Environment Fundamental Of Physics Solution Manual Pdf


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Request PDF on ResearchGate | Fundamentals of Physics, Student's Solutions Manual | No other book on the market today can match the success of Halliday. Chapter 1 – Student Solutions Manual 3. Using the given conversion factors, we find (a) the distance d in rods to be d. Get instant access to our step-by-step Fundamentals Of Physics solutions manual . better than downloaded Fundamentals of Physics PDF solution manuals?.

Fundamental Of Physics Solution Manual Pdf

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Fundamentals of Physics 7th Edition: Instructor's Manual. Pages·· FUNDAMENTALS OF PHYSICS 10TH EDITION SOLUTION MANUAL. Instructor's Solution Manual for Fundamentals of Physics, 6/E by Halliday, Resnick, and Wa. Fundamentals of Physics 7th Edition: Instructor's Manual. Pages·· Strength Hibbeler Solution manual, 8th edition- Authors: Halliday, Resnick, Walker. Find your textbook below for step-by-step solutions to every problem. Fundamentals of Physics, 9th Edition.

Solutions- Fundamentals of physics, 8th edition Home Solutions- Fundamentals of physics, 8th edition. Chapter 1 — Student Solutions Manual 3. Various geometric formulas are given in Appendix E. None of the clocks advance by exactly 24 h in a h period but this is not the most important criterion for judging their quality for measuring time intervals. What is important is that the clock advance by the same amount in each h period.

Over a cycle, the internal energy is the same at the beginning and end, so the heat Q absorbed equals the work done: The CA portion of the cycle is at constant volume, so no work is done.

This means the gas loses 30 J of energy in the form of heat. The thermal conductivity is found in Table of the text. Recall that a change in Kelvin temperature is numerically equivalent to a change on the Celsius scale. Let h be the thickness of the slab and A be its area. Thus, 1.

This is similar to Sample Problem As discussed in part a of that sample problem, there are three steps to the total process: And we convert the volume to SI units: Now, according to the ideal gas law, 1.

Suppose the gas expands from volume Vi to volume Vf during the isothermal portion of the process. The final pressure is atmospheric pressure: The gas starts in a state with pressure pf, so this is the pressure throughout this portion of the process.

We also note that the volume decreases from Vf to Vi. According to Table , the molar mass of molecular hydrogen is 2. This value can be used in Eq. Here we illustrate the latter approach, using v for vrms: Thus Evaluate the integral by calculating the area under the curve in Fig. The area of the triangular portion is half the product of the base and altitude, or 12 av0. The area of the rectangular portion is the product of the sides, or av0.

Thus the number of particles with speeds in the given range is N a 2. Two formulas other than the first law of thermodynamics will be of use to us.

It is straightforward to show, from Eq. Further, Eq. In this case, the work done consists of that done during the constant pressure part the horizontal line in the graph plus that done during the constant volume part the vertical line: In this solution we will use non-standard notation: This is very similar to Sample Problem and we use similar notation here except for the use of Eq.

The negative value implies that the heat transfer is from the sample to its environment. Chapter 20 Student Solutions Manual 5.

Now the internal energy of an ideal gas depends only on the temperature and not on the pressure and volume. The two energies are the same in magnitude since no energy is lost. Since the temperatures in the equation must be in Kelvins, the temperature in the denominator is converted to the Kelvin scale.

Vb and pb are given. We need to find pa. Now pa is the same as pc and points c and b are connected by an adiabatic process. The coefficient of performance is the energy QL drawn from the cold reservoir as heat divided by the work done: Here QH is the energy ejected to the hot reservoir as heat. There are N! These rearrangements do not produce a new configuration. WB 33! Except for the phase change which just uses Eq. In this case, using Eq. Thus, 3!

Solutions- Fundamentals of physics, 8th edition - PDF Free Download

Thus, 5! Let q1 and q2 be the original charges and choose the coordinate system so the force on q2 is positive if it is repelled by q1. Take the distance between the charges to be r. After the wire is connected, the spheres, being identical, have the same charge. Since charge is conserved, the total charge is the same as it was originally. Since the spheres are identical, the solutions are essentially the same: Let the charge on the third particle be q0. Thus the y coordinate of the particle must be zero.

