Find: ΔPAF [Pa] T=20 C kerosine benzene 40 [cm] air A F 20 [cm] 8 [cm] mercury water Find the change in pressure, between point F and point A, in pascals. [pause] In this problem, --- A) -8,900 B) -900 C) 900 D) 8,900

Find: ΔPAF [Pa] T=20 C kerosine benzene 40 [cm] air A F 20 [cm] 8 [cm] mercury water a monometer contains 5 different fluids, which include, --- A) -8,900 B) -900 C) 900 D) 8,900

Find: ΔPAF [Pa] T=20 C kerosine benzene 40 [cm] air A F 20 [cm] 8 [cm] mercury water benzene, --- A) -8,900 B) -900 C) 900 D) 8,900

Find: ΔPAF [Pa] T=20 C kerosine benzene 40 [cm] air A F 20 [cm] 8 [cm] mercury water mercury, ---- A) -8,900 B) -900 C) 900 D) 8,900

Find: ΔPAF [Pa] T=20 C kerosine benzene 40 [cm] air A F 20 [cm] 8 [cm] mercury water kerosine, --- A) -8,900 B) -900 C) 900 D) 8,900

Find: ΔPAF [Pa] T=20 C kerosine benzene 40 [cm] air A F 20 [cm] 8 [cm] mercury water water, --- A) -8,900 B) -900 C) 900 D) 8,900

Find: ΔPAF [Pa] T=20 C kerosine benzene 40 [cm] air A F 20 [cm] 8 [cm] mercury water and air. [pause] A) -8,900 B) -900 C) 900 D) 8,900

Find: ΔPAF [Pa] T=20 C kerosine benzene 40 [cm] air A F 20 [cm] 8 [cm] mercury water The vertical distance between adjacent liquid interfaces, --- A) -8,900 B) -900 C) 900 D) 8,900

Find: ΔPAF [Pa] T=20 C kerosine benzene 40 [cm] air A F 20 [cm] 8 [cm] mercury water is either given, or, can be calculated from the given distances. Points A and F, --- A) -8,900 B) -900 C) 900 D) 8,900

Find: ΔPAF [Pa] T=20 C kerosine benzene 40 [cm] air A F 20 [cm] 8 [cm] mercury water are located at the two ends of the manometer, and, the temperature is, --- A) -8,900 B) -900 C) 900 D) 8,900

Find: ΔPAF [Pa] T=20 C kerosine benzene 40 [cm] air A F 20 [cm] 8 [cm] mercury water 20 degrees, celsius. [pause] The problem asks for --- A) -8,900 B) -900 C) 900 D) 8,900

Find: ΔPAF [Pa] T=20 C kerosine benzene 40 [cm] air A F 20 [cm] 8 [cm] mercury water delta P, of A, F. Which is the same as, the pressure at point F, --- A) -8,900 B) -900 C) 900 D) 8,900

Find: ΔPAF [Pa] ΔPAF=PF - PA kerosine T=20 C air 40 [cm] benzene H O A minus the pressure at point A. This pressure difference can be computed using the general equation, --- benzene H O 2 A F Hg 14 [cm] 20 [cm] 8 [cm]

Find: ΔPAF [Pa] ΔPAF=PF - PA ΔP=-ρ* g *Δh kerosine T=20 C air 40 [cm] for the change in pressure across a fluid medium, delta p, equals negative rho, g, delta h. Where, delta p, equals, --- benzene H O 2 A F Hg 14 [cm] 20 [cm] 8 [cm]

Find: ΔPAF [Pa] ΔPAF=PF - PA ΔP=-ρ* g *Δh pressure difference kerosine T=20 C air 40 [cm] the change in pressure, rho, --- benzene H O 2 A F Hg 14 [cm] 20 [cm] 8 [cm]

Find: ΔPAF [Pa] density ΔPAF=PF - PA ΔP=-ρ* g *Δh pressure difference o kerosine T=20 C air 40 [cm] is the density of the fluid, g, --- benzene H O 2 A F Hg 14 [cm] 20 [cm] 8 [cm]

