CH 13 HW - Chapter 13 Physics Homework for Mastering PDF

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Chapter 13 Physics Homework for Mastering

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CH 13 HW Due: 3:00pm on Monday, December 9, 2019 You will receive no credit for items you complete after the assignment is due. Grading Policy

An Iron Rod

Part A On a part-time job, you are asked to bring a cylindrical iron rod of density 7800 room to a machinist. Calculate the weight of the rod,

, length 79.6

. The acceleration due to gravity,

= 9.81

and diameter 2.25

from a storage

.

Express your answer numerically in newtons.

Hint 1. General approach To calculate the weight of a body you need to know its mass. If the mass is unknown, and instead the problem gives you information on the volume of the body and the density of the material of which it is made, you can calculate the mass of that body by applying the definition of density. Hint 2. Weight of a body A body of mass

has weight with magnitude

given by ,

where

= 9.81

is the acceleration due to gravity.

Hint 3. Find the mass of the rod From the length and the cross-sectional area of the rod and the density of iron, find the mass

of the rod.

Express your answer numerically in kilograms.

Hint 1. Finding mass from volume and density Consider a mass

of homogeneous material, whose volume is

. The density .

Hint 2. Volume of a cylinder The volume

of a cylinder of length

and cross-sectional area

If the cross section is a circle with radius , then

ANSWER: = 2.47

ANSWER: = 24.2

Correct

P

Typesetting math: 78%

is given by

of that material is defined as

Now that you know the weight of the rod, do you think that you will be able to carry the rod without a cart?

Hint 1. Weight and mass A weight measured in newtons may not be very familiar to you; therefore it is more convenient to determine the corresponding mass measured in kilograms, or in pounds. Hint 2. Unit conversion factors The unit convertion factors for newtons, kilograms, and pounds are , .

ANSWER: yes no

Correct

Pressure in the Ocean The pressure at 10

below the surface of the ocean is about 2.00×105

.

Part A Which of the following statements is true?

Hint 1. How to approach the problem Since pressure is defined as force per unit area, the pressure at a given depth below the surface of the ocean is the normal force on a small area at that depth divided by that area. Thus, given the pressure at 10 you can determine the normal force on a unit area at that depth. Also recall that, in general, the pressure in a fluid varies with height, but it also depends on the external pressure applied to the fluid. In the case of the ocean, the external pressure applied at its surface is the atmospheric pressure.

ANSWER:

The weight of a column of seawater 1

in cross section and 10

high is about 2.00×105

The weight of a column of seawater 1

in cross section and 10

high plus the weight of a column of air with the same cross

section extending up to the top of the atmosphere is about 2.00×105 The weight of 1

of seawater at 10

.

.

below the surface of the ocean is about 2.00×105

.

The density of seawater is about 2.00×105 times the density of air at sea level.

Correct The pressure at a given level in a fluid is caused by the weight of the overlying fluid plus any external pressure applied to the fluid. In this case, the external pressure is the atmospheric pressure. Tropical storms are formed by low-pressure systems in the atmosphere. When such a storm forms over the ocean, both the atmospheric pressure and the fluid pressure in the ocean decrease.

Part B Typesetting math: 78%

Now consider the pressure 20

below the surface of the ocean. Which of the following statements is true?

Hint 1. Variation of pressure with height If the pressure at a point in a fluid is

where is the density of the fluid and height of pressure is greater than height and 1 in cross section.

, then, at a distance

below this point, the pressure

in the fluid is

, is the acceleration due to gravity. This means that the pressure at a depth below a reference by an amount . Note that the quantity is related to the weight of a column of fluid of

ANSWER:

The pressure is twice that at a depth of 10

.

The pressure is the same as that at a depth of 10 The pressure is equal to that at a depth of 10 The pressure is equal to the weight per 1

.

plus the weight per 1

cross sectional area of a column of seawater 10

cross sectional area of a column of seawater 20

high.

high.

