Cunninham, Drew Homework 3 Due: Apr 1 006, 4:00 am Inst: Florin 1 This print-out should have 10 questions. Multiple-choice questions may continue on the next column or pae find all choices before answerin. The due time is Central time. 001 part 1 of 1 10 points A.3 cm thick bar of soap is floatin on a water surface so that 1.64 cm of the bar is underwater. Bath oil specific ravity 0.6 is poured into the water and floats on top of the water. in water B water W soap W soap B water +B oil Before the oil is added: After the oil is added: in water and oil W soap B water 1 ρ water A y. 1.64 cm.3 cm water W soap B water + B oil ρ water [A h y oil ] + ρ oil A y oil, since y oil is the depth of the oil layer. Settin Eq. 1 equal to Eq., we have y oil.3 cm oil water What is the depth of the oil layer when the top of the soap is just level with the upper surface of the oil? Correct answer: 1.65 cm. ρ water A y ρ water [A h y oil ] + ρ oil A y oil ρ water y ρ water h ρ water y oil + ρ oil y oil ρ water ρ oil y oil ρ water h y y oil h y 1 ρ 3 oil ρ water.3 cm 1.64 cm 1 0.6 1.65 cm Given : h.3 cm, y 1.64 cm, specific ravity ρ oil 0.6. ρ water Let A be the surface area of the top or bottom of the bar. The weiht of the soap bar is equal to the buoyant force when it floats in water alone: F net B W soap 0 B m fluid ρ m V V A y 00 part 1 of 10 points A block of volume 0.35 m 3 floats with a fraction 0.58 of its volume submered in a liquid of density 1130 k/m 3, as shown in the fiure below. The acceleration of ravity is 9.8 m/s. V B liquid Find the manitude of the buoyant force on the block. Correct answer: 48.0 N.
Cunninham, Drew Homework 3 Due: Apr 1 006, 4:00 am Inst: Florin Let : V L volume of displace liquid V B 0.35 m 3, ρ L 1130 k/m 3, and V L 0.58. V B Basic Concepts: Archimedes Principle: Buoyant Force on ObjectWeiht of Fluid Displaced by Object. Solution: From Archimedes Principle, we have ρ L V L ρ B V B, so V L ρ B 0.58 V B ρ L F Buoyant,P art1 Weiht of displaced liquid ρb ρ L V B ρ L 1130 k/m 3 0.58 0.35 m 3 9.8 m/s 48.0 N. 003 part of 10 points A block of density ρ and volume V is placed on top of the first block. 3. V correct ρ ρ L ρl + ρ 4. V ρ L + ρ B1 5. V ρ L ρ ρl ρ 6. V ρ B1 ρ L In equilibrium F y 0, which means: F Buoyant,P art m B1 + m ρ B1 + ρ V, where ρ B1 is the density of the first block. But we also know that: F Buoyant,P art Weiht of displaced liquid ρ L + V. Therefore ρ L +V ρ B1 +ρ V. 1 Solvin for V : V. ρ ρ L We can find ρ B1 usin the equilibrium condition for Part 1: V liquid If the volume is iven, find V such that the two blocks are just submered, as shown in fiure above. ρl + ρ B1 1. V ρ L + ρ ρl ρ. V ρ L ρ B1 Weiht of block 1 F Buoyant,P art1 ρ B1 ρ L X, which ives us ρ B1 X ρ L. Substitutin this into Eq. : ρl X ρ L V ρ ρ L ρ 1 X 1 ρ L 1 ρ B 1 ρ L ρ 1 ρ L. ρ ρ L
Cunninham, Drew Homework 3 Due: Apr 1 006, 4:00 am Inst: Florin 3 004 part 1 of 4 10 points Assume: The fluid is incompressible and nonviscous. Shown below is a cross-section of a vertical view of a pipe discharin a fluid into the atmosphere at its hihest elevation. The pipe diameter increases and then remains constant. P i is the pressure and v i is the speed of the fluid, at locations i u, y, z, and w. u y Bernoulli s equation P + 1 ρ v + ρ y constant. At a lower elevation the pressure P ρ y is reater. Thus, P z > P y. 006 part 3 of 4 10 points position u and w is 1. P w > P u. z w. indeterminable, not enouh information correct 3. P w P u. The relationship between the manitude of the velocity v v at position u and y is 1. v y v u.. indeterminable, not enouh information 3. v y < v u. correct 4. v y > v u. The fluid is incompressible, so v u > v y. 005 part of 4 10 points position y and z is 1. indeterminable, not enouh information. P z < P y. 3. P z P y. 4. P z > P y. correct 4. P w < P u. The pressure at u and w may be comparable. Not enouh information is available to determine which is larer, if they are not equal. Both P u < P y and P w < P y, see Parts 4 &. 007 part 4 of 4 10 points position u and y is 1. P y P u.. P y < P u. 3. indeterminable, not enouh information 4. P y > P u. correct Usin Bernoulli s equation, we have P + 1 ρ v + ρ y constant. At a reater velocity the pressure P 1 ρ v is smaller. Since v u > v y, then P y > P u.
