OA01 Basics and Fluids - Mastering Physics assignment PDF

Preview

Summary

Mastering Physics assignment OA01 Basics and Fluids...

Description

OA01 Basics and Fluids Due: 11:59pm on Sunday, August 23, 2020 You will receive no credit for items you complete after the assignment is due. Grading Policy

Converting between Different Units Unit conversion problems can seem tedious and unnecessary at times. However, different systems of units are used in different parts of the world, so when dealing with international transactions, documents, software, etc., unit conversions are often necessary. Here is a simple example. The inhabitants of a small island begin exporting beautiful cloth made from a rare plant that grows only on their island. Seeing how popular the small quantity that they export has been, they steadily raise their prices. A clothing maker from New York, thinking that he can save money by "cutting out the middleman," decides to travel to the small island and buy the cloth himself. Ignorant of the local custom of offering strangers outrageous prices and then negotiating down, the clothing maker accepts (much to everyone's surpise) the initial price of 400 tepizes /m 2. The price of this cloth in New York is 120 dollars /yard 2.

Part A If the clothing maker bought 500 m 2 of this fabric, how much money did he lose? Use 1 tepiz = 0.625 dollar and 0.9144 m = 1 yard. Express your answer in dollars using two significant figures.

Hint 1. How to approach the problem To find how much money the clothing maker loses, you must find how much money he spent and how much he would have spent in New York. Furthermore, since the problem asks how much he lost in dollars, you need to determine both in dollars. This will require unit conversions. Hint 2. Find how much he paid If the clothing maker bought 500 m 2 at a cost of 400 tepizes /m 2, then simple multiplication will give how much he spent in tepizes. Once you've found that, convert to dollars. How much did the clothing maker spend in dollars? Express your answer in dollars to three significant figures.

Hint 1. Find how much he paid in tepizes If the clothing maker bought 500 m 2 at a cost of 400 tepizes /m 2, then how much did he pay in total, in tepizes? Express your answer in tepizes. ANSWER: 2.00×105 tepizes

ANSWER: 1.25×105 dollars

Hint 3. Find the price in New York You know that the price of the fabric in New York is 120 dollars /yard 2. Thus, you need only to find the number of square yards that the clothing maker purchased and then multiply to find the price in New York. What would it have cost him to buy the fabric in New York? Express your answer in dollars to three significant figures.

Hint 1. Determine how much cloth he bought in yard 2 You are given that 0.9144 m = 1 yard. Squaring both sides, you would get that 0.8361 m 2 = 1 yard 2. How much is 500 m2? Express your answer in yard2 to three significant figures. ANSWER: 598 yard 2

ANSWER: 7.18×104 dollars

ANSWER: 5.3×104 dollars

Correct

Still think that unit conversion isn't important? Here is a widely publicized, true story about how failing to convert units resulted in a huge loss. In 1998, the Mars Climate Orbiter probe crashed into the surface of Mars, instead of entering orbit. The resulting inquiry revealed that NASA navigators had been making minor course corrections in SI units, whereas the software written by the probe's makers implicitly used British units. In the United States, most scientists use SI units, whereas most engineers use the British, or Imperial, system of units. (Interestingly, British units are not used in Britain.) For these two groups to be able to communicate to one another, unit conversions are necessary. The unit of force in the SI system is the newton (N), which is defined in terms of basic SI units as 1 N = 1 kg m /s 2. The unit of force in the British system is the pound (lb), which is defined in terms of the slug (British unit of mass), foot (ft), and second (s) as 1 lb = 1 slug ft /s 2.

Part B Find the value of 15.0 N in pounds. Use the conversions 1 slug = 14.59 kg and 1 ft = 0.3048 m. Express your answer in pounds to three significant figures.