Find: ΔPAF [Pa] density ΔPAF=PF - PA ΔP=-ρ* g *Δh pressure constant difference acceleration o kerosine T=20 C air 40 [cm] is the gravitational acceleration constant, and delta h, --- benzene H O 2 A F Hg 14 [cm] 20 [cm] 8 [cm]

Find: ΔPAF [Pa] density ΔPAF=PF - PA ΔP=-ρ* g *Δh change in height pressure constant difference acceleration o kerosine T=20 C air 40 [cm] is the change height in the fluid, between the two points whose pressure difference is being calculated. This equation is only valid if, --- benzene H O 2 A F Hg 14 [cm] 20 [cm] 8 [cm]

Find: ΔPAF [Pa] constant density ΔPAF=PF - PA ΔP=-ρ* g *Δh change in height pressure constant difference acceleration o kerosine T=20 C air 40 [cm] the density is constant. If we want to find the pressure difference between 2 points, --- benzene H O 2 A F Hg 14 [cm] 20 [cm] 8 [cm]

Find: ΔPAF [Pa] constant density ΔPAF=PF - PA ΔP=-ρ* g *Δh change in height pressure constant difference acceleration o kerosine T=20 C air 40 [cm] which are separated by multiple fluids, having different densities, then, we must, --- benzene H O 2 A F Hg 14 [cm] 20 [cm] 8 [cm]

Find: ΔPAF [Pa] constant density ΔPAF=PF - PA ΔP=-ρ* g *Δh change in height ΔP=-g * Σ ρ * Δh for multiple densities o kerosine T=20 C air 40 [cm] compute the pressure difference for each density, and sum those pressure differences together. In our problem, --- benzene H O 2 A F Hg 14 [cm] 20 [cm] 8 [cm]

Find: ΔPAF [Pa] constant density ΔPAF=PF - PA ΔP=-ρ* g *Δh change in height ΔP=-g * Σ ρ * Δh for multiple densities o kerosine T=20 C air 40 [cm] there are 5 different fluids in the manometer, so we’ll use ---- benzene H O 2 A F Hg 14 [cm] 20 [cm] 8 [cm]

Find: ΔPAF [Pa] constant density ΔPAF=PF - PA ΔP=-ρ* g *Δh change in height ΔP=-g * Σ ρ * Δh for multiple densities o kerosine T=20 C air 40 [cm] the second eqauation. [pause] For convenience sake, we’ll define the points, B, --- benzene H O 2 A F Hg 14 [cm] 20 [cm] 8 [cm]

Find: ΔPAF [Pa] constant density ΔPAF=PF - PA ΔP=-ρ* g *Δh change in height ΔP=-g * Σ ρ * Δh for multiple densities o kerosine T=20 C air 40 [cm] D C, D, and E, at the fluid boundaries. This makes it easier to write out, --- benzene H O 2 A F Hg C E B 14 [cm] 20 [cm] 8 [cm]

Find: ΔPAF [Pa] ΔPAF=-g * Σ ρ * Δh ΔP=-g * Σ ρ * Δh for multiple densities o kerosine T=20 C air 40 [cm] D our equation for delta P, of A, F. [pause] After expanding the summation term, --- benzene H O 2 A F Hg C E B 14 [cm] 20 [cm] 8 [cm]

Find: ΔPAF [Pa] +ρkerosine*(hD-hC)+ρH O* (hE-hD) +ρair*(hF-hE) } ΔPAF=-g * Σ ρ * Δh ΔPAF=-g * { ρbenzene*(hB-hA)+ρHg* (hC-hB) +ρkerosine*(hD-hC)+ρH O* (hE-hD) 2 +ρair*(hF-hE) } o kerosine T=20 C air 40 [cm] D we see all the density and height terms we need to determine, in order to solve the problem. [pause] First of all we know, g, --- benzene H O 2 A F Hg C E B 14 [cm] 20 [cm] 8 [cm]