Cunninham, Drew Homework 3 Due: Apr 1 006, 4:00 am Inst: Florin 4 008 part 1 of 10 points Note: P atm 101300 Pa. The viscosity of the fluid is neliible and the fluid is incompressible. A liquid of density 153 k/m 3 flows with speed 1.0 m/s into a pipe of diameter 0.1 m. The diameter of the pipe decreases to 0.05 m at its exit end. The exit end of the pipe is 8.85 m lower than the entrance of the pipe, and the pressure at the exit of the pipe is 1 atm. The acceleration of ravity is 9.8 m/s. P 1 Therefore, v A 1 A π π d1 d1 d d 0.1 m 1.0 m/s 0.05 m 17.998 m/s. 1.0 m/s 0.1 m P v 1 atm 0.05 m 8.85 m What is the velocity v of the liquid flowin out of the exit end of the pipe? Correct answer: 17.998 m/s. 009 part of 10 points Applyin Bernoulli s principle, what is the pressure P 1 at the entrance end of the pipe? Correct answer: 194799 Pa. Applyin Bernoulli s principle to the fluid flow at the entrance and exit of the pipe ives P 1 + ρ y 1 + 1 ρ v 1 P + ρ y + 1 ρ v Let : 1.0 m/s, d 1 0.1 m, d 0.05 m. and Basic Concepts The continuity equation for incompressible fluids is A v constant. Bernoulli s equation for fluid flow alon streamlines is P + ρ h + 1 ρ v constant. Solution: The continuity equation tells us that the volume of liquid enterin the pipe durin a certain time interval must equal the volume of liquid leavin the pipe durin that time interval, so A 1 A v. P 1 P + ρ y y 1 + 1 ρ v. We also have y y 1 h, since the entrance heiht y 1 is reater than the exit heiht y. Therefore P 1 P ρ h + 1 ρ [v ] 101300 Pa 153 k/m 3 9.8 m/s 8.85 m + 1 153 k/m3 [17.998 m/s 1.0 m/s ] 194799 Pa. 010 part 1 of 1 10 points A jet of water squirts horizontally from a hole near the bottom of the tank.
Cunninham, Drew Homework 3 Due: Apr 1 006, 4:00 am Inst: Florin 5 The hole is 1.3 m above the floor and the stream of water extend a distance of 0.55 m before hittin the floor, as shown below. h 1.3 m 0.55 m If the hole has a diameter of 3.09 mm and the top of the tank is open, what is the heiht of the water in the tank? Correct answer: 0.0581731 m. Let : x 0.55 m and y 1.3 m Basic Concepts: P 1 + 1 ρ v 1 + ρ h 1 P + 1 ρ v + ρ h For the motion of the water after leavin the tank, Horizontally: since a x 0 m/s. Vertically: x v x t y t since v y,i 0 m/s. Solution: From the vertical equation y t, and from the horizontal equation v x x t x y Bernouli s equation simplifies to 1 ρ v 1 + ρ h 1 ρ h, since, for the open top of the tank and the hole, P 1 P P 0 P 0 atmospheric pressure, and v 0 m/s because the hole is small. since v x. ρ h h 1 1 ρ v x h h h 1 v x [ x y x 4 y 0.55 m 4 1.3 m 0.0581731 m ] 1 x y.