Hint 1. How to approach the problem

When doing a unit conversion, you should begin by comparing the units you are starting with and the units you need to finish with. In this problem, we have the following: Starting units Final units kg m

slug ft

s2

s2

Notice that both have seconds squared in the denominator. You will only have to change the units in the numerator. Match up the units that measure the same quantity (e.g., kilograms and slugs both measure mass). Once you've done this, create a fraction (e.g., 1 hour/60 minutes) based on conversion factors such that the old unit is canceled out of the expression and the new unit appears in the position (i.e., numerator or denominator) of the old unit. In this problem, there are two pairs within the starting and final units that must be converted in this way (i.e., kilograms/slugs and meters/feet). Hint 2. Calculate the first conversion The first step is to eliminate kilograms from the expression for newtons in favor of slugs. What is the value of 15 kg m /s 2 in slug m /s 2? Express your answer in slug-meters per second squared to four significant figures. ANSWER: 1.028

ANSWER: 15.0 N = 3.37 lb

Correct Thus, if the NASA navigators believed that they were entering a force value of 15 N (3.37 lb), they were actually entering a value nearly four and a half times higher, 15 lb. Though these errors were only in tiny course corrections, they added up during the trip of many millions of kilometers. In the end, the blame for the loss of the 125-million-dollar probe was placed on the lack of communication between people at NASA that allowed the units mismatch to go unnoticed. Nonetheless, this story makes apparent how important it is to carefully label the units used to measure a number.

Problem 1.47 An average family of four uses roughly 1200 L (about 300 gallons) of water per day ( 1 L = 1000 cm3).

Part A How much depth would a lake lose per year if it uniformly covered an area of 40 km2 and supplied a local town with a population of 70000 people? Consider only population uses, and neglect evaporation and so on. Express your answer using one significant figure. ANSWER: d = 20 cm / y

Correct

18-Karat Gold vs. 14-Karat Gold Gold in its pure form is too soft to be used for most jewelry. Therefore, the gold is mixed with other metals to produce an alloy. The composition of gold alloys are always calculated by mass, using the karat (kt) as a unit of measure. A karat represents a proportion by mass of one part in twenty-four. The higher the karat value, the higher the proportion of gold in relation to the total metal content. Pure gold is therefore 24 karat, while an 18-karat gold alloy contains (at least) 18 parts (by mass) of gold out of 24 parts total.

Part A In a sample of 18-karat gold, 75 percent of the total mass is pure gold, while the rest is typically 16 percent silver and 9 percent copper. If the density of pure gold is ρ gold = 19.3 g /cm3, while the densitites of silver and copper are respectively ρ silver = 10.5 g / cm3 and ρ copper = 8.90 g / cm3, what is the overall density ρ 18ktrho_18kt of this alloy of 18-karat gold? Express your answer in grams per cubic centimeter to three significant figures.

Hint 1. How to approach the problem To compute the density of a material, you need to know its mass per unit volume, or alternatively, the volume of a given mass of that material. In this problem you are given information based on percentages by weight of each metal content of 18-karat gold. Therefore, in order to determine the density of the alloy, it is most convenient to find the total volume of a given mass of 18-karat gold, say 1 gram. The total volume is simply the sum of the volumes of each metal component in the alloy; however, before evaluating the volume of each metal, first you need to determine each metal's mass in one gram of alloy. Hint 2. Find the volume of pure gold in 1 gram of alloy What is the volume V goldV_gold of pure gold in 1 gram of 18-karat gold? Express your answer numerically in cubic centimeters.

Hint 1. How are mass, volume, and density related? Consider a piece of a homogeneous material of mass mm whose volume is VV. The density ρrho of that material is given by ρ=

m V

.

If the mass mm and the density ρrho of a material are known, you can then determine the volume VV. Write an expression for volume in terms of mass and density. Express your answer in terms of ρrho and mm. ANSWER: VV =

m ρ

Hint 2. Find the mass of pure gold in 1 gram of alloy

Consider a sample of 18-karat gold; 75 percent of the total mass is pure gold. What is the mass m goldm_gold of pure gold contained in 1 gram of 18-karat gold? Express your answer numerically in grams. ANSWER: m goldm_gold = 0.75 g

ANSWER: V goldV_gold = 3.89×10−2 cm 3

ANSWER: ρ 18ktrho_18kt = 15.6 g /cm 3

Correct

Part B The percentage composition by mass of the less expensive 14-karat gold is typically 58.5 percent gold, 4.0 percent silver, 31.2 percent copper, and 6.3 percent zinc. Find the percentage composition by volume of 14-karat gold. The density of zinc is ρ zinc = 7.14 g /cm 3. Enter four answers as percentages, separated by commas. Round each to the nearest integer.