Find: ΔPAF [Pa] +ρkerosine*(hD-hC)+ρH O* (hE-hD) +ρair*(hF-hE) } g = 9.81 [m/s2] ΔPAF=-g * Σ ρ * Δh ΔPAF=-g * { ρbenzene*(hB-hA)+ρHg* (hC-hB) +ρkerosine*(hD-hC)+ρH O* (hE-hD) 2 +ρair*(hF-hE) } o kerosine T=20 C air 40 [cm] D equals 9.81 meters per second squared. Next, we’ll use the figure to determine, --- benzene H O 2 A F Hg C E B 14 [cm] 20 [cm] 8 [cm]

Find: ΔPAF [Pa] +ρkerosine*(hD-hC)+ρH O* (hE-hD) +ρair*(hF-hE) } g = 9.81 [m/s2] ΔPAF=-g * Σ ρ * Δh ΔPAF=-g * { ρbenzene*(hB-hA)+ρHg* (hC-hB) +ρkerosine*(hD-hC)+ρH O* (hE-hD) 2 +ρair*(hF-hE) } o kerosine T=20 C air 40 [cm] D the change in heights, between adjacent fluid interfaces. Height B, minus, --- benzene H O 2 A F Hg C E B 14 [cm] 20 [cm] 8 [cm]

Find: ΔPAF [Pa] +ρkerosine*(hD-hC)+ρH O* (hE-hD) +ρair*(hF-hE) } g = 9.81 [m/s2] -20 [cm] ΔPAF=-g * Σ ρ * Δh ΔPAF=-g * { ρbenzene*(hB-hA)+ρHg* (hC-hB) +ρkerosine*(hD-hC)+ρH O* (hE-hD) 2 +ρair*(hF-hE) } o kerosine T=20 C air 40 [cm] D height A, equals negative 20 centimeters. [pause] Height C, minus, height B, --- benzene H O 2 A F Hg C E B 14 [cm] 20 [cm] 8 [cm]

Find: ΔPAF [Pa] +ρkerosine*(hD-hC)+ρH O* (hE-hD) +ρair*(hF-hE) } g = 9.81 [m/s2] 8 [cm] -20 [cm] ΔPAF=-g * Σ ρ * Δh ΔPAF=-g * { ρbenzene*(hB-hA)+ρHg* (hC-hB) +ρkerosine*(hD-hC)+ρH O* (hE-hD) 2 +ρair*(hF-hE) } o kerosine T=20 C air 40 [cm] D equals 8 centimeters. [pause] Height D minus height C, --- benzene H O 2 A F Hg C E B 14 [cm] 20 [cm] 8 [cm]

Find: ΔPAF [Pa] +ρkerosine*(hD-hC)+ρH O* (hE-hD) +ρair*(hF-hE) } g = 9.81 [m/s2] 8 [cm] -20 [cm] ΔPAF=-g * Σ ρ * Δh ΔPAF=-g * { ρbenzene*(hB-hA)+ρHg* (hC-hB) +ρkerosine*(hD-hC)+ρH O* (hE-hD) 2 +ρair*(hF-hE) } 40 [cm]-8 [cm] o kerosine T=20 C air 40 [cm] D equals, 40 centimeters, minus 8 centimeters, which equals, --- benzene H O 2 A F Hg C E B 14 [cm] 20 [cm] 8 [cm]

Find: ΔPAF [Pa] +ρkerosine*(hD-hC)+ρH O* (hE-hD) +ρair*(hF-hE) } g = 9.81 [m/s2] 8 [cm] -20 [cm] ΔPAF=-g * Σ ρ * Δh ΔPAF=-g * { ρbenzene*(hB-hA)+ρHg* (hC-hB) +ρkerosine*(hD-hC)+ρH O* (hE-hD) 2 +ρair*(hF-hE) } 32 [cm] o kerosine T=20 C air 40 [cm] D 32 centimeters. [pause] Height E, minus height D, --- benzene H O 2 A F Hg C E B 14 [cm] 20 [cm] 8 [cm]

