1. A 2.61-m-tall container is filled to the brim, partway with mercury and the r

Question

1. A 2.61-m-tall container is filled to the brim, partway with mercury and the rest of the way with water. The container is open to the atmosphere. What must be the depth of the mercury so that the absolute pressure on the bottom of the container is twice the atmospheric pressure?

2. A 80.8-kg person puts on a life jacket, jumps into the water, and floats. The jacket has a volume of 3.02 x 10-2 m3 and is completely submerged under the water. The volume of the person's body that is underwater is 6.05 x 10-2 m3. What is the density of the life jacket?

3. A patient recovering from surgery is being given fluid intravenously. The fluid has a density of 1020 kg/m3, and 6.71 x 10-4 m3 of it flows into the patient every 6 hours. Find the mass flow rate in kg/s.

4. Prairie dogs are burrowing rodents. They do not suffocate in their burrows, because the effect of air speed on pressure creates sufficient air circulation. The animals maintain a difference in the shapes of two entrances to the burrow, and because of this difference, the air ( = 1.29 kg/m3) blows past the openings at different speeds, as the drawing indicates. Assuming that the openings are at the same vertical level, find the difference in air pressure between the openings.
Va=8.5m/s and Vb=1.1m/s

5. The blood speed in a normal segment of a horizontal artery is 0.146 m/s. An abnormal segment of the artery is narrowed down by an arteriosclerotic plaque to 1/4 of its normal cross-sectional area. What is the difference in blood pressure between the normal and constricted segments of the artery?

6. Water is circulating through a closed system of pipes in a two-floor apartment. On the first floor, the water has a gauge pressure of 4.6 × 105 Pa and a speed of 1.5 m/s. However, on the second floor, which is 4.1 m higher, the speed of the water is 4.3 m/s. The speeds are different because the pipe diameters are different. What is the gauge pressure of the water on the second floor?

Explanation / Answer

1)

It's easiest to use pressure units of "mm Hg" for this problem, and keep in mind that:

1 atmosphere of pressure = 760 mm Hg

Since the bottom pressure is 2 atmospheres, we have:
P_bottom = 2 atm = 2 x 760 mm Hg = 1520 mm Hg

Also,

P_bottom = 1 atm + height_Hg
+ height_H2O x (density_H2O / density_Hg)

Put it all together and you get:

1520 mmHg = 760 mmHg + height_Hg
+ height_H2O x (density_H2O / density_Hg)

You'll need to look up the densities of Hg and water, and also use the fact that
height_Hg + height_H2O = 1 m = 1000 mm

2)

First I will define terms.

v1 = volume of life jacket = 0.0302 m^3
v2 = volume of body under water = 0.0605 m^3
m1 = mass of the life jacket
m2 = mass of the person = 80.8 kg
g = acceleration of gravity
p = density of water = 1000 kg/m^3

The buoyancy force on an object is given by:

B = M*g - V*p*g = g*(M - V*p)

This must be applied to both the lifejacket and the person and then the sum of these must be 0 in order for there to be no net force acting on the person and they will float.

Lifejacket: BL = g*(m1 - v1*p)
Person: BP = g*(m2 - v2*p)

BP + BL = 0 = g*(m2 - v2*p) + g*(m1 - v1*p)
0 = m2 - v2*p) + (m1 - v1*p)
0 = (m2 + m1) - p*(v2 + v1)
m1 = p*(v2 + v1) - m2

m1 = 1000*(0.0605 + 0.0302) - 80.8 = 1000(0.0907) - 80.8 = 90.7 - 80.8
m1 = 9.9 kg

So the lifejacket has a mass of 9.9 kg
Density is mass divided by volume = 9.9/0.0302 = 327.81 kg/m^3

So the density of the life jacket is 327.81 kg/m^3

3)

The mass of the flow rate = density*volume flowing per sec
= 1020*6.71x10^-4/(6*60*60)
= 3.17*10^-5 kg/sec

4)

Velocity at opening 'a' = Va =8.5 m/s (to the right)

Velocity at opening 'b' = Vb =1.1 m/s (to the right)

Density of air = p = 1.29 kg/m^3

Using Bernoullie's theorem,

Pa+(1/2)pVa^2 = Pb+(1/2)pVb^2

Pb - Pa = (1/2)pVa^2 - (1/2)pVb^2

Pb - Pa = (1/2)p [ Va^2 - Vb^2 ]


Pb - Pa = (1/2)*1.29 [ 8.5^2 - 1.1^2 ]

Pb - Pa = 0.645 [ 72.25 - 1.21 ]

Pb - Pa = 0.645 [ 71.04 ]

Pb - Pa = 45.821 pascal

5)

  as cross-secton become 1/4, the velocity will become 4 times. Kinetic energy density will become 16 times.
If P1 is he pressure in normal region and P2 in plaque region we have
P1 + k1 = P2 +k2 now k2 = 16k1 or
P1-P2 = 15k1 = 15*0.5*d*0.146^2 , where d = density of blood which can be taken as 1060 kg/m^3
So P1 - P2 = 15*0.5*1060*(0.146^2) = 169.46 Pa
P2 will be less by 169.46 Pa

6)

given ;
P1 = 4.6 x 10^5 Pa
V1 = 1.5 m/s

V2 = 4.3 m/s
h = 4.1 m

P1 + ½ v1² + g h1 = P2 + ½ v2² + g h2

4.6 x 10^5 + ½ (1000)1.5² + 0 = P2 + ½ (1000) 4.3² + (1000) (9.8)(4.1)

calculate for P2 from here

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