Suppose the third particle is a distance x from the particle with charge q, as shown on the diagram to the right. Solve this equation for x. The force on the particle with charge 4: Solve for the charge: Each force is a force of attraction and is directed toward the cesium ion that exerts it, along the body diagonal of the cube. We can pair every cesium ion with another, diametrically positioned at the opposite corner of the cube.

Since the two ions in such a pair exert forces that have the same magnitude but are oppositely directed, the two forces sum to zero and, since every cesium ion can be paired in this way, the total force on the chlorine ion is zero. This neutralizes the ion and, as far as the electrical force on the chlorine ion is concerned, it is equivalent to removing the ion.

The forces of the eight cesium ions at the cube corners sum to zero, so the only force on the chlorine ion is the force of the added charge. The chlorine ion is pulled away from the site of the missing cesium ion. Atomic numbers numbers of protons and numbers of electrons and molar masses combined numbers of protons and neutrons can be found in Appendix F of the text. One of the neutrons is freed in the reaction. Equate these forces to each other and solve for d.

The result is s s 1: Particles 2 and 3 repel each other. The later angle is associated with a vector that has negative x and y components and so is the correct angle. Since electrostatic forces are along the lines that join the particles, particle 3 must be on the x axis. Its y coordinate is zero. Particle 3 is repelled by one of the other charges and attracted by the other. As a result, particle 3 cannot be between the other two particles and must be either to the left of particle 1 or to the right of particle 2.

Since the magnitude of q1 is greater than the magnitude of q2 , particle 3 must Chapter 21 be closer to particle 2 than to particle 1 and so must be to the right of particle 2. Let x be the coordinate of particle 3. At the center of the square, the electric fields produced by the particles at the lower left and upper right corners are both along the x axis and each points away from the center and toward the particle that producespit.

Ex It is upward in the diagram, from the center of the square toward the center of the upper side. The moments point in opposite directions and produce fields in opposite directions at points on the quadrupole axis.

Consider the point P on the axis, a distance z to the right of the quadrupole center and take a rightward pointing field to be positive. Thus F 3: The force is downward. Such a line has slope 3: Since the field is downward, the flux through the upper face is negative and the flux through the lower face is positive.

Neglect fringing. Symmetry can be used to show that the electric field is radial, both between the rod and the shell and outside the shell. It is zero, of course, inside the rod and inside the shell since they are conductors. The flux through the ends is zero. The negative sign indicates that the field points inward. The charge enclosed by the Gaussian surface is only the charge Q1 on the conducting rod.

The electric field is zero at all points on the curved surface and is parallel to the ends, so the total electric flux through the Gaussian surface is zero and the net charge within it is zero. The charge is distributed uniformly over both sides of the original plate, with half being on the side near the field point. Thus 8: The gravitational force has magnitude mg, where m is the mass of the ball; the electrical force has magnitude qE, where q is the charge on the ball and E is the electric field at the position of the ball; and the tension in the thread is denoted by T.

The electric field produced by the plate is normal to the plate and points to the right. Since the ball is positively charged, the electric force on it also points to the right.

The field points inward, toward the sphere center, so the charge is negative: The electric field, if it exists, is radial and so is normal to the surface. Since there is no charge in the cavity the charge on the inner surface of the smaller shell is zero. The total charge on the smaller shell is qa and this must reside on the outer surface. Now consider a spherical Gaussian surface with radius slight larger than the inner radius of the larger shell.

This surface also encloses zero net charge, which is the sum of the charge on the outer surface of the smaller shell and the charge on the inner surface of the larger shell. Since an electron is negatively charged the electrical force on it is opposite to the electric field, so the electric field must be downward.

The field is normal to the sheet and is uniform. Place the origin of a coordinate system at the sheet and take the x axis to be parallel to the field and positive in the direction of the field. The equipotential surfaces are surfaces of constant x; that is, they are planes that are parallel to the plane of charge. The charge is twice the charge of original drop: This means that when the full disk is present each quadrant contributes equally to the electric potential at P , so the potential at P due to a single quadrant is one-fourth the potential due to the entire disk.

First find an expression for the potential at P due to the entire disk. Number the particles 1, 2, 3, and 4, in clockwise order starting with the particle in the upper left corner of the arrangement. The potential energy of the interaction of particles 1 and 2 is 2 8: This is equal to the work that must be done to assemble the system from infinite separation.