Find: ΔPAF [Pa] +ρkerosine*(hD-hC)+ρH O* (hE-hD) +ρair*(hF-hE) } g = 9.81 [m/s2] 8 [cm] -20 [cm] ΔPAF=-g * Σ ρ * Δh ΔPAF=-g * { ρbenzene*(hB-hA)+ρHg* (hC-hB) +ρkerosine*(hD-hC)+ρH O* (hE-hD) 2 +ρair*(hF-hE) } 32 [cm] 14 [cm]-40 [cm] o T=20 C kerosine 40 [cm] D air equals, 14 centimeters minus 40 centimters, which equals, --- benzene H O 2 A F Hg C E B 14 [cm] 20 [cm] 8 [cm]

Find: ΔPAF [Pa] +ρkerosine*(hD-hC)+ρH O* (hE-hD) +ρair*(hF-hE) } g = 9.81 [m/s2] 8 [cm] -20 [cm] ΔPAF=-g * Σ ρ * Δh ΔPAF=-g * { ρbenzene*(hB-hA)+ρHg* (hC-hB) +ρkerosine*(hD-hC)+ρH O* (hE-hD) 2 +ρair*(hF-hE) } 32 [cm] -26 [cm] o T=20 C kerosine 40 [cm] D air negative 26 centimeters. [pause] And we currently don’t know the change in height, --- benzene H O 2 A F Hg C E B 14 [cm] 20 [cm] 8 [cm]

Find: ΔPAF [Pa] +ρkerosine*(hD-hC)+ρH O* (hE-hD) +ρair*(hF-hE) } g = 9.81 [m/s2] 8 [cm] -20 [cm] ΔPAF=-g * Σ ρ * Δh ΔPAF=-g * { ρbenzene*(hB-hA)+ρHg* (hC-hB) +ρkerosine*(hD-hC)+ρH O* (hE-hD) 2 +ρair*(hF-hE) } 32 [cm] -26 [cm] o T=20 C kerosine 40 [cm] D air between points F and E. [pause] Next we’ll look up the densities, --- benzene H O 2 A F Hg C E B 14 [cm] 20 [cm] 8 [cm]

Find: ΔPAF [Pa] +ρkerosine*(hD-hC)+ρH O* (hE-hD) +ρair*(hF-hE) } g = 9.81 [m/s2] 8 [cm] -20 [cm] ΔPAF=-g * Σ ρ * Δh ΔPAF=-g * { ρbenzene*(hB-hA)+ρHg* (hC-hB) +ρkerosine*(hD-hC)+ρH O* (hE-hD) 2 +ρair*(hF-hE) } 32 [cm] -26 [cm] o T=20 C kerosine 40 [cm] D air of these 5 fluids, at standard temperature, in units of, --- benzene H O 2 A F Hg C E B 14 [cm] 20 [cm] 8 [cm]

Find: ΔPAF [Pa] +ρkerosine*(hD-hC)+ρH O* (hE-hD) +ρair*(hF-hE) } -20 [cm] 8 [cm] ΔPAF=-g * { ρbenzene*(hB-hA)+ρHg* (hC-hB) +ρkerosine*(hD-hC)+ρH O* (hE-hD) m 2 +ρair*(hF-hE) } 32 [cm] 9.81 s2 -26 [cm] fluid densitiy [g/cm3] benzine 0.88 grams per cubic centimeter. Then we’ll mutiply these values by 1,000, --- mercury 13.55 kerosine 0.80 water 1.00 air 0.0012

