ANSWER THE GUIDE QUESTIONS USING THE GIVEN DATA. Data and Observations:Table1. Buoyancy on WaterMass ofWeight Difference(N)displacedWeight ofdisplacedWeight inVolume of displacedWeight in Air(N)waterMaterialWaterwater(Weight in air -Weight in water)(kg)|(mass= volumex Puater )m= 0.000054m x 1000kg/ m3water(N)(m³)(N)54mlStandard Cork 0.71 NON0.71 N0.71 N=0.000054 mm= 0.054 kgm= 0.000104m x 1000kg/ m?m= 0.104 kgDisc ofAluminium|104 ml=0.000104 m3.70 N2.33 N1.37 N1.37 Nm= 0.000404m3 x 1000kg/ m?m= 0.404 kgDish ofAluminium404 ml4.41 N0.1 N4.31 N4.31 Nattached to a=0.000404 m3piece of cork Guide Questions:1. What is the relationship of mass, volume and density?2. At what material does the object float? Why?3. What causes the object float or sink?Conclusion and Recommendation:

Name: ANSWER THE GUIDE QUESTIONS USING THE GIVEN DATA. Data and Observations
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Description: ANSWER THE GUIDE QUESTIONS USING THE GIVEN DATA. Data and Observations:Table1. Buoyancy on WaterMass ofWeight Difference(N)displacedWeight ofdisplacedWeight inVolume of displacedWeight in Air(N)waterMaterialWaterwater(Weight in air -Weight in water)(

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Answered: Guide Questions: 1. What is the…

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Guide Questions: 1. What is the relationship of mass, volume and density? 2. At what material does the object float? Why? 3. What causes the object float or sink?

College Physics

1st Edition

ISBN:9781938168000

Author:Paul Peter Urone, Roger Hinrichs

Publisher:Paul Peter Urone, Roger Hinrichs

Chapter11: Fluid Statics

Section: Chapter Questions

Problem 58PE: Figure 11.35(a) shows the effect of tube radius on the height to which capillary action can raise a...

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Oil having a density of 930 kg/m3 floats on water. A rectangular block of wood 4.00 cm high and with a density of 960 kg/m3 floats partly in the oil and partly in the water. The oil completely covers the block. How far below the interface between the two liquids is the bottom of the block?
Referring to Figure 14.20, prove that the buoyant force on be cylinder is equal to the weight of the fluid displaced (Archimedes' principle). You may assume that be buoyant force is F2F1 and that the ends of be cylinder have equal areas A. Note Eat be volume of be cylinder (and that of be fluid displaces) equals (h2h1)A .
What effect does capillary action have on the reading of a manometer with uniform diameter? Explain your answer.
Fluid originally flows through a tube at a rate of 100 cm3/s. To illustrate the sensitivity of flow rate to various factors, calculate be new flow rate for following changes with all other factors remaining the same as in original conditions. (a) Pressure difference increases by a factor of 1.50. (b) A new fluid wit 3.00 times greater viscosity is substituted. (c) The tube is replaced by one having 4.00 times the length. (d) Another tube used with a 0.100 times the original. (e) Yet another tube is substituted with a radius 0.100 times the original and half length, and pressure difference is increased by factor of 1.50.
Figure 11.35(a) shows the effect of tube radius on the height to which capillary action can raise a fluid. (a) Calculate the height h for water in a glass tube with a radius of 0.900 cm—a rather large tube like the one on the left. (b) What is the radius of the glass tube on the right if it raises water to 4.00 cm?
Find the gauge and absolute pressures in the balloon and peanut jar shown in Figure 11.16, assuming the manometer connected to the balloon uses water whereas the manometer connected to the jar contains mercury. Express in units of centimeters of water for the balloon and millimeters of mercury for the jar, taking h=0.0500 m for each. Figure 11.16 An open-tube manometer has one side open to the atmosphere. (a) Fluid depth must be the same on both sides, or the pressure each side exerts at the bottom will be unequal and there will be flow from the deeper side. (b) A positive gauge pressure Pg=hg transmitted to one side of the manometer can support a column of fluid of height h. (c) Similarly, atmospheric pressure is greater than a negative gauge pressure Pg by an amount hg .The jar's rigidity prevents atmospheric pressure from being transmitted to the peanuts.
The hydraulic system of a backhoe is used to lift a load as shown in Figure 11.45. (a) Calculate the force F the slave cylinder must exert to support the 400-kg load and the 150-kg brace and shovel. (b) What is the pressure in the hydraulic fluid if the slave cylinder is 2.50 cm in diameter? (c) What force would you have to exert on a lever with a mechanical advantage of 5.00 acting on a master cylinder 0.800 cm in diameter to create this pressure? Figure 11.45 Hydraulic and mechanical lever systems are used in heavy machinery such as this back hoe.
What must be the contact area between a suction cup (completely evacuated) and a ceiling if the cup is to support the weight of an 80.0-kg student?
(a) Using Bernoulli's equation, show that the measured fluid speed v for a pitot tube, like the one in Figure 12.7(b), is given by v=( 2gh)1/2 , where h is the height of the manometer fluid, is the density of the manometer fluid, is the density of the moving fluid, and g is the acceleration due to gravity. (Note that v is indeed proportional to the square root of h, as stated in the text.) (b) Calculate v for moving air if a mercury manometer's h is 0.200 m. Figure 12.7 Measurement of fluid speed based on Bernoulli's principle. (a) A manometer is connected to two that are close together and small enough not to disturb the flow. Tube 1 is open at the end facing the flow. A dead spot having zero speed is created there. Tube 2 has an opening on the side, and so the fluid has a speed V across the opening; thus, pressure there drops. The difference in pressure at the manometer is 12v22 , and so h is proportional to 12v22 . (b) This type of velocity measuring device is a Prandtl tube, also known as a pitot tube.
A table-tennis ball has a diameter of 3.80 cm and average density of 0.084 0 g/cm3. What force is required to hold it completely submerged under water?
Is there a limit to the height to which an entrainment device can raise a fluid? Explain your answer.
Show that the total force on a rectangular dam due to the water behind it increases with the square of the water depth. In particular, show that this force is given by F=gh2L/2 , where is the density of water, h is its depth at the dam, and L is the length of the dam. You may assume the face of the dam is vertical. (Hint: Calculate the average pressure exerted and multiply this by the area in contact with the water. (See Figure 11.42.)