physics volume 1 5th edition

Physics 5Th Edition David Halliday, Robert Resnick, Kenneth S. Krane

Chapter 1 Multiple Choice 1

Step 1 of 4

Solution:

Length of the sheet,

Width of the sheet,

The number of significant figures in the given length and width is three.

Thus, the surface area of the sheet should be expressed in three significant figures.

The surface area of the given sheet is equal to its length times its width.

Thus, the surface area is as follows:

Substitute for and for in above equation,

Round off to three significant figures, the surface area of the sheet is .

Therefore, the option (D) is correct.

Step 2 of 4

The value of area has five significant figures. Thus, it cannot be the right answer. So, option (A) is not correct.

Step 3 of 4

The value of area has four significant figures. Thus, it cannot be the right answer. Thus, option (B) is not correct.

Step 4 of 4

The value of area has three significant figures. Thus, it cannot be the right answer. Hence, option (C) is not correct.


Step 1 of 5

Solution:

Thickness of the stack of 80 sheets measured by vernier calipers, .

Number of sheets. .

The number of significant digits in the given length of the stack of sheets is three.

The number of significant digits in the given number of sheets is one.

The thickness of the single sheet should be expressed in least number of significant digits.

Thus, the thickness of the single sheet should be expressed in one significant digit.

Step 2 of 5

If there is no gap in between the sheets, the thickness of the individual sheet is equal to the thickness of the stack of sheets divided by the number of sheets.

Thus, the thickness of a single sheet is as follows:

Substitute for and for in above equation,

Round off to one significant digit, to get the thickness of the single sheet that is .

Therefore, the option (D) is correct.

Step 3 of 5

The given value of the thickness of single sheet is expressed in five significant digits. So, option (A) is not correct.

Step 4 of 5

The given value of the thickness of single sheet is expressed in three significant digits. Thus, option (B) is not correct.

Step 5 of 5

The given value of the thickness of single sheet is expressed in two significant digits.

So, option (C) is not correct.


Step 1 of 5

Solution:

It is given that dimensions of mass , force constant , and amplitude are as follows:

Step 2 of 5

The period of oscillation of the nonlinear oscillator is as follows:

…… (1)

Here are the exponents.

It is known that the dimension of period is time.

Equation (1) gives

Step 3 of 5

Exponents of M:

…… (2)

Exponents of L:

…… (3)

Exponents of T:

Or,

Use equation (2), the value of is as follows:

Use equation (3), to get the value of as follows:

Step 4 of 5

Put the values of in equation (1),

This represents the oscillation of the nonlinear oscillator.

Therefore, the option (C) is correct.

Step 5 of 5

Other options are ruled out.

Chapter 1 Exercise 1

Step 1 of 12

Solution:

a) The prefix used for factor is mega.

It is given that there are phones.

Thus, there are .

Step 2 of 12

b) The prefix used for factor is micro.

It is given there are phones.

Thus, there are .

Step 3 of 12

c) The prefix used for factor is deka.

It is given there are cards.

Thus, there are .

Step 4 of 12

d) The prefix used for factor is giga.

It is given there are lows.

Thus, there are .

Step 5 of 12

e) The prefix used for factor is tera.

It is given there are bulls.

Thus, there are .

Step 6 of 12

f) The prefix used for factor is deci.

It is given there are mates.

Thus, there are .

Step 7 of 12

g) The prefix used for factor is centi.

It is given there are pedes.

Thus, there are .

Step 8 of 12

h) The prefix used for factor is nano.

It is given there are Nannettes.

Thus, there are .

Step 9 of 12

i) The prefix used for factor is pico.

It is given there are boos.

Thus, there are .

Step 10 of 12

j) The prefix used for factor is atto.

It is given there are boys.

Thus, there are .

Step 11 of 12

k) The prefix used for factor is hecto.

It is given there are withits.

Thus, there are .

Step 12 of 12

l) The prefix used for factor is kilo.

It is given there are mocking birds.

Thus, there are .


Step 1 of 4

Solution:

a) The number of week in a year is roughly about 52 weeks.

It is given that the annual salary is $36K.

Now, the weekly salary is as follows:

Round off to three significant figures, to get the weekly equivalent of annual salary of 36K that

is .

Step 2 of 4

b) 1 year has 12 months.

It is given that the lottery awards of 10 megabucks are payable over 20 years.

In one year, the prize award is about

Step 3 of 4

In one month, the prize award is as follows:

1 buck is equivalent to 1 $.

Round off to three significant figures, to get the amount of money received through the lottery

Awards per month that is about .

Step 4 of 4

c) It is given that the hard disk of the computer has a capacity of 30 GB.

Also, it is given that for a single word 8 bytes are required.

Thus, the number of words that can be stored in 30 GB hard disk can be calculated as follows:

Therefore, the given 30 GB hard disk can store .


Step 1 of 4

Solution:

A century is equal to 100 years.

Thus,

Step 2 of 4

1 micro is equal to .

Thus, 1 microcentury is equal to

Step 3 of 4

Therefore, 1 microcentury is about .

Step 4 of 4

The duration of the lecture of Enrico Fermi is 50 minutes.

The percentage difference from Fermi’s approximation is as follows:

Therefore, the percentage difference from Fermi’s approximation is .


Step 1 of 3

Solution:

Distance between New York and Los Angeles,

Time difference between New York and Los Angeles,

The Earth is divided into 24 time zones. Thus, each zone is equal to1 hour.

The distance covered per 1 hour between the given two cities is as follows:

Step 2 of 3

This means that one time zone hour is equal to 1000 mi.

Step 3 of 3

The circumference of the Earth is equal to the distance covered in one time zone hour times the 24 time zones.

Now, the circumference of the Earth is as follows:

Therefore, the circumference of the Earth is .


Step 1 of 3

Solution:

It is given that the number for seconds in a year is times .

Take the value of up to four significant figures, so the number of seconds in a year is as follows:

Step 2 of 3

One year is equal to 365.25 days. 1 day is equal to 24 hours. 1 hour is equal to 3600 seconds.

Thus, the calculated value of second in a year is as follows:

Round off to five significant figures, to get the calculated number of seconds in a year that is:

.

Step 3 of 3

The percentage error for the number of seconds in a year as taken from the value of and calculated value is as follows:

Round off to two significant figures, to get the percentage error that is: .