The potential energy when the moving particle is at any coordinate y is qV , where V is the electric potential produced at that place by the two fixed particles. The factor 2 appears since the two fixed particles produce the same potential at points on the y axis.

Let q1 be the charge on one, q2 be the charge on the other, and d be their separation. The potential of either one of the spheres is due to the charge on that sphere and the charge on the other sphere. The factor 2 appears since the potential Chapter 24 energy is the same for the interaction of the movable particle and each of the fixed particles. The Gaussian surface is a cylindrical surface that is concentric with the cylinder and has a radius r that is greater than the radius of the cylinder.

All points on the ring are the same distance from the point. The potential at infinity was taken to be zero. Take the z axis to be the dipole axis and consider a point with z positive on the positive side of the dipole. The charge on the sphere when the potential reaches V is 0: Let R be the decay rate and t be the time for the potential to reach it final value.

The total charge on the combination is the net charge on either pair of connected plates. The potential difference is 2: After the plates are pulled apart, their separation is d0 and the potential difference is V 0. Solve for A: Note that the field due to the induced charge is opposite the field due to the free charge, so the fields tend to cancel.

Capacitors 1, 2, and the equivalent capacitor that replaced 3 and 4 are all in series, so the sum of their potential differences must equal the potential difference across the battery. Since all of these capacitors have the same capacitance the potential difference across each of them is one-third the battery potential difference or 3: The charge induced on the dielectric surfaces of the upper region has the same magnitude but opposite sign on the two surfaces and so produces a net field of zero in the lower region.

Similarly, the electric field in the upper region is due to the charge on the plates and the charge induced on the upper and lower surfaces of dielectric in that region. According to Table 26—1, the resistivity of copper is 1: Solve for i: Solve for R: Chapter 26 53 a and b Calculate the electrical resistances of the wires. The current is the charge that moves through the cross section per unit time. Now find the particle speed. Here V is the electric potential through which the particles are accelerated.

Chapter 26 Chapter 27 7 a Let i be the current in the circuit and take it to be positive if it is to the left in R1. The battery is discharging. The current in battery 2 is opposite the direction of the emf, so this battery absorbs energy from the circuit. It is charging. The negative terminal is at B. The current is taken to be positive if it is Only the potential difference across the battery with the larger internal resistance can be made to vanish with the proper choice of R.

Here d and D are diameters. The circuit now consists of the two emf devices and four resistors. Take the current to be upward in the right-hand emf device.

To find the potential at point 2 continue the path through the lowest resistor on the digram. Chapter 27 c Consider the copper wire. Let i1 be the current in R1 and take it to be positive if it is to the right.

Let i2 be the current in R2 and take it to be positive if it is downward. Let i3 be the current in R3 and take it to be positive if it is downward. Since the resistances are all the same, you can simplify the mathematics by replacing R1 , R2 , and R3 with R. The solution to the three simultaneous equations is 2 1: Chapter 27 The solution is 1: V2 i The graph of V2 versus t is shown to the right.

This equation is solved for the time constant, with result t: The resistivity of wire A is RA A 0: The cable is defective. The motor is not defective. Then the resistance of the combination will be independent of the temperature. Apply the loop rule to the right-hand loop to find that the potential difference across R2 must also be zero.

Thus FB 6: This is 1: Use the kinetic energy to find the speed: The situation is shown in the left diagram above. The y axis is along the hinge and the magnetic field is in the positive x direction. A torque around the hinge is associated with the wire opposite the hinge and not with the other wires. The right diagram shows the view from above. Chapter 28 Use the right-hand rule to show that the torque is directed downward, in the negative y direction.

The magnitude of the net dipole moment is the sum of the magnitudes of the individual moments: The net dipole moment is directed into the page. The net dipole moment is again into the page. For maximum torque, orient the plane of the loops parallel to the magnetic field, so the dipole moment is perpendicular to the field.

The moment is in the negative y direction, as you can tell by wrapping the fingers of your right hand around the coil in the direction of the current. Your thumb is then in the negative y direction. Since the electron moves with constant velocity you know that the net force must vanish.

E 75 a and b Suppose the particles are accelerated from rest through an electric potential difference V. The solid curve on the diagram is the path. Suppose it subtends Since the currents are the same, the net field is zero along the line that runs halfway between the wires.

