1. 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

2. 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

3. 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

4. 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

5. 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

6. 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

7. 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

8. 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

9. 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

10. 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

11. 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

12. 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

13. 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]

14. 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]

15. 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]

16. Find: ΔPAF [Pa] density ΔPAF=PF - PA ΔP=-ρ* g *Δh pressure differenceo
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]

17. 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]

18. 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]

19. 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]

20. 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]

21. 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]

22. 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]

23. 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]

24. 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]

25. 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]

26. 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]

27. 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]

28. 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]

29. 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]

30. 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]

31. 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]

32. 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]

33. 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]

34. 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]

35. 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]

36. 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]

37. 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

38. 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

39. 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

40. 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

41. 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

42. 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

43. 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

44. 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, ----

45. 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

46. 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

47. 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

48. 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]

49. 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]

50. 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]

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