Hint 1. How to approach the problem Find the volume of each metal present in a gram of alloy and add the volumes to obtain the total volume. The percentage composition by volume of a metal is just the volume of that metal in the alloy divided by the total volume of the alloy. Hint 2. Find the volume of pure gold in 1 gram of alloy What is the volume V goldV_gold of pure gold in 1 gram of 14-karat gold? Express your answer numerically in cubic centimeters, to three significant figures.

Hint 1. How are mass, volume, and density related? Consider a piece of a homogeneous material of mass mm whose volume is VV. The density ρrho of that material is given by ρ=

m V

.

If the mass mm and the density ρrho of a material are known, you can then determine the volume VV. Write an expression for volume in terms of mass and density. Express your answer in terms of ρrho and mm.

ANSWER: VV =

m ρ

Hint 2. Find the mass of pure gold in 1 gram of alloy Consider a sample of 14-karat gold; 58.5 percent of the total weight is pure gold. What is the mass m goldm_gold of pure gold contained in 1 gram of 14-karat gold? Express your answer numerically in grams. ANSWER: m goldm_gold = 0.585 g

ANSWER: V goldV_gold = 3.03×10−2 cm 3

ANSWER: V %gold, V %silver, V %copper, V %zinc = 39,5,45,11 %

Correct In a sample of 14-karat gold, less than 50 percent of the volume is pure gold!

Video Tutor: Pressure in Water and Alcohol First, launch the video below. You will be asked to use your knowledge of physics to predict the outcome of an experiment. Then, close the video window and answer the question on the right. You can watch the video again at any point.

Part A

Rank, from smallest to largest, the pressures in the tank of motionless fluid shown in the figure.

To rank items as equivalent, overlap them.

Hint 1. How to approach the problem Recall that the pressure at any point in an open container of fluid is due to the weight of the overlying fluid plus the weight of the overlying air. Also, think about the fact that this fluid is motionless. What does that tell you about the pressure at any two points on the same horizontal level, such as points B and C?

ANSWER:

Reset











Help



B A

C

E

F

D

The correct ranking cannot be determined.

Correct Pressure depends on vertical distance from the fluid surface. As we descend deeper in the fluid, the pressure increases.

Fluid Pressure in a U-Tube

A U-tube is filled with water, and the two arms are capped. The tube is cylindrical, and the right arm has twice the radius of the left arm. The caps have negligible mass, are watertight, and can freely slide up and down the tube.

Part A A one-inch depth of sand is poured onto the cap on each arm. After the caps have moved (if necessary) to reestablish equilibrium, is the right cap higher, lower, or the same height as the left cap?

Hint 1. Pressure at the surface The pressure at the base of each arm depends on the pressure at the surface of each arm, not the weight at the surface of each arm. Hint 2. Evaluate the weight of the sand Compare the weight of the sand on the two caps. Which of the following is true? ANSWER: greater than The weight of the sand on the right cap is

less than equal to

the weight of the sand on the left cap.

Hint 3. Evaluate the pressure applied by the sand Now compare the pressure exerted by the sand on the two caps. Which of the following is true? ANSWER: greater than The pressure exerted by the sand on the right cap is

less than equal to

the pressure exerted by the sand on the left cap.

ANSWER: higher lower the same height

Correct Although one inch of sand on the right cap is much heavier than one inch of sand on the left cap, the pressures exerted by the sand are the same on both caps. Since the pressures exerted by the sand are equal, the pressures at the base of each arm due to the water must be equal. This requires equal heights of water in the two arms.

Part B The sand is removed and a 1.0-kg-mass block is placed on each cap. After the caps have moved (if necessary) to reestablish equilibrium, is the right cap higher, lower, or the same height as the left cap?

Hint 1. Evaluate the pressure applied by the 1.0-kg blocks Compare the pressure exerted on the water from the two 1-kg blocks. Which of the following is true? ANSWER:

greater than The pressure exerted on the right cap is

less than

the pressure exerted on the left cap.

equal to

ANSWER: higher lower the same height

Correct Although the masses of the blocks are equal, the pressures exerted by them on the caps are not equal. There is a greater pressure on the left cap, which results in a greater pressure at the base of the left arm. To compensate for this increased pressure, the height of the water column in the right arm will have to be greater than in the left arm.

Part C If a 1.0-kg-mass block is on the left cap, how much total mass must be placed on the right cap so that the caps equilibrate at equal height? Express your answer in kilograms.

