Quiz 10: 4, 7, 9, 12, 13, 14, 16, 22, 24, 29, 30, 38, 41, 46, 48 (48 is ec) , 72, 77
Turn in: 9, 14, 24, 29, 30, 38, 41, 46
9.
(a) Pressure = P = F/A. Pressure =PATM = 1.013 x 10 5 N/m2. Compute the area A from the given dimensions, then find the force magnitude F.
(b) Same force, neglecting slightly lowered height ( and slightly higher pressure) from the small  thickness of the table.
14.
(a) P = PATM + (density)*g*h. where density = 1000 kg/m3.  To get the force magnitude F, multiply by the area of the bottom.
(b) Same pressure since the pressure is the same in all directions at the same depth.
24. The first step is to get the volume V of the hull. That is easy since m = 1800 kg = (Steel density) *V. Note: Steel density = 7800 kg/m3. Find V.
(a) We assume the crane lifts the  hull at constant velocity, so the acceleration is zero.  Now, when the hull is completely submerged, we have: 0 = pos - neg
0 = T' + (water density)*g*V - mg . Find T' = upward cable tension  magnitude when hull is under water. Note : water density = 1000 kg/m3.  Note also that FB = buoyant force magnitude = (water density)*g*V.
(b) When the hull is above water, we still assume the acceleration upward is zero, so
0 = T - mg, where T = tension. Find T = cable tension when hull is above water.  
29. I did this in class ! Were you there ? Here we go again:
Note that the tension force now acts downward to hold the object down.
(a) FB = buoyant for magnitude = (water density)*g*V, where V = external volume of the chamber. Get V from the formula for the volume of a sphere of radius r = 2.6 m. Use water density = 1025 kg/m3 since we are talking about sea water that has a slightly higher value.  
(b) 0 = pos - neg
0 =  FB - mg - T.
You know the buoyant force from part (a). You know mg. Find  T. 
30.
(a) FB = (water density)*g*V, where the volume V displaced by the scuba person is given. Note: you must convert L to m3.
(b) Compute the diver density and compare with water density = 1025 kg/m3 (seawater.)  Note: diver density =
(diver mass)/V = m/V. If diver density > water density, then she will sink; otherwise she will rise and eventually float after she breaks the water surface. You could also compare mg with FB. 
38.  See the example associated with fig 10-24. Use equation 10-6. Read section 10-10 carefully, integrating these ideas with your lecture notes.
41. First of all,  you must know that the volume rate of flow is A*v, where A = crossectional area of the pipe and v = speed of water at a certain location in the pipe. A*v = constant, so if you know it at one location, you know it at all other locations.

P1 +(1/2)(1000)v12 + (1000)gy1 = P2 +(1/2)(1000)v22 + (1000)gy2 . This is equation 10-5 with the water density used.

Also: A1*v1 = A2*v2. 

Let point 1 be the thick section and point 2 be the narrow section of the pipe.

Now, the two values of y are equal , so they cancel.
P1 +(1/2)(1000)v12  = P2 +(1/2)(1000)v22 .

For P1 you may substitute 32.0 kPa and for P2 you may substitute  24.0 kPa   . Note 1 Pa = 1 N/m2 and kPa = 1000Pa.
You want to get one of the speeds so you can get A*v for that speed and location. Thus make this substitution:
v1 = (A2/A1)v2  = (42/62)v2  = 4v2/9 . Thus:
P1 +(1/2)(1000)v22 *(4/9)2 = P2 +(1/2)(1000)v22 .  Solve for v2 and compute  A2*v2.   .

46.  First of all,  you must know that the volume rate of flow is A*v, where A = crossectional area of the pipe and v = speed of water at a certain location in the pipe. A*v = constant, so if you know it at one location, you know it at all other locations.

P1 +(1/2)(1000)v12 + (1000)gy1 = P2 +(1/2)(1000)v22 + (1000)gy2 . This is equation 10-5 with the water density used.

Also: A1v1 = A2v2. 

Let point 1 be at the bottom (street) and point 2 be at the top. Thus: v1  = 0.60 m/s. You can immediately find v2 from A1*v1 = A2*v2  since you can easily compute the pipe area at the bottom and the top.

Finally:
Let y1 = 0, so that y2 = 18 m.   You may use P1 = 3.8*(1.013 x 10 5 N/m2).  Solve:

P1 + (1/2)(1000)v12  = P2 +(1/2)(1000)v22 + (1000)gy2 

for P2.