Find: ΔPAF [Pa] +ρkerosine*(hD-hC)+ρH O* (hE-hD) +ρair*(hF-hE) } * -20 [cm] 8 [cm] ΔPAF=-g * { ρbenzene*(hB-hA)+ρHg* (hC-hB) +ρkerosine*(hD-hC)+ρH O* (hE-hD) m 2 +ρair*(hF-hE) } 32 [cm] 9.81 s2 -26 [cm] fluid densitiy [g/cm3] benzine 0.88 to convert the densities to units of, kilograms per --- mercury 13.55 kg*cm3 kerosine 0.80 1,000 * g * m3 water 1.00 air 0.0012

Find: ΔPAF [Pa] +ρkerosine*(hD-hC)+ρH O* (hE-hD) +ρair*(hF-hE) } -20 [cm] 8 [cm] ΔPAF=-g * { ρbenzene*(hB-hA)+ρHg* (hC-hB) +ρkerosine*(hD-hC)+ρH O* (hE-hD) m 2 +ρair*(hF-hE) } 32 [cm] 9.81 s2 -26 [cm] fluid densitiy [kg/m3] benzine 880 metered cubed. [pause] mercury 13,550 kg*cm3 kerosine 800 1,000 g * m3 water 1,000 air 1.2

Find: ΔPAF [Pa] +ρkerosine*(hD-hC)+ρH O* (hE-hD) +ρair*(hF-hE) } -20 [cm] 8 [cm] ΔPAF=-g * { ρbenzene*(hB-hA)+ρHg* (hC-hB) +ρkerosine*(hD-hC)+ρH O* (hE-hD) m 2 +ρair*(hF-hE) } 32 [cm] 9.81 s2 -26 [cm] fluid densitiy [kg/m3] benzine 880 And after we plug these densities into --- mercury 13,550 kerosine 800 water 1,000 air 1.2

Find: ΔPAF [Pa] +ρkerosine*(hD-hC)+ρH O* (hE-hD) +ρair*(hF-hE) } -20 [cm] 8 [cm] ΔPAF=-g * { ρbenzene*(hB-hA)+ρHg* (hC-hB) +ρkerosine*(hD-hC)+ρH O* (hE-hD) m 2 +ρair*(hF-hE) } 32 [cm] 9.81 s2 -26 [cm] fluid densitiy [kg/m3] benzine 880 their appropriate variables, we notice the density of air is --- mercury 13,550 kerosine 800 water 1,000 air 1.2

Find: ΔPAF [Pa] +ρkerosine*(hD-hC)+ρH O* (hE-hD) +ρair*(hF-hE) } -20 [cm] 8 [cm] ΔPAF=-g * { ρbenzene*(hB-hA)+ρHg* (hC-hB) +ρkerosine*(hD-hC)+ρH O* (hE-hD) m 2 +ρair*(hF-hE) } 32 [cm] 9.81 s2 -26 [cm] fluid densitiy [kg/m3] benzine 880 so much smaller than the other densities, that we can neglect this final term, --- mercury 13,550 kerosine 800 water 1,000 air 1.2

Find: ΔPAF [Pa] +ρkerosine*(hD-hC)+ρH O* (hE-hD) +ρair*(hF-hE) } -20 [cm] 8 [cm] ΔPAF=-g * { ρbenzene*(hB-hA)+ρHg* (hC-hB) +ρkerosine*(hD-hC)+ρH O* (hE-hD) m 2 +ρair*(hF-hE) } 32 [cm] 9.81 s2 -26 [cm] fluid densitiy [kg/m3] benzine 880 from, the calculation. [pause] After plugging in all the variables, --- mercury 13,550 kerosine 800 water 1,000 air 1.2

Find: ΔPAF [Pa] +ρkerosine*(hD-hC)+ρH O* (hE-hD) -20 [cm] -26 [cm] 13,550 [kg/m3] 880 [kg/m3] -20 [cm] 8 [cm] ΔPAF=-g * { ρbenzene*(hB-hA)+ρHg* (hC-hB) +ρkerosine*(hD-hC)+ρH O* (hE-hD) m 2 9.81 32 [cm] -26 [cm] s2 800 [kg/m3] 1,000 [kg/m3] delta P, of A, F, equals, ----

