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prelab09

Course: PHYS 152, Fall 2011
School: IUPUI
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Word Count: 142

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FLUIDS Name Due Pre-Lab09: at the beginning of the lab period Consider a beaker filled partially with water sitting on a scale. The scale reads wb, the total weight of the beaker and its water in air (Figure 1a). An object of mass m and volume V attached to a massless string is then lowered into the water until it is fully submerged (Figure 1b). The scale now reads wb* &gt; wb. wb Figure wb* 1a Figure...

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FLUIDS Name Due Pre-Lab09: at the beginning of the lab period Consider a beaker filled partially with water sitting on a scale. The scale reads wb, the total weight of the beaker and its water in air (Figure 1a). An object of mass m and volume V attached to a massless string is then lowered into the water until it is fully submerged (Figure 1b). The scale now reads wb* > wb. wb Figure wb* 1a Figure 1b 1. Show that the buoyant force B acting on the object is given by B = wb* wb. 2. Show that the volume V of the object is given by V = wb * wb , where water is the density of g water the water and g is the acceleration of gravity. 3. Show that the density of the object is given by = IUPUI PHYS 15200 mg . wb * wb water Page 1 of 1
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IUPUI - PHYS - 152
EXAM 1A SOLUTIONS PHYS 15200 Part I. Multiple Choice Questions [2 pts each]1) C2) E3) B4) E5) CPart II. Word Problems [30 pts each]1) (A) Set v(t) = 0 and solve for t t = 4 sPlug into x(t) = 6t2 t3 x = 32 m = maximum(B) Plug 3 s into FNet = ma
IUPUI - PHYS - 152
EXAM 1B SOLUTIONS PHYS 15200 Part I. Multiple Choice Questions [2 pts each]1) A2) C3) C4) A5) DPart II. Word Problems [30 pts each]1) (A) a(t ) = 36 12t = 0 t = 3 s, so v(3 s) = 54 m/s = maximum(B) FNet = ma(2 s) = 120 N2) v0y = 1.5 m/s; vx = 4
IUPUI - PHYS - 152
EXAM 2A SOLUTIONS Part II. Word Problems [30 pts each]1) (A) At the top of the loop, FNet = N + mg =mv 2.rSet N = 0 to obtain minimum speed vmin = gr = 7.80 m/sN mg (B) Use conservation of total mechanical energy:21kx 2 = 1 mvmin + mg ( 2r
IUPUI - PHYS - 152
EXAM 2B SOLUTIONS Part II. Word Problems [30 pts each]1) (A) Use conservation of total mechanical energy:1mv 2 = mgh2h=L L 2v= 3.26 m = L(1 cos) = 54.42gh mv 2(B) At the bottom of the rope, FNet = T mg =.LT = m( g + v 2 /L) = 901
IUPUI - PHYS - 152
EXAM 3A SOLUTIONSPart I Multiple Choice Questions1) A 2) B 3) D 4) A 5) BPart II Word Problems1) 0 = 1200 rpm = 40 rad/s; = 5500 rpm = 550/3 rad/s 0(A) == 180 rad/s2t(B) = 1 ( 0 + )t = 140 rev2(C) = I = 1.62 N-m2)(A) I = 1/3ML2 + mr2 = 0.627
IUPUI - PHYS - 152
EXAM 3A SOLUTIONSPart I Multiple Choice Questions1) A 2) B 3) D 4) A 5) BPart II Word Problems1) 0 = 1200 rpm = 40 rad/s; = 5500 rpm = 550/3 rad/s 0(A) == 180 rad/s2t(B) = 1 ( 0 + )t = 140 rev2(C) = I = 1.62 N-m2)(A) I = 1/3ML2 + mr2 = 0.627
IUPUI - PHYS - 152
EXAM 3B SOLUTIONSPart I Multiple Choice Questions1) B 2) D 3) A 4) D 5) BPart II Word Problems1) 0 = 5000 rpm = 500/3 rad/s; = 440 rpm = 44/3 rad/s 0(A) == -133 rad/s2t(B) = 1 ( 0 + )t = 163 rev2(C) I = / = (-58 N-m) / (-133 rad/s2) = 0.437 kg
IUPUI - PHYS - 152
EXAM 3B SOLUTIONSPart I Multiple Choice Questions1) B 2) D 3) A 4) D 5) BPart II Word Problems1) 0 = 5000 rpm = 500/3 rad/s; = 440 rpm = 44/3 rad/s 0(A) == -133 rad/s2t(B) = 1 ( 0 + )t = 163 rev2(C) I = / = (-58 N-m) / (-133 rad/s2) = 0.437 kg
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N.C. State - CHEM - 223
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N.C. State - CHEM - 223
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N.C. State - CHEM - 223
N.C. State - CHEM - 223
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N.C. State - CHEM - 223
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