Hint 1. Meaning of equal water levels Equal water levels in the two arms require equal pressures on the two caps. Remember that pressure is force divded by area and that area is proportional to the square of the radius.

ANSWER: 4.0 kg

Correct

Part D The locations of the two caps at equilibrium are now as given in this figure. The dashed line represents the level of the water in the left arm. What is the mass of the water located in the right arm between the dashed line and the right cap? Express your answer in kilograms.

Hint 1. Pressure at the dashed line The pressure at the location of the dashed line must be equal in the two arms. Therefore, the total mass above the line in the left arm, divided by the area of the left arm, must equal the total mass above the line in the right arm, divided by the area of the right arm.

ANSWER: 3.0 kg

Correct

± Buoyant Force Conceptual Question A rectangular wooden block of weight WW floats with exactly one-half of its volume below the waterline.

Part A What is the buoyant force acting on the block?

Hint 1. Archimedes' principle The upward buoyant force on a floating (or submerged) object is equal to the weight of the liquid displaced by the object. Mathematically, the buoyant force F buoyantF_buoyant on a floating (or submerged) object is F buoyant = ρgV, where ρrho is the density of the fluid, VV is the submerged volume of the object, and gg is the acceleration due to gravity.

Hint 2. What happens at equilibrium The block is in equilibrium, so the net force acting on it is equal to zero.

ANSWER:

2W WW 1 2

W

The buoyant force cannot be determined.

Correct

Part B The density of water is 1.00 g /cm 3. What is the density of the block?

Hint 1. Applying Archimedes' principle In Part A, you determined that the buoyant force, and hence the weight of the water displaced, is equal to the weight of the block. Notice, however, that the volume of the water displaced is one-half of the volume of the block. Hint 2. Density The density ρrho of a material of mass mm and volume VV is ρ=

ANSWER:

m V

.

2.00 g /cm 3 between 1.00 and 2.00 g /cm 3 1.00 g /cm 3 between 0.50 and 1.00 g /cm 3 0.50 g /cm 3 The density cannot be determined.

Correct

Part C Masses are stacked on top of the block until the top of the block is level with the waterline. This requires 20 g of mass. What is the mass of the wooden block?

Hint 1. How to approach the problem When you add the extra mass on top of the block, the buoyant force must change. Relating the mass to the new buoyant force will allow you to write an equation to solve for mm. Hint 2. Find the new buoyant force With the 20-g mass on the block, twice as much of the block is underwater. Therefore, what happens to the buoyant force on the block? ANSWER: The buoyant force doubles. The buoyant force is halved. The buoyant force doesn't change.

Hint 3. System mass Before the 20-g mass is placed on the block, the mass of the system is just the mass of the block (mm), and this mass is supported by the buoyant force. With the 20-g mass on top, the total mass of the system is now m + 20 g. Hint 4. The mass equation The original buoyant force was equal to the weight WW of the block. If the new buoyant force F bF_b is twice the old buoyant force, then F b = 2W = 2mg, where mm is the mass of the block. The new buoyant force must support the weight of the block and the mass, so according to Newton's 2nd law F b = (m + 20)g.

By setting these two expressions for F bF_b equal to each other, you can solve for mm.

ANSWER: 40 g 20 g 10 g

Correct

Part D The wooden block is removed from the water bath and placed in an unknown liquid. In this liquid, only one-third of the wooden block is submerged. Is the unknown liquid more or less dense than water? ANSWER: more dense less dense

Correct

Part E What is the density of the unknown liquid ρ unknownrho_unknown? Express your answer numerically in grams per cubic centimeter.

Hint 1. Comparing densities From Part C, the mass of the block is 20 g. In water, one-half of the block is submerged, so one-half of the volume of the block times the density of the water must be equivalent to 20 g. In the unknown substance, one-third of the block is submerged, so one-third of the volume of the block times the density of the unknown substance must be equivalent to 20 g. You can use these observations to write an equation for the density of the unknown substance. Hint 2. Setting up the equation The buoyant force on the block in water F b , waterF_{b,water} is the same as the buoyant force in the unknown liquid F b , unknownF_{b,unknown}, F b , water = F b , unknown, so

ρ waterg

() V 2

= ρ unknowng

() V 3

,

where ...