Step 1 of 3

Solution:

a) It is given that one shake equals to .

Thus, the number of shake in one second is as follows:

Step 2 of 3

One year is equal to 365.25 days. 1 day is equal to 24 hours. 1 hour is equal to 3600 seconds.

Thus, the calculated value of second in a year is as follows:

Round off to five significant figures, to get the calculated number of seconds in a year that is

.

Therefore, there are more shakes in a second than seconds in a year.

Step 3 of 3

b) It is given that humans have existed for about and the universe is about .

Now, the fraction of time that humans existed when compared to that of the universe is as follows:

If the age of the universe is taken to be 1 day, then the number of seconds that humans have existed is as follows:

Therefore, the duration of time that human existed in the time scale is .


Step 1 of 3

Solution:

Maximum tolerable error in the distances of the winners can be determined as follows.

The speed of the runner is

Here, distance covered by the runner is and time taken by the runner is .

Now, the speed of the runner with longer time is

Here, distance covered by the runner with longer time is and time taken by the runner with

longer time is .

Also, the speed of the runner with shorter time is

Here, distance covered by the runner with shorter time is and time taken by the runner with

shorter time is .

Step 2 of 3

The speed of the runner with shorter time should be greater than the speed of the runner with longer time.

Substitute for and for in the above inequalities

Step 3 of 3

Consider the longer runner covered exactly one mile.

Substitute for in the inequality

This means that the runner with shorter time was indeed the winner.

The maximum tolerable error is

Therefore, the maximum tolerable error in laying out the distance between the runners is .


Step 1 of 1

Solution:

The number of minutes in a day is

It is given that the pendulum clock gains 1 min per day.

Therefore, it is needed to wait roughly to indicate the correct time.


Step 1 of 3

Solution:

Given data:

Age of the universe,

Time lasted for the shortest light that produced in the laboratory, .

Step 2 of 3

The logarithm average of the given time interval is

Use to get

Step 3 of 3

Use ,

Or

This represents the physically meaningful time interval approximately halfway between the two given time intervals.

Rounding off to three significant figures, the physical meaningful time interval is .


Step 1 of 2

Solution:

1 century is equal to 100 years.

It is given that the length of the day uniformly increases by 0.001 s in a century.

Thus, the increased in the day after 20 centuries is

This represents the increase in time in a day.

Step 2 of 2

The cumulative effect would be that half the product of the increased in time in a day and the number of days in 200 centuries. Thus, the cumulative effect is:

Therefore, the cumulative effect is that there would be an increase of in 200 centuries.


Step 1 of 3

Solution:

The total angle traced by the moon in one complete revolution around the Earth is . The Earth moves around the Sun and the Moon moves around the Earth. So, the Moon has to move extra path along its orbit to complete its phase. This makes the lunar month larger than a sidereal month.

Step 2 of 3

The angle made by the Moon with respect to the center of the Earth in a day is

The Moon takes to complete a complete phase around the Earth. So, the angle made by the moon in one complete phase is

This means the Moon needs an angle of further to move around the Earth.

Thus, the time taken by the moon to go along with the Earth is

But in the above time, the Earth will move further and the Moon has to catch up further. So, the angle made by the Moon to go further is

Step 3 of 3

The time taken by the Moon when moved through is

Thus, the extra total time taken by the Moon to move along with the Earth is

This much of time is the lunar month longer than a sidereal month.

Rounding off to three significant figures, the lunar month is longer than a sidereal month by.


Step 1 of 2

Solution:

In British units, the unit of length is expressed in foot or ft.

Height of the Pierre, .

Step 2 of 2

The relation between meter (m) and foot (ft) is given by

Thus, the height of the Pierre in British unit is

Round off to two significant figures. The height of the Pierre is .


Step 1 of 4

Solution:

(a) The relation between yard and feet is

Now, 100 yards is equal to

The relation between feet and meter is

Now, 100 m is equal to

Step 2 of 4

The relation between yard and meter is

Now, 100 yard is equal to

Therefore, is longer than the 100 yard.

Step 3 of 4

The difference between 100 yard and 100 m in meters is

Therefore, 100 m is longer than the 100 yard by .

The difference between 100 yard and 100 m in ft is:

Thus, 100 m is longer than 100 yard by .

Step 4 of 4

(b) The relation between mile and meter is

Thus, 1 mile is equal to 1609 m.

And 1 metric mile is equal to1500 m.

Thus, the difference between them is

Convert the difference from meter into feet.

Therefore, the 1 mi is or longer than the metric mile.


Step 1 of 2

Solution:

One year equals 365.25 days. 1 day is equal to 24 hours. 1 hour is equal to 3600 seconds.

Thus, the calculated value of second in a year is

The value of seconds in 300,000 years is

It is given that the two cesium clocks will lose 1 s with respect to each other in about 300,000

years or . This shows that the error is 1 part in .

Step 2 of 2

The distance between New York and San Francisco is 2572 mi.

The relation between meter and mile is

When the same precision, like time1 part in is applied to distance, then the accuracy in distance is

Round off to three significant figures, the successive measurements of the given distance will

differ by a distance of .


Step 1 of 2

Solution:

The Antarctica is roughly semicircular in shape, so find the area of the semicircular curve. The volume of Antarctica will be equal to the surface area times its thickness.

Radius of the Antarctica is 2000 km.

Convert the radius from km into m.

Step 2 of 2

The area of the Antarctica will be equal to half the area of a circle with radius .

Thus, the surface area of the Antarctica is

The amount of ice in the Antarctica will be equal to its volume. It is equal to the surface area times the thickness.

Thus, the amount of ice in the Antarctica will be

Substitute for radius , and for average thickness of the Antarctica in the above equation to calculate the volume.

Convert the volume from cubic meter into cubic centimeters.

Therefore, the amount of ice in the Antarctica is .


Step 1 of 2

Solution:

Volume of the earth removed is equal to the surface area of the land consumed by the open-pit coal mine times the depth.

The unit1 hectare is equal to .

Thus, Surface area of land consumed by the open-pit coal mine is .

Convert the area from hectare into square meters.

Depth of the surface area of land consumed by the open-pit coal mine, .

Step 2 of 2

The volume of the earth that is removed due to the coal mine is equal to the surface area consumed by the open-pit coal mine times the depth.