Find: ΔPAF [Pa] +ρkerosine*(hD-hC)+ρH O* (hE-hD) -20 [cm] -26 [cm] 13,550 [kg/m3] 880 [kg/m3] -20 [cm] 8 [cm] ΔPAF=-g * { ρbenzene*(hB-hA)+ρHg* (hC-hB) +ρkerosine*(hD-hC)+ρH O* (hE-hD) m 2 9.81 32 [cm] -26 [cm] s2 800 [kg/m3] 1,000 [kg/m3] negative 886,824 kilograms centimeters, per second squared meter squared. [pause] Next we’ll convert, --- kg*cm ΔPAF=-886,824 s2 * m2

Find: ΔPAF [Pa] +ρkerosine*(hD-hC)+ρH O* (hE-hD) * -20 [cm] -26 [cm] 13,550 [kg/m3] 880 [kg/m3] -20 [cm] 8 [cm] ΔPAF=-g * { ρbenzene*(hB-hA)+ρHg* (hC-hB) +ρkerosine*(hD-hC)+ρH O* (hE-hD) m 2 9.81 32 [cm] -26 [cm] s2 800 [kg/m3] 1,000 [kg/m3] centimeters to meters, and then convert to units of pascals, --- kg*cm ΔPAF=-886,824 1 m * s2 * m2 100 cm

Find: ΔPAF [Pa] +ρkerosine*(hD-hC)+ρH O* (hE-hD) * * -20 [cm] -26 [cm] 13,550 [kg/m3] 880 [kg/m3] -20 [cm] 8 [cm] ΔPAF=-g * { ρbenzene*(hB-hA)+ρHg* (hC-hB) +ρkerosine*(hD-hC)+ρH O* (hE-hD) m 2 9.81 32 [cm] -26 [cm] s2 800 [kg/m3] 1,000 [kg/m3] and our final pressure difference between points F and A, --- kg*cm 1 m Pa*s2*m ΔPAF=-886,824 * * s2 * m2 100 cm kg

Find: ΔPAF [Pa] +ρkerosine*(hD-hC)+ρH O* (hE-hD) * * -20 [cm] -26 [cm] 13,550 [kg/m3] 880 [kg/m3] -20 [cm] 8 [cm] ΔPAF=-g * { ρbenzene*(hB-hA)+ρHg* (hC-hB) +ρkerosine*(hD-hC)+ρH O* (hE-hD) m 2 9.81 32 [cm] -26 [cm] s2 800 [kg/m3] 1,000 [kg/m3] equals, -8,868 pascals. [pause] kg*cm 1 m Pa*s2*m ΔPAF=-886,824 * * s2 * m2 100 cm kg ΔPAF=-8,868 [Pa]

Find: ΔPAF [Pa] +ρkerosine*(hD-hC)+ρH O* (hE-hD) * * A) -8,900 B) -900 ΔPAF=-g * { ρbenzene*(hB-hA)+ρHg* (hC-hB) +ρkerosine*(hD-hC)+ρH O* (hE-hD) when reviewing the possible solutions, --- kg*cm Pa*s2*m 1 m ΔPAF=-886,824 * * s2 * m2 100 cm kg ΔPAF=-8,868 [Pa]

Find: ΔPAF [Pa] +ρkerosine*(hD-hC)+ρH O* (hE-hD) * * A) -8,900 B) -900 answerA ΔPAF=-g * { ρbenzene*(hB-hA)+ρHg* (hC-hB) +ρkerosine*(hD-hC)+ρH O* (hE-hD) the answer is A. [fin] kg*cm Pa*s2*m 1 m ΔPAF=-886,824 * * s2 * m2 100 cm kg ΔPAF=-8,868 [Pa]

ΔP13=-ρA* g * (hAB-h1)+… 3 liquids, A, B, and C. The specific gravity of each of these 3 liquids, ---