There is no possible current for which the field does not vanish. If there is to be a field on the bisecting line the currents must be in opposite directions. Then the fields are in the same direction in the region between the wires. Look at the derivation of the expression for the field of a long straight wire, leading to Eq. Since the wires we are considering are infinite in only one direction, the field of either of them is half the field of an infinite wire.

It is the radius of the arc. The fields of both wires are out of the page at the center of the arc. Now find an expression for the field of the arc at its center. Divide the arc into infinitesimal segments. Each segment produces a field in the same direction. For the arc of the diagram, the field is into the page. The fields due to the wires at the upper left and lower right corners both point toward the upper right corner of the square.

The Chapter 29 fields due to the wires at the upper right and lower left corners both point toward the upper left corner.

Treat this as a long straight wire. Since the currents are in opposite directions the wires repel each other so the force on wire 2 is along the line that joins the wires and is away from wire 1. The field of the solenoid at the point is parallel to the solenoid axis and the field of the wire is perpendicular to the solenoid axis. Each of the two fields is a vector component of the net field, so the magnitude of the net field p is the square root of the sum of the squares of the individual fields: Chapter 29 65 a Take the magnetic field at a point within the hole to be the sum of the fields due to two current distributions.

The first is the solid cylinder obtained by filling the hole and has a current density that is the same as that in the original cylinder with the hole. The second is the solid cylinder that fills the hole. It has a current density with the same magnitude as that of the original cylinder but it is in the opposite direction. Notice that if these two situations are superposed, the total current in the region of the hole is zero.

Here R is the radius of the cylinder. The field points upward in the diagram if the current is out of the page. This is correct for the field on the axis of a cylindrical shell carrying a uniform current. P is a point within the hole, A is on the axis of the cylinder, and C is on the axis of the hole. Let dy be an infinitesimal length of wire at coordinate y. All four sides of the square produce magnetic fields that are into the page at P, so we sum their magnitudes. Reverse the direction of the current.

According to the Biot-Savart law, the field reverses, so it will be as in the second diagram. The field, of course, will rotate with it and end up in the direction The current distribution is now But it is not. Only if the field is parallel to the sheet will be final direction of the field be the same as the original direction. If the current is out of the page, The field must be as drawn in Fig. The upper and lower edges are the same distance from the current sheet and each has length L.

B same magnitude along these edges. The total current is zero because the field is perpendicular to the sides. The current through the Amperian path is the current through the region outside B the circle of radius b and inside the circle of radius r.

A point on the axis of the loop is also on a perpendicular bisector of each of the loop sides. The diagram shows the field due to one of the loop sides, the one on the left. The field due to the side is therefore P B current through the loop.

Instructor’s Solution Manual for Fundamentals of Physics

We conclude that the geometry shown for the magnetic field lines is in error. The lines actually bulge outward and their density decreases gradually, not precipitously as shown.

Since the magnetic field is zero outside the solenoid, this is also the flux through the coil. The current changes linearly by 3: The field is upward in the diagram. As the smaller loop moves away, the flux through it decreases. The induced current is directed so as to produce a magnetic field that is upward through the smaller loop, in the same direction as Chapter 30 the field of the larger loop. It is counterclockwise as viewed from above, in the same direction as the current in the larger loop.

The resistivity can be found in Table 26—1.

That is, 6: This means 0: Let i1 be the current in the resistor and take it to be downward. Let i2 be the current in the inductor and also take it to be downward. Solve the current equation for the time constant. V is calculated as the product of the cross-sectional area and the length. First consider the left-hand coil. The magnetic field due to the current in that coil points to the left.

So does the magnetic field due to the current in coil 2. The two sources of emf are again in the same direction and the emf in coil 2 is di: Then the field produced by coil 2 at the site of coil 1 is opposite the field produced by coil 1 itself.

The fluxes have opposite signs. An increasing current in coil 1 tends to increase the flux in that coil but an increasing current in coil 2 tends to decrease it. The emf across coil 1 is di: Chapter 30 b The average current in the loop is the emf divided by the resistance of the loop: Then according to the right-hand rule we must take the normal to the loop to be into the page, so the flux is negative if the magnetic field is out of the page and positive if it is into the page.