Thus, the volume of the Earth removed is

Substitute for area , and 26 m for depth of the surface area of land consumed by the open-pit coal mine in the above equation to calculate the volume.

Round off to two significant figures, the amount of land removed from the earth is.


Step 1 of 4

Solution:

The Earth is considered as a solid sphere. Using the formula for the circumference, surface area and the volume of a sphere, the corresponding values of the Earth can be determined.

Radius of the Earth is .

Step 2 of 4

(a) The circumference of the Earth is

Substitute for radius in the above equation to calculate the circumference.

Round off to three significant figures, the circumference of the Earth is .

Step 3 of 4

(b) The surface area of the Earth is equal to the surface area of a sphere with radius .

Thus, the surface area of the Earth is

Substitute for radius in the above equation to calculate the surface area.

Round off to three significant figures, the surface area of the Earth is .

Step 4 of 4

(c) The volume of the Earth is equal to the volume of a sphere with radius .

Thus, the volume of the Earth is

Substitute for radius in the above equation to calculate the volume.

Round off to three significant figures, the volume of the Earth is .


Step 1 of 5

Solution:

The maximum speed of the squirrel is .

Convert the speed from km/h into m/s.

Round off to two significant figures, the speed of the squirrel is .

Step 2 of 5

The maximum speed of the rabbit is .

Round off to two significant figures, the speed of the rabbit is .

The maximum speed of the snail is .

Round off to two significant figures, the speed of the snail is .

Step 3 of 5

The maximum speed of the spider is .

Round off to two significant figures, the speed of the spider is .

The maximum speed of the cheetah is .

Round off to two significant figures, the speed of the cheetah is .

Step 4 of 5

The maximum speed of the human is .

Therefore, the speed of the human is .

The speed of the fox is

Round off to two significant figures, the speed of the fox is .

Step 5 of 5

The maximum speed of the lion is .

Round off to two significant figures, the speed of the lion is .

It is found that the maximum speeds of the given animals are in the following order:

.

Therefore, the animals in order of increasing speed are Snail, Spider, Squirrel, Human, Rabbit, Fox, Lion, and Cheetah.


Step 1 of 2

Solution:

The speed of the spaceship can be converting into light-year per century by using proper unit conversion factors.

The relation between miles (mi) and meter (m) is

The relation between 1 year and second is

The relation between one light year and meter is

One century is equal to 100 years.

Step 2 of 2

Now, the speed of the spaceship is .

Convert the speed from mi/h into light-year/century.

Round off to three significant figures, the speed of the spaceship is .


Step 1 of 1

Solution: Now, the SI unit of is its inverse in L/m. Use the above conversion factor to get the SI reading as

Rounding off to three significant figures, the SI reading corresponding to is

.


Step 1 of 4

Solution:

A light year is the distance traveled by light in a year. It is equal to the velocity of light times

one year. Light travels at the speed of . Thus,

Step 2 of 4

Given that 1 AU is equal to and 1 second of arc is equal to .

A parsec is defined as the distance at which one astronomical unit subtends an angle of 1 second of arc.

Thus,

Step 3 of 4

(a) The distance from the earth between the Sun in parsec is

Rounding off to three significant figures, the distance between the Earth and the Sun is

.

Also, the distance from earth between the Sun in light year is

Rounding off to three significant figures, the distance between the Earth and the Sun is

.

Step 4 of 4

(b) The required conversion parsec into kilometer and light-year into kilometers are shown above.

Therefore, 1 parsec is equal to . Also, 1 light year is equal to .


Step 1 of 2

Solution:

The physical meaningful distance between the given radii is equal to the average of the logarithm of the given radii.

Thus, the logarithm average of the given radii is

Substitute for effective radius of the proton is , and for radius of the

observable universe in the above equation.

Step 2 of 2

Use the logarithmic expression , thus

Use another equation, thus

Thus, the radii is

Rounding off to two significant figures, the physically meaningful distance that is approximately

halfway between the given extremes on a logarithmic scale is .


Step 1 of 2

Concept:

Consider the mass of a single hydrogen atom. Use it to find the number of hydrogen atoms in 1 kg of hydrogen.

Step 2 of 2

Solution:

Mass of a single hydrogen atom is

This shows that mass of hydrogen consist of only one hydrogen atom.

Therefore, the number of hydrogen atoms in 1.00 kg of hydrogen is

Round off to three significant figures, the number of hydrogen atoms in 1.00 kg of hydrogen is

.


Step 1 of 2

Solution:

(a) The mass of the water molecule can be calculated as follows.

Here, is the mass of the oxygen atom and is the mass of hydrogen atom.

Substitute 16.0 u for , 1.0 u for in the above equation, and solve for .

This shows that mass of a single water molecule is about . Thus, the number of water molecules in 1 kg of water is

Round off to two significant figures. The number of water molecules in 1kg of water is .

Step 2 of 2

(b) The number of water molecules in the entire ocean is equal to the number of water molecules in 1 kg times the total mass of the ocean. Thus, the number of water molecules in the ocean is

Round off to three significant figures, the number of water molecules in the entire ocean is

.


Step 1 of 2

Solution:

It is given that one pound is half a kilogram in continental Europe. Paris is in Continental Europe. As coffee costs $9.00 in Paris, it means half kilogram costs $9.00 only. Thus, the cost of one kilogram of coffee in Paris is

Step 2 of 2

The relation between pound and kilogram is

This shows that 0.4536 kg of coffee will cost $7.20 in New York. Thus, the cost of one kilogram of coffee in New York is

It is found that the cost of one pound of coffee in New York is lower than that in Paris. Hence,

one should buy .


Step 1 of 2

Concept:

The mass of a substance is equal to density times its volume. Use it to find the mass of the air contained in the given volume of the air. It can be expressed as follows.

Here, m is the mass of the substance and is the density of the substance.

Step 2 of 2

Solution:

Volume of the air in the room is

The mass of the air is its density at room temperature times the density. Thus, the mass of the air in the room is

Substitute for V, for in the above equation, and solve for .

Hence, the mass of the air in the given room is .


Step 1 of 3

Solution:

One mole of a substance has Avogadro number . It is the number of particles in 1 mole of

substance. The number of particle present in Avogadro number is .