Correct

Crown of Gold? According to legend, the following challenge led Archimedes to the discovery of his famous principle: Hieron, king of Syracuse, was suspicious that a new crown that he had received from the royal goldsmith was not pure gold, as claimed. Archimedes was ordered to determine whether the crown was in fact made of pure gold, with the condition that only a nondestructive test would be allowed. Rather than answer the problem in the politically most expedient way (or perhaps extract a bribe from the goldsmith), Archimedes thought about the problem scientifically. The legend relates that when Archimedes stepped into his bath and caused it to overflow, he realized that he could answer the challenge by comparing the volume of water displaced by the crown with the volume of water displaced by an amount of pure gold equal in weight to the crown. If the crown was made of pure gold, the two volumes would be equal. If some other (less dense) metal had been substituted for some of the gold, then the crown would displace more water than the pure gold. A similar method of answering the challenge, based on the same physical principle, is to compute the ratio of the actual weight of the crown, and the apparent weight of the crown when it is submerged in water, . See whether you can follow in Archimedes' footsteps. The figure shows what is meant by weighing the crown while it is submerged in water.

Part A Take the density of the crown to be

. What is the ratio of the crown's apparent weight (in water)

Express your answer in terms of the density of the crown

and the density of water

Hint 1. Find an expression for the actual weight of the crown Typesetting math: 78%

e crown has volume

. Find the actual weight

of the crown.

.

to its actual weight

?

,

Express

in terms of

,

(the magnitude of the acceleration due to gravity), and

.

ANSWER: =

Hint 2. Find an expression for the apparent weight of the crown Assume that the crown has volume submerged in water.

, and take the density of water to be

Express your answer in terms of

,

. Find the apparent weight

(the magnitude of the acceleration due to gravity),

, and

of the crown .

Hint 1. How to approach the problem The apparent weight of the crown when it is submerged in water will be less than its actual weight (weight in air) due to the buoyant force, which opposes gravity. Hint 2. Find an algebraic expression for the buoyant force. Find the magnitude

of the buoyant force on the crown when it is completely submerged in water.

Express your answer in terms of

,

, and the gravitational acceleration .

ANSWER: =

ANSWER: =

ANSWER:

=

Correct

Part B Imagine that the apparent weight of the crown in water is , and the actual weight is pure (100%) gold? The density of water is grams per cubic centimeter. The density of gold is

. Is the crown made of grams per cubic centimeter.

Hint 1. Find the ratio of weights for a crown of pure gold Given the expression you obtained for

, what should the ratio of weights be if the crown is made of pure gold?

Express your answer numerically, to two decimal places. ANSWER:

= 0.95

ANSWER: Typesetting math: 78%

Yes No

Correct For the values given,

, whereas for pure gold,

. Thus, you can

conclude that the the crown is not pure gold but contains some less-dense metal. The goldsmith made sure that the crown's (true) weight was the same as that of the amount of gold he was allocated, but in so doing was forced to make the volume of the crown slightly larger than it would otherwise have been.

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:

Typesetting math: 78%

greater than The weight of the sand on the right cap is

less than

the weight of the sand on the left cap.

equal to

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

the pressure exerted by the sand on the left cap.

equal to

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- -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-

blocks

Compare the pressure exerted on the water from the two 1ANSWER:

Typesetting math: 78%

blocks. Which of the following is true?

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-

-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

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.

Typesetting math: 78%

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

Correct

Force on a Goldfish Vector Drawing A fishbowl contains a single goldfish and is filled with water to the level indicated.

Part A At each of the designated points, rotate the given vector to indicate the direction of the force exerted by the water on either the inside of the fishbowl (for points A and B) or the outside of the goldfish (for points C, D, and E). The orientation of your vectors will be graded.

Hint 1. Direction of the force due to the water The water in the fishbowl will exert a force perpendicular to every surface it contacts. The pressure of the water, like the pressure in any fluid, pushes the water outward, equally in all directions, exerting forces on whatever it comes into contact with.

ANSWER:

Typesetting math: 78%

No elements selected

Select the elements from the list and add them to the canvas setting the appropriate attributes.

Correct

Testing Experiment Table 13.3 It was known in Torricelli's time that a suction pump could not lift water higher than 10.3 . To test why this is, he filled a meter long glass tube closed at one end with mercury, which is 13.6 times denser than water. He then put his finger over the open end and placed it upside down in a dish filled with mercury, as shown in from Testing Experiment Table 13.3.