Assume the field in region 1 is out of the page. We will obtain a negative result for the field if the assumption is incorrect. Let x be the distance that the front edge of the loop is into region 1. Let x now be the distance the front end of the loop is into region 2 as the loop enters that region.

The first two terms are: L 33 a The generator emf is a maximum when sin! Furthermore, since there is only one element in the circuit, the amplitude of the potential difference across the element must be the same as the amplitude of the generator emf: The phasor diagram is drawn to scale on the right.

To find the maximum, set the derivative with respect to! The last equation is a quadratic equation for! The circuit in the box is predominantly capacitive. This means the box must contain a capacitor and a resistor. The inductive reactance may be zero, so there need not be an inductor. If there is an inductor, its reactance must be less than that of the capacitor at the operating frequency.

If values are given for R, L, and C, then the value of the frequency would also be needed to compute the power factor. Similarly, p 3A cos! All of these are sinusoidal functions of! This means! Actually UE has the same value at the beginning and end of each half cycle. Now the integral of sin2! The circle. If a is the area enclosed by the dashed lines and A is the area of a plate, then the displacement current through the dashed path is 0: Subtract the magnitude of the total dipole moment per unit volume of the antialigned moments from the total dipole moment per unit volume of the aligned moments.

See Appendix E for the power series expansion of the exponential function. The figure agrees with these predictions. The mean kinetic energy of translation at room temperature is about 0: Thus if dipole-dipole interactions were responsible for aligning dipoles, collisions would easily randomize the directions of the moments and they would not remain aligned.

We wish to find the radius of an iron sphere with N iron atoms. The mass of such a sphere is N m, where m is the mass of an iron atom. There are eight of these: The frequency is the same as thep frequency of oscillation of the current in the LC circuit of the generator.

The factor 2 enters the first expression because the momentum of the reflected portion is reversed. The total radiation pressure is the sum of the two contributions: The intensity and energy density are inherently positive, regardless of the propagation direction.

Since the total momentum of the spaceship and light is conserved, this is the magnitude of Chapter 33 the momentum acquired by the spaceship. It can be done with two sheets. You want the smallest integer value of n for which this is greater than 0: If the result is greater than 0: If it is less, increase n by 1 and try again. Now look at the next diagram and consider the triangle formed by the two normals and the ray in the interior.

Here ng is the index of refraction for the glass and na is the index of refraction for air. You want to find the smallest value of the index of refraction n for which this inequality holds. Solve for n: Let n be the index of refraction of the glass.

The emerging ray is parallel to the incident ray. See Fig. In this case r is the distance from the source to the aircraft. The refracted light is white.

Red light is refracted but blue light is not. The refracted light is reddish. Neither is refracted. A similar analysis shows that the function for B also satisfies the wave equation.

Chapter 33 Chapter 34 The light bulb is labeled O and its image is la Consider the This ray is perpendicular to the wa The second ray leaves the It is reflected from the mir At C the angle mirror At D the angles of incidence and reflection.. Light appears to come from I. We want to.

Since the image here is real it is on the same side of the mirror as the object.

Since the image here is virtual it is on the opposite side of the mirror from the object. Since the image here is virtual it is on the opposite side of the mirror from the object 27 Since the mirror is convex the radius of curvature is negative.

Since the image distance is negative the image is virtual and on the opposite side of the mirror from the object.

Fundamentals of Physics Solutions Manual

Since the magnification is positive the image is not inverted. The solve for p. Since the focal length and radius of curvature are negative the mirror is convex.

The object distance is 1: Since the image distance is negative the image is virtual and appears on the same side of the lens as the object. Since the image distance is positive the image is real and appears on the opposite side of the lens from the object.

Since the magnification is negative the image is inverted. Since the image distance is positive the image is real and on the opposite side of the lends from the object. This means that the image is virtual and the image distance is negative. The negative sign indicates that the image is behind the second lens. The lens equation is still valid. When the eye focuses on closer objects, the image distance i remains the same but the object distance and focal length change. For the lens pictured in Fig.

This can be accomplished by decreasing the magnitudes of either or both radii. The final image is produced by two lenses, with the image of the first lens being the object for the second. The negative sign must be used since the image formed by the first lens is beyond the second lens if i1 is positive.