Step 2 of 3

It is given that the sugar cube is having an edge length of 1 cm. Thus, the volume of such sugar cube is

Therefore, the volume of 1 mole of sugar is

Let be the length of the volume of 1 mole of sugar. Then, the volume of 1 mole of sugar is

Compare the above both equations.

Step 3 of 3

Substitute for in the equation , and solve for L.

Rounding off to two significant figures, the edge length of 1 mole of sugar cube is .


Step 1 of 2

Solution:

Convert the unit of the mass of the diet loss by the person from kilograms to milligrams.

Convert the unit of the time interval from weeks to seconds.

Step 2 of 2

The mass loss rate is equal to the amount of the diet lost divided by the time interval.

Round off to two significant figures. The mass loss rate in milligram per second is

.


Step 1 of 2

Meter is defined as 1,650,763.73 wavelength of the certain orange-red light which is emitted by krypton atoms.

Let be the wavelength of the orange-red light. According to definition, one meter is,

.

Step 2 of 2

The distance corresponding to one wavelength of the orange red light is equal to the meter per wavelength.

From the expression , the distance of one wavelength is

The definition on one meter is given in nine significant figures. So, the distance of the corresponding one wavelength should be given in nine significant figures.

Rounding off to nine significant figures, the distance corresponding to one wavelength of the

orange-red light emitted by krypton atoms is .


Step 1 of 2

Solution:

(a) The given expression is .In this expression, the number of least

Significant is three in the number 0.132. Thus, the result of the given expression should be

rounded off in three significant figures. Therefore, the final answer is .

Step 2 of 2

(b) The given expression is given by . In this expression, the number of least significant is five in the number 16.264. Thus, the result of the given expression should

be rounded off in five significant figures. Therefore, the final answer is .


Step 1 of 1

Solution:

The volume of water in time moves through a cross section of area of the aquifers is

Here, is the vertical drop of the aquifer over the horizontal distance and K is the hydraulic conductivity of the aquifer.

Rearrange the above equation for K.

Substitute for the unit of volume (V), m for the unit of the distance (L), m for the unit of the vertical distance (H), for unit of the cross-sectional area (A), in the above equation, and determine the SI unit of K.

Hence, the SI unit of the hydraulic conductivity of the aquifer is .


Step 1 of 5

CONCEPT:

The Planck length depends on the given constants.

Here, are the exponents to be determined, and A is a constant.

Step 2 of 5

SOLUTION:

Use the units given in the above three constants, the dimensions of constant are

Step 3 of 5

The dimensions of the above expression are

Equating power on both sides gives the following results.

Exponents of M:

Exponents of L:

Exponents of T:

Step 4 of 5

From equation (1),

Put the above value in equation (3), and solve for i.

Use the above value in equation (2) and solve for k.

This gives that .

Now, the value of i is calculated as follows.

Step 5 of 5

Put the value of i, j and k in the given formula for Planck length and remove the proportionality sign.

Substitute for , for h, for c in the above equation.

Round off to three significant figures; the Planck length is .


Step 1 of 5

CONCEPT:

The Planck length depends on the given constants.

Here, are the exponents to be determined, and A is a constant.

Step 2 of 5

SOLUTION:

Use the units given in the above three constants, the dimensions of constant are

Step 3 of 5

The dimensions of the above expression are

Equating power on both sides gives the following results.

Exponents of M:

Exponents of L:

Exponents of T:

Step 4 of 5

From equation (1),

Substitute the above value in equation (3),

Use the value of k and i in equation (2),

Now, the value of k is calculated as follows.

And the value of i is calculated as follows.

Step 5 of 5

Put the value of i, j and k in the given formula for Planck length and remove the proportionality sign.

Substitute for , for h, for c in the above equation.

Rounding off to three significant figures, the Planck mass is .

Chapter 1 Question 1

Step 1 of 1

The meaning of invariable in picking a standard is that it should not change at any condition. But it is found that most of the standard is not invariant under certain circumstances. The very example can be cited when taking the standard kilogram. This amount of mass can be found variable when one takes its measurement in other reference frame. So, it can be inferred that a standard is not invariable.


Step 1 of 1

A physical standard should be easily available and could be easily stored under any circumstances. The maintenance of physical standard should be very easy. The standard should be indestructible and reproducible.


Step 1 of 1

Yes, we can imagine a system of base units without time.

Consider speed as the base unit instead of time. It is the distance travelled by a body in unit time. Dimensionally, it the length divided by time. So, it can be expressed as:

Now, time can be expressed as

Time becomes a derived unit of length and speed.

Therefore, the base unit of time is replaced by the base unit speed.


Step 1 of 1

Yes, we can imagine a system of base units without time.

Consider speed as the base unit instead of time. It is the distance travelled by a body in unit time. Dimensionally, it the length divided by time. So, it can be expressed as:

Now, time can be expressed as

Time becomes a derived unit of length and speed.

Therefore, the base unit of time is replaced by the base unit speed.


Step 1 of 2

It would not be wise to redefine the mass of platinum-iridium cylinder at the International Bureau of Weights and Measures as 1 g rather than 1 kg. The reason is that certain units are assigned taken into consideration the unit of mass as 1 kg. Replacing kg by g will make a lot of discrepancies in defining certain units which involve kg.

Step 2 of 2

For example, newton is the SI unit of force. One newton is the amount of force when a body of one kg is accelerated one meter second squared. If kilogram is replaced by gram, there would be a change in the definition of one newton. Correspondingly, every unit involve newton will get affected by such replacement.


Step 1 of 2

Solution:

The “micro-” used in the microwave signifies the that has been used to irradiate the food kept inside it. Microwave radiation is an electromagnetic wave whose wavelength is in the range of 1 mm to 1 m.

Step 2 of 2

Gamma rays are electromagnetic radiation whose wavelength is less than . This shows that its wavelength is in the pico range.

Thus, the oven that used gamma rays to irradiate food that has been kept inside is named picowaved. This means it used a radiation of wavelength in the order of pico.


Step 1 of 2

The physical quantities that can define extrasensory perception are , , and

.

Step 2 of 2

Extrasensory perception is due to the vibration in the sense organs. As vibrations are involved in the phenomenon, it can be described by its period of vibration. It can also be inferred that length of the sense organs changes when vibrations take place. Thus, it can also be described by length. Also, the phenomenon can be described as due to electric signals. This means the signals travel through electric field. Thus, it can be described by electric current.


