=\Delta P A=605(2.5)=1.51\times10^3\,\text{N}$.
$1.5\times10^3\,\text{N}$ upward.
At the constriction, flow speed is greater by continuity. For a non-viscous steady flow, Bernoulli's principle then requires pressure to be lower at the constriction. The piezometer level at the narrow section should therefore be lower, as in Fig. 9.20(b), not higher as in Fig. 9.20(a).
Figure 9.20(a) is incorrect.
Tube area is $A_1=8.0\,\text{cm}^2=8.0\times10^{-4}\,\text{m}^2$ and tube speed is $v_1=1.5/60=0.025\,\text{m s}^{-1}$. Total hole area is $A_2=40\pi(0.0005)^2=3.14\times10^{-5}\,\text{m}^2$. Continuity gives $A_1v_1=A_2v_2$, so $v_2=(8.0\times10^{-4})(0.025)/(3.14\times10^{-5})=0.64\,\text{m s}^{-1}$.
$0.64\,\text{m s}^{-1}$ approximately.
A soap film has two surfaces, so the upward force is $2Sl$. Thus $S=W/(2l)=1.5\times10^{-2}/[2(0.30)]=2.5\times10^{-2}\,\text{N m}^{-1}$.
$2.5\times10^{-2}\,\text{N m}^{-1}$.
For a liquid drop with one surface, excess pressure is $2S/r$. Thus $\Delta P=2(4.65\times10^{-1})/(3.00\times10^{-3})=310\,\text{Pa}$. Pressure inside $=1.01\times10^5+310=1.0131\times10^5\,\text{Pa}$.
Excess pressure is $310\,\text{Pa}$; total pressure inside is $1.013\times10^5\,\text{Pa}$ approximately.
A soap bubble in air has two surfaces, so $\Delta P=4S/r=4(2.50\times10^{-2})/(5.00\times10^{-3})=20\,\text{Pa}$. An air bubble inside liquid has one interface, so excess pressure is $2S/r=10\,\text{Pa}$. At depth $h=0.40\,\text{m}$ in liquid of density $1.20\times10^3\,\text{kg m}^{-3}$, outside pressure is $P_0+\rho gh=1.01\times10^5+1200(9.8)(0.40)$. Adding 10 Pa gives $P_{inside}=1.057\times10^5\,\text{Pa}$.
Soap bubble excess pressure is $20\,\text{Pa}$. The pressure inside the air bubble at 40.0 cm depth is about $1.057\times10^5\,\text{Pa}$.