Part A Typesetting math: 78%

He makes the hypothesis that the atmospheric pressure limits the height of the liquid in this arrangement. Which of the following predictions are consistent with that hypothesis? Select all that apply. ANSWER:

When the tube is placed upside down, mercury will not leak out as atmospheric pressure will keep it within the tube. When the tube is placed upside down, mercury will completely leak out as this is what happens when we turn containers with liquids upside down. Mercury will not leak out of the tube as the length of the tube is less than 10.3

, same as water.

Some of mercury will leak out but most will stay. The length of the remaining column should be less than 0.1 of the length of the water column in the suction pump experiments. The height of the remaining mercury in the tube should always be 760

.

All attempts used; correct answer displayed This is because the mercury has a higher density compared to water and so the atmosphere can only hold up approximately 1/13.6 as much. The exact amount depends on atmospheric conditions and so the height will not always be 760 .

Block Suspended in Water Conceptual Question A flask of water rests on a scale that reads 100 . Then, a small block of unknown material is held completely submerged in the water. The block does not touch any part of the flask, and the person holding the block will not tell you whether the block is being pulled up (keeping it from falling further) or pushed down (keeping it from bobbing back up). Assume that no water spills out of the flask.

Part A What is the new reading on the scale?

Hint 1. Archimedes' principle The upward buoyant force on a submerged object is equal to the weight of the liquid displaced by the object. Hint 2. Newton’s 3rd law applied to the buoyant force According to Newton's 3rd law, for each force that acts on the block, a force equal in magnitude and opposite in direction acts on its surroundings. Since the water exerts an upward buoyant force on the block, by Newton’s 3rd law a downward force is exerted on the water by the block.

ANSWER: Typesetting math: 78%

greater than 100 100 less than 100 It is impossible to determine.

Correct

The experiment is repeated with the six different blocks listed below. In each case, the blocks are held completely submerged in the water. Mass ( )

Volume (

A

100

50

B

100

200

C

200

50

D

50

100

E

200

100

F

400

50

)

Part B Rank these blocks on the basis of the scale reading when the blocks are completely submerged. Rank from largest to smallest. To rank items as equivalent, overlap them.

Hint 1. Buoyant force on a submerged block The buoyant force

on a submerged object is ,

where

is the density of the fluid and

ANSWER:

Typesetting math: 78%

is the volume of the submerged object.

Reset

Help

C D B

F E A

The correct ranking cannot be determined.

Correct

Part C If the blocks were released while submerged, which, if any, would sink to the bottom of the flask? Enter the correct letters from the table in alphabetical order without commas or spaces (e.g., ABC).

Hint 1. Density of water The density of water is 1.0

. Only objects of lesser or equal density can float in water.

ANSWER: ACEF

Correct

A Water Tank on Mars You are assigned the design of a cylindrical, pressurized water tank for a future colony on Mars, where the acceleration due to gravity is 3.71 . The pressure at the surface of the water will be 105 , and the depth of the water will be 13.5 . The pressure of the air outside the tank, which is elevated above the ground, will be 93.0 {\rm kPa} .

Part A Find the net downward force on the tank's flat bottom, of area 1.60 {\rm m^2} , exerted by the water and air inside the tank and the air outside the tank. Assume that the density of water is 1.00 \rm g/cm^3. Express your answer in newtons. Typesetting math: 78%

Hint 1. The net force The net force on the tank's flat bottom is the sum of the (downward) force exerted on the bottom of the tank by the water and the (upward) force exerted on the bottom of the tank by the air outside the tank. Hint 2. What is a \rm Pa (pascal)? The SI unit of pressure, a Pascal, is a force per unit area: 1 \rm Pa = 1 \rm N/m^2. Hint 3. Find the force exerted on the tank's bottom by the air outside the tank Write an expression for the force exerted on the tank's bottom by the air outside the tank. Express your answer in newtons.

Hint 1. Pressure and force The pressure \texttip{p}{p} in a gas is defined as the normal force \texttip{F}{F} exerted by the gas on a surface in contact with it. If the force is the same at all points of a finite plane surface with area \texttip{A}{A}, then the pressure is uniform and given by \large{p=\frac{F}{A}}.

ANSWER: 1.49×105

{\rm N}

Hint 4. Find the force exerted on the tank's bottom by the water Write an expression for the force exerted on the t...