Step 1 of 1

Solution:

Several repetitive phenomena occurring in nature that could serve as reasonable time standards are as follows:

a) Oscillating pendulum.

b) Oscillation of mass-spring system.

c) Oscillation of a quartz crystal.

d) The rotation of the Earth on its axis

e) The revolution of the Earth around the Sun.


Step 1 of 1

The definition of second based on the atomic clock is much better than one pulse beat of the president of the American Association of Physics.

The reason is that the atomic clock gives the definition of second based on the duration of the vibration of the radiation emitted by a specific isotope. Such duration of time never change from any external disturbances.

On the other hand, the duration of pulse might change depending on the physical condition of the human. So, such duration due to pulse cannot be standardized to give the correct definition of a second.


Step 1 of 1

Solution:

A good clock should give standard time. It should not change its standardization with factors such as temperature and physical dimensions. Any time standard should answer the fact that at what time any event occurs and how long the events last. Any good clocks should correctly go with some astronomical observations like the time taken by the rotation of the earth around the Sun.


Step 1 of 2

It is conventionally assumed that the Sun is overhead at equinox. On this day, the clock should show 12:00:00. The clock will have a symmetric increase or decrease in the time. So, the given table is arranged in order of the clocks’ relative value as good timekeepers.

Clock D gives decreasing time with time interval of 57 s between successive days.

Clock C gives increasing time with time interval of 58 s between successive days.

Clock E gives decreasing time; but the time interval is increasing between successive days.

Clock A gives increasing time; but the increasing time interval between successive days is not symmetric.

Clock B does not give symmetric time.

Thus, clock D gives the accurate time at noon.

Step 2 of 2

Therefore, the following table shows the five clocks in the order of their relative values as good timekeepers.

ClockSundayMondayTuesdayWednesdayThursdayFridaySaturday

D12:03:5912:02:5212:01:4512:00:3811:59:3111:58:2411:57:17

C15:50:4515:51:4315:52:4115:53:3915:54:3715:55:3515:56:33

E12:03:5912:02:4912:01:5412:01:5212:01:3212:01:2212:01:12

A12:36:4012:36:5612:37:1212:37:2712:37:4412:37:5912:38:14

B11:59:5912:00:0211:59:5712:00:0712:00:0211:59:5612:00:03


Step 1 of 1

Solution:

The period of the oscillation of the pendulum depend on the length of the string of the

pendulum and acceleration due to gravity at that place of observation.

Thus,

The period of the repeated oscillations of the pendulum is taken as the time standard.

But the value of acceleration due to gravity is not fixed quantity and changes from place to place. Again, the length of the string is also affected by temperature and other factors. Thus, the overall effect is that standardization of time due to the period of the pendulum will have certain drawbacks.


Step 1 of 1

The period of the swinging pendulum is

Here, l is the length of the string, and g is the acceleration due to gravity of the Earth at the point of observation.

The frequency of the swinging pendulum is the reciprocal of its period. Thus, the frequency of the swinging pendulum is,

Substitute

for .

In the above expression, the frequency of the pendulum does not depend on the amplitude of the pendulum. It solely depends on the length of the string and the acceleration due to gravity of the Earth. Keeping fixed the length of the string, the frequency of the pendulum can be maintained at constant value. Thus, Galileo knew this fact that the pendulum swings at the same frequency regardless of the amplitude.


Step 1 of 3

The egg based sand timer is based on the period of the rate of flow of particulate materials from one reservoir to the other. The particulate materials used in such timer are grain of similar size with smooth and regular size. The uncertainty is that the rate of flow of grains is reduced with the change in temperature. This in fact gives inaccurate results.

Step 2 of 3

For hour glass, the period of flow of grains from one reservoir to the other is one hour. Due to temperature change, the rate of flow is also reduced.

Step 3 of 3

For candle light, the measurement of time is based on the rate of burning. If the rate of burning is not maintained constant, then there would be uncertainty in given the accurate time. This might happen as the rate of burning of the candle might change due to its internal factors such as chemical composition.


Step 1 of 1

It is desirable to readjust our clocks because a small error in the time will lead to a large error in the season after long years. Leap second is occasionally introduced so that the time given by our clocks after many years remains the same. This will not change the seasons. If such error is not adjusted in the clocks, the season in a year will not be the same in the next few years.


Step 1 of 2

Solution:

The number 89.5 means to operate the radio receiver at the frequency of .

Step 2 of 2

The advertisement made by the radio station that it is “at 89.5 on your FM dial” means the receiver of the radio should tune to 89.5 kHz frequency. At this frequency, the receiver can received signals sent by the transmitter of the radio station.


Step 1 of 1

Solution:

The SI base unit of length is meter .

The SI unit of area is meter squared . Thus, the unit for area is a derived unit.

The SI unit of volume is meter cube . Thus, the unit for volume is a derived unit.

Thus, units for area and volume are derived units

Therefore, there are no SI base units for area or volume.


Step 1 of 1

The standard meter bar is an alloy of platinum and iridium. Meter was defined in the bar as the distance between two lines in it which are measured at the melting point of ice.

The definition of meter as the distance of one ten millionth of the meridian line from North Pole to the equator that passes through Paris disagrees with the meter bar by 0.023%. It means that the comparison of the definition of meter in both the cases is inaccurate. It depends on which standard definition is chosen.

If meter is chosen as given by a standard meter bar, then there will be no error in the length of one meter.

Therefore, it does not mean that the meter bar was not inaccurate.


Step 1 of 1

As per the definition of meter, one meter is equal to one ten-millionth of the meridian line from-

-the North Pole to the equator passing through the Paris.

The measurement of the distance between North Pole to the equator was not completely done. Half of the distance is measured. The measurement was done through expedition between Dunkerque belfry and Montjuic castle, Barcelona. Thus, it serves as the distance of half the distance connecting between North Pole and equator. So, twice the length will give the distance between North Pole to the equator.

There is a certainty in the experimental measurement of the distance between the North Pole and equator. The shape of the Earth is not regular but is an oblate spheroid. Such shape is undesirable for defining standard definition of length at the level of required precession.


Step 1 of 1

Yes, the length can be measured along a curved line.

Make a standard thread having standard length by marking its end points by a meter scale. Now, the measurement is done by keeping the thread along the curved line.