This means the object for the second lens is virtual and the object distance is negative. If i1 is negative, the image formed by the first lens is in front of the second lens and p2 is positive. The angular frequencies are the same since the waves have the same wavelength in air and the frequency of a wave does not change when the wave enters another medium. The value of L that makes this 5: The interference is therefore intermediate, neither completely constructive nor completely destructive.

It is, however, closer to completely constructive than to completely destructive. You want all values of m positive and negative for which j0: There are sixteen different angles in all and therefore sixteen maxima.

Then 1: The resultant amplitude Em is given by the trigonometric law of cosines: If L is the thickness of the coating, the wave reflected from the back surface travels a distance 2L farther than the wave reflected from the front.


Here m is an integer. The result is 4Ln2 2 nm 1: Other wavelengths are shorter. This is in the visible range. Other values of m are associated with wavelengths that are not in the visible range. One wave travels a distance 2L further Chapter 35 than the other, where L is the thickness of the oil.

The result is 2no L 2 1: This wave is reflected twice, once from the back surface and once from the front surface. Since n2 is greater than n3 there is no change in phase at the back-surface reflection. There are, therefore, bright fringes in all. As the geometry of Fig. Solve for r. These are different because the wavelength in air is different from the wavelength in vacuum. The factor 2 arises because the light traverses the tube twice, once on the way to a mirror and once after reflection from the mirror.

Since the index of refraction of water is greater than that of air this last wave suffers a phase change on reflection of half a wavelength. Look at the diagram on the right. Use the binomial theorem to approximate this expression: Solve for x. Assume their amplitudes are the same.

According to Eq. To be sure that 1: To resolve two point sources, the central maximum of the diffraction pattern of one must lie at or beyond the first minimum of the diffraction pattern of the other. The angular width of the central maximum is twice this, or 6: Notice that for a given Chapter 36 order, the line associated with a long wavelength is produced at a greater angle than the line associated with a shorter wavelength.

The crystal should be turned For clockwise turns the smaller value is For counterclockwise turns the smaller value is 3: Only the last three are in the visible range, so the longest wavelength in the visible range is nm, the next longest is nm, and the third longest is nm.

There are 13 such values: Thus 13 interference maxima appear in the central diffraction envelope. That is, there are 6 interference maxima in the first diffraction envelope.

Chapter 36 93 If you divide the original slit into N strips and represent the light from each strip, when it reaches the screen, by a phasor, then at the central maximum in the diffraction pattern you add N phasors, all in the same direction and each with the same amplitude.

The intensity there is proportional to N 2. If you double the slit width, you need 2N phasors if they are each to have the amplitude of the phasors you used for the narrow slit. The intensity at the central maximum is proportional to 2N 2 and is, therefore, four times the intensity for the narrow slit.

The energy reaching the screen per unit time, however, is only twice the energy reaching it per unit time when the narrow slit is in place. The energy is simply redistributed. For example, the central peak is now half as wide and the integral of the intensity over the peak is only twice the analogous integral for the narrow slit. The negative sign means that an increase in frequency corresponds to a decrease in wavelength.

The lens does not affect the travel time. The full Lorentz transformation must be used: The reading on the clock at the beginning of the interval is zero, so the reading at the end is t 1: Let S 0 be the reference frame for which the data is given and attach frame S to the spaceship.

Suppose the micrometeorite is going in the positive x direction and the spaceship is going in the negative x direction, both as viewed from S 0.

Then, in Eq. Notice that v in the equation is the velocity of S 0 relative to S. This provides the same energy as 2: The distance traveled is 8: Let v1 be the initial speed and v2 be the final speed. The lamp emitting light with the longer wavelength the nm lamp emits more photons per unit time. The energy of each photon is less so it must emit photons at a greater rate. Replace K with eV , where V is the accelerating potential and e is the fundamental charge, to obtain 6: A little thought should convince you that there are no other possible values for the energy less than 5.

The possible frequencies are 0: The highest is 3: The neutron does not have sufficient kinetic energy to excite the hydrogen atom, so the hydrogen atom is left in its ground state and all the initial kinetic energy of the neutron ends up as the final kinetic energies of the neutron and atom.

The collision must be elastic. You can check your reasoning as you tackle a problem using our interactive solutions viewer. Plus, we regularly update and improve textbook solutions based on student ratings and feedback, so you can be sure you're getting the latest information available.

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