If the curve line is much longer than the thread, the measurement of the curved line can be done by successively measuring how many times the curve line is longer than the thread. So, the length of the curved line is equal to the number of the length of the standard thread.


Step 1 of 2

Solution:

Yes, length can be called a fundamental property if another physical quantity such as temperature be specified in choosing a standard.

Step 2 of 2

The material of the meter bar is platinum-iridium alloy. One meter is defined as the distance between two fine lines engraved near the ends of the bar, when the bar is held at a temperature of . This is due to the fact that any change in temperature will change the dimension of the meter bar. It is also supported mechanically in a prescribed way. This makes the position of the meter bar intact in its original position.

Keeping at the constant temperature and supported mechanically in a prescribed way will make the length of the meter bar unchanged. Thus, the length can be standardized using the meter bar. Any measurable length can be compared with the length of the meter bar.


Step 1 of 3

The accepted experimental value of the speed of light in vacuum is about . So, the standard length of meter is defined as the distance travelled by light in vacuum during a

time interval of of a second.

The general approximate value of the speed of light in vacuum is about . Defining the standard value of meter in approximate value will give discrepancies in the standard definition of meter. Thus, the delegates to the 1983 General Conference on Weights and Measures did

not simplify the definition of meter by taking the speed of light to be .

Step 2 of 3

The value of the speed of light is a constant quantity. Taking into consideration the speed of light to be will give the desired result of one meter. But it is convenient to standardize from some known values which never changes.

Step 3 of 3

The delegates of the conference rejected both the possibilities because the exact value of the speed of light is known. It is easy and most accurate to define the length of one meter from the experimental value of the speed of light in vacuum.


Step 1 of 2

The value of is equal to the ratio of the circumference of the circle to its diameter. This ratio always remains constant. is an irrational number and the problem rises in calculating the constant.

Most recent value of is about for most calculations. But in some cases it is assigned to for easy calculation. The value of is about and is not matching with the value Therefore, metric version of cannot be redefined so that it is exactly equal to .

Step 2 of 2

When the value of is defined to 3, there would be an error in most calculations. This is due to the fact that is an irrational number and cannot de defined in rational number as 3. Using

in calculations will give unexpected results.

For example, the circumference of a given circle will give much error if is used equal to 3.


Step 1 of 3

(a)

Radius of the Earth can be determined as follows.

Use shadow cast by the sun’s rays at two different places on Earth at noon to calculate the radius of Earth. This method was used by Greek philosopher Eratosthenes. Let us chose two cities, Alexandria and Syene, where he measured the circumference of the Earth. It was found to be 25,000 miles or .

The relation between the circumference and the radius of Earth is

Rearrange the

Therefore, the radius of Earth is approximately .

Step 2 of 3

(b)

Distance between Sun and the Earth can be determined as follows.

The distance between the Sun and the Earth can be measured by knowing the time it takes for the light coming from the Sun to reach the Earth. The distance is equal to the speed of the light times the time taken. In fact, the light rays from the Sun take approximately 8 minutes to reach

the Earth. The speed of the light is roughly . Thus, the required distance is

Here speed of light is and time is .

Therefore, the distance between the Sun and the Earth is .

Step 3 of 3

(c)

Radius of the Sun can be determined as follows.

The radius of the Sun can be measured using the parallax method. In this method, the angle made by a point to the diameter of the Sun is measured. The radius of the Sun is given by

Here, distance between the Sun and the Earth is and the angle made by the distance between the Sun and the Earth and the diameter of the Sun is .


Step 1 of 3

(a)

The thickness of a sheet of paper can be measured by direct method. Consider a stack of paper. Let be the number of papers in it. If be the thickness of such stack of paper, then the thickness of single sheet of paper is

The thickness x can be measured directly with a ruler.

Step 2 of 3

(b)

The thickness of a soap bubble can be measured by using the interference of thin films. Consider a monochromatic light having wavelength incident on the soap film having thickness . The light will undergo interference pattern.

For constructive interference,

Here, refractive index of the soap film is and is the integer. From the above equation, the thickness of the soap film is

For , the thickness of the soap film is

Therefore, the thickness of the paper can be measured by knowing the refractive index of the soap film and wavelength of the light.

Step 3 of 3

(c)

The diameter of the atom can be measured by the x-ray diffraction of crystals.

When x-ray is incident on the atomic crystals, it undergoes diffraction. The Bragg’s equation for such diffraction is

Here, atomic spacing inside the crystal is , diffraction angle is , wavelength of the x-ray is and is an integer.

From the above equation, the atomic spacing is

This gives the spacing between the similar atoms. Such spacing is equal to the diameter of the atoms.

Therefore, the diameter of the atom is measured by diffraction of x-ray.


Step 1 of 1

The reduction in the dimension of every object to half its original value is not really happen. The change in the dimension of the object is minute and gives inaccurate result. But this does not mean that much of the dimension will reduced in a single overnight. Such reduction might take a lot of time even year but will not happen in a single overnight.


Step 1 of 3

The current standard kilogram is and .

Step 2 of 3

The SI standard of kilogram is the platinum-iridium cylinder which is kept at the International Bureau of Weights and Measures. This standard can be moved from one place to another for checking the values of tertiary standards. So, it is easily accessible. It is also invariance as the value is used a standard sample.

It can be destructible as it is made from destructible material. It is not reproducible.

Step 3 of 3

, an atomic standard would not be better in any respect.

Mass is a fundamental quantity. It gives the amount of matter present in it. An atomic standard cannot be adopted as for length and time. The adoption of atomic standard is based on the wavelength of light emitted by certain atoms. Mass cannot be defined from the wavelength. Thus, the atomic standard cannot be used for mass.


Step 1 of 2

ANSWER:

It is useful to have two standards of mass, the kilogram and the atom because present laboratory techniques allow us to compare atomic masses with each other to greater precision than we can presently compare them with the standard kilogram.

Step 2 of 2

One kilogram is the mass of the platinum-iridium cylinder which is carefully kept at the International Bureau of Weights and Measures. It is assigned by the international agreement. It is the SI unit of mass.

Unified atomic mass unit (u) is the atomic mass of carbon 12 atom which is used to measure in atomic scale. It is assigned by the international agreement. It is not the SI unit of mass.


Step 1 of 3

SOLUTION:

The SI unit of the quantity of a substance is mole. The mass of one mole of atoms has exactly 12 grams.

The SI unit of mass is kilogram. The relation between kilogram and gram is expressed as follows.

Mass of 1 mole of atoms is converted from g to kg as follows.

This shows that the mass of 1 mole of atoms is about .

Step 2 of 3

It is also known that one mole of atoms is equal to Avogadro constant .

The number of atoms present in the Avogadro constant is as follows.

The mass of a single atom is equal to the mass of 1 mole of atom divided by the Avogadro constant.

Step 3 of 3

Now, 1 unified atomic mass unit is equal to 1/12 the mass of the carbon atom.

Thus, the value of 1 unified atomic mass unit is as follows:

Round off to four significant figures, to get the relation between unified atomic mass unit and

kilogram that is .


Step 1 of 14

Known Universe:

The mass of known Universe can be estimated by knowing the number of stars present in it. If the mass of all the stars are approximately same as the Sun, then the mass of the galaxy can be equal to the number of the stars times the mass of the Sun.

Step 2 of 14

Our Galaxy:

The mass of our galaxy can be estimated through the number of stars present in it. If the masses of all stars are approximately same as the Sun, then, number of the starts multiplied with the mass of the Sun gives mass of the galaxy.

Step 3 of 14

Sun:

The mass of the Sun is measured by knowing the period of the Earth and the distance from the Sun.

If be the time period of the Earth and be distance from the Sun, then the mass of the Sun is calculated as

Here, universal gravitational constant is .

Step 4 of 14

Earth:

The mass of the Earth is measured by knowing the period of the Moon and the distance from the Earth.

If be the time period of the Moon and be distance from the Earth, then the mass of the Earth is calculated as

Step 5 of 14

Moon:

The mass of the Moon is measured by sending a satellite around the Moon.

If be the period of the satellite and be its distance from the Moon, then the mass of the Moon is calculated as

Step 6 of 14

Ocean liner:

The mass of the ocean liner is measured by using Bernoulli’s principle.

If be the volume of water displaced in the water, then the mass of the ocean liner is given as

Here, acceleration due to gravity of the Earth is , volume of the ocean liner displaced is and density of the water is .

Step 7 of 14

Elephant:

The mass of the elephant can be measured by using Archimedes’ principle.

If the elephant is placed in the wood block and allow to partially submerged in the water, then the sum of the weight of the wood block and elephant is equal to the weight of the water displaced by them.

Thus,

Here, mass of the elephant is , mass of the wood is , volume of the water displaced is and density of the water is .

From the above equation, the mass of the elephant is

Step 8 of 14

Person:

The mass of the person can be measured by using spring balance

Step 9 of 14

Grape:

The mass of the grape can be measured by using spring balance.

Step 10 of 14

Speck of dust:

The mass of the speck of dust can be measured by using spring balance.

Virus:

The mass of the virus can be measured by using mass spectrograph.

Step 11 of 14

Penicillin molecule:

The mass of the penicillin can be measured by using mass spectrograph.

Step 12 of 14

Uranium atom:

The mass of the Uranium atom is equal to the mass of its nucleus. The nucleus is composed of 92 protons and 146 neutrons. Knowing the mass of the protons and neutrons, the mass of the Uranium atom is measured.

Its mass can also be directly measured by using mass spectrograph.

Step 13 of 14

Proton:

The mass of the proton is roughly equal to the hydrogen nucleus. So, knowing the mass of the hydrogen nucleus, the mass of the proton is measured.

Step 14 of 14

Electron:

The mass of the electron is measured by knowing charge to mass ratio. Such ratio is measured in Millikan oil drop experiment. As the charge of the electron is known, the mass of the electron is measured.


Step 1 of 1

Solution:

The mass of the moon from table 1-5 is and the mass of ocean liner is .

The object whose mass lies in between Moon and ocean liner is given in the table shown below:

ObjectEstimated mass in Kilograms

Seawise giant (Largest known ship)

Great pyramid of Giza

Amount of concrete present in the three gorges dam

Average World population

Approximate amount of world crude oil production in 2009

Earth’s atmosphere

Rings of Saturn

Ceres(Dwarf planet)

Earth’s ocean

Charon (Dwarf planet)


Step 1 of 1

ANSWER:

The SI unit of mass is kilogram (kg). In British unit, the unit of mass if pound (lb).

The relation between kilogram and pound is as follows:

Thus, buying of 1 lb of butter is just equivalent to buying 0.456 kg of butter. It will not change the amount of butter bought. The only thing is that it depends on which countries are you buying butter. Different countries have different metric system for measuring units.

Therefore, implication of different metric system will not make life more complicated.

Chapter 1 Problem 1

Step 1 of 3

SOLUTION:

It is given that 1 day is equal to10 decimal hours. 1 decimal hour is equal to 100 decimal minutes.

Step 2 of 3

It is known that 1 day is equal to 24 hours. 1 hour is equal to 60 minutes.

The relation between decimal minute and minute is calculated as follows.

Step 3 of 3

Now, the value of 8 decimal hours, 22.8 decimal minutes is

The integer part is in hour. But the decimal part is still in decimal minutes. It can be converted into minutes as

Therefore, the clock shows or .

Chapter 1 Problem 2

Step 1 of 6

SOLUTION:

It is given that the average distance of the Sun from the Earth is 390 times the average distance of the Moon from the Earth.

Here, is the average distance between the Earth and the Moon, is the average distance between the Sun and the Earth.

Step 2 of 6

SKETCH:

The figure which presents the Moon, Earth and the Sun is shown below:

Step 3 of 6

(a) From the figure shown above, the diameters of the Earth and the Sun are, respectively,

and .The triangles and are similar triangles. The corresponding sides of the triangle are proportional to each other.

Therefore, the ratio is as:

The distance and are the diameters of the Earth and the Sun. Therefore, the ratio of the

diameter of the Earth and the Sun is .

Step 4 of 6

(b) The volume of a sphere having radius is

Consider the shape of the Earth and the Sun to be uniform sphere. Thus, the volume of the Earth and Sun are given by

Step 5 of 6

Now, the ratio of the volume of the Sun and the Earth is

Substitute all the values.

Rounding off to three significant figures, the ratio of the volume of the Sun to the Earth is

.

Step 6 of 6

(c) It is given that the angle intercepted at the eye by the Moon is

The angle in radian is given by

Consider the diameter of the Earth is a section of an arc having radius of the average distance between the Earth and the Moon.

From the figure given above in part (a), the corresponding arc is equal to the diameter of the Earth.

Rounding off to two significant figures, the diameter of the Earth is .

Chapter 1 Problem 3

Step 1 of 4

The north-south error relation is,

Here, of an arc with respect to the circumference of the Earth is and is circumference of the Earth.

Determine the uncertainty in tankers position by the north-south line relation,

The uncertainty in the tanker’s position measured along the east-west line is,

Here, is latitude of the east west line.

Step 2 of 4

(a)

Approximately the circumference of the Earth is,

Now, of an arc with respect to the circumference of the Earth is,

Substitute for in the equation

Then, the range of the error is,

Rounding off to two significant figures, the uncertainty in the tanker’s position measured along

the north-south line is .

Step 3 of 4

(b)

The east-west range is smaller because the distance measured along latitude is smaller than the circumference by a factor of the cosine of the latitude.

Substitute for and for in the equation .

Rounding off to two significant figures, the uncertainty in the tanker’s position measured along

the east-west line is .

Step 4 of 4

(c)

From the positions of latitude and longitude the tanker is situated in Lake Ontario. It is about 20 km off the coast of Hamlin.

Chapter 1 Problem 4

Step 1 of 3

(a)

The rate at which the Sun sweep above the equator is,

The correction factor, since latitude at England is equal to cosine of the latitude, so the sun curves above the England is approximately .

A accuracy requires an error in time of no more than

Trip takes about 6 months. Thus, the accuracy required by the clock is

Step 2 of 3

Therefore, the correctness that would be required by the clock in daily is .

Step 3 of 3

(b)

Now, an accuracy requires an error in time of no more than,

Trip takes about 1 year. Thus, the accuracy required by the clock is

Therefore, the correctness of the clock to fix position within after 1 year at sea in daily

is .

Chapter 1 Problem 5

Step 1 of 3

SOLUTION:

A single breathing contains about 0.3 L of oxygen. Thus, the mass of the oxygen inhaled in a single breathing is

It is given that single breathing contains about 0.3 L of carbon dioxide. Thus, the mass of the carbon dioxide exhaled in a single breathing is

Thus, the net loss of the mass is about

Step 2 of 3

If we consider the number of complete breathing in a minute to be 20, then the amount of weight loss from our body in a minute is

Step 3 of 3

Thus, the weight of our body loss due to the breathing is

Round off to one significant figure; the loss in the weight in 8 hours is .

Chapter 1 Problem 6

Step 1 of 3

Let us assume that the container is completely filled with water. So, the volume of the water will be equal to the volume of the container. The mass of the water that flows from the container is,

Here, volume of the water in the container is , and density of the water is .

Step 2 of 3

The mass flow of the water in the container is equal to the amount of water that is flowing from the container divided by the time it takes to drain completely the container. Thus, the mass flow rate is,

Step 3 of 3

Substitute for in the equation

Substitute for the time taken , for the volume of the container, for the density of the water in the equation

Rounding off to three significant figures, the mass flow rate of water from the container is

.

Chapter 1 Problem 7

Step 1 of 5

SOLUTION:

The average radius of the grain is . Convert the units of radius from to m.

Step 2 of 5

It is given that of volume has a mass of 2600 kg. The density of the cube is equal to the mass divided by volume.

The surface area of the cube is calculated as follows.

Step 3 of 5

Consider the grain to a sphere. The surface area of the grain is calculated as follows.

Let be the total number of grains that would have the total surface area equal to the surface area of the cube having edge 1 m.

Step 4 of 5

The volume of a single grain is calculated as follows.

The volume of the cube is calculated as follows.

Substitute all values in the above equation.

Step 5 of 5

The mass of the sand grains in the volume of the cube is

Therefore, the mass of the sand grains that would have a total surface area equal to the

surface area of the cube exactly 1 m on an edge is .

Chapter 1 Problem 8

Step 1 of 3

CONCEPT:

The volume of the circular cylinder is expressed as follows.

Rearrange for h.

Here, and be the radius and height of the circular cylinder, respectively.

Step 2 of 3

The total surface area of the circular cylinder is

Using the value of from equation 1,

Step 3 of 3

SOLUTION:

The condition of minimum of the surface area is that the derivative of should be equal to zero.

Rearrange for V.

This represents the volume for the minimum surface area.

The above value of surface area is obtained when the radius is equal to the height of the cylinder.

It is proven that for a circular cylinder of fixed volume, the smallest surface area is obtained when the radius and height of the circular cylinder are equal.

Chapter 1 Problem 9

Step 1 of 4

SOLUTION:

(a) The volume of an iron atom is equal to mass of the iron atom divided by its density. The volume of the iron atom is expressed as follows.

Consider the iron atom to be a perfect sphere having radius . The volume of the corresponding sphere of iron sphere is

Substitute all values in the above equation.

Step 2 of 4

The radius of the iron atom is calculated as follows.

The spacing between the neighboring iron atoms is equal to two times the radius of the single iron atom. Thus, the spacing between the neighboring iron atoms is calculated as follows.

Rounding off to three significant figures, the spacing between the neighboring iron atoms is

.

Step 3 of 4

(b) The volume of a sodium atom is equal to mass of the sodium atom divided by its density. The volume of the sodium atom is calculated as follows.

Consider the sodium atom to be a perfect sphere having radius . The volume of the corresponding sphere of sodium sphere is

Step 4 of 4

The radius of the sodium atom is calculated as follows.

The spacing between the neighboring sodium atoms is equal to two times the radius of the single sodium atom. Thus, the spacing between the neighboring sodium atoms is

Rounding off to three significant figures, the spacing between the neighboring sodium atoms is

.

Chapter 1 Problem 10

Step 1 of 2

SOLUTION:

(a) The area of the rectangular metal plate is equal to its length times the width. Thus, the area of the plate is

The number of significant figures in the area of the plate should be expressed in three significant figures.

Rounding off to three significant figures, the area of the rectangular metal plate is .

Step 2 of 2

(b) The area of the circular plate is

The number of significant figures in the area of the plate should be expressed in two significant figures.

Rounding off to two significant figures, the area of the circular plate is .