Concept explainers
(a)
The work function of the metal.
(a)

Answer to Problem 94P
The work function of the metal is
Explanation of Solution
Given that the
Write the expression for the magnitude of the magnetic force on the electron when it enters perpendicular into the field.
Here,
The magnetic force is providing the necessary centripetal force for the motion of the electron. Thus, the magnetic force can be equated to the centripetal force on electron.
Here,
Solve equation (II) for
Write the expression for the maximum kinetic energy of the ejected electron.
Here,
Write the expression for the maximum kinetic energy of the ejected electron in terms of the work function of the metal.
Here,
Equate the right hand sides of equations (IV) and (V) and solve for
Use equation (III) in (VI).
Conclusion:
Substitute
Therefore, the work function of the metal is
(b)
Whether the electrons with maximum kinetic energy follow a path with the maximum or minimum radius.
(b)

Answer to Problem 94P
The electrons with maximum kinetic energy follow a path with the maximum radius.
Explanation of Solution
Equation (IV) indicates that the maximum kinetic energy of the electron is directly proportional to the square of the speed of the electrons.
Equation (III) indicates that the speed of the electron is directly proportional to the radius of the circular path.
Hence, it can be summarized that the maximum kinetic energy of the electron corresponds to the large value of the radius. Thus, The electrons with maximum kinetic energy follow a path with the maximum radius.
Conclusion:
Therefore, the electrons with maximum kinetic energy follow a path with the maximum radius.
Want to see more full solutions like this?
Chapter 27 Solutions
PHYSICS
- Find the net torque on the wheel in the figure below about the axle through O perpendicular to the page, taking a = 9.00 cm and b = 23.0 cm. (Indicate the direction with the sign of your answer. Assume that the positive direction is counterclockwise.) N.m 10.0 N 30.0% 12.0 N 9.00 Narrow_forwardAn automobile tire is shown in the figure below. The tire is made of rubber with a uniform density of 1.10 × 103 kg/m³. The tire can be modeled as consisting of two flat sidewalls and a tread region. Each of the sidewalls has an inner radius of 16.5 cm and an outer radius of 30.5 cm as shown, and a uniform thickness of 0.600 cm. The tread region can be approximated as having a uniform thickness of 2.50 cm (that is, its inner radius is 30.5 cm and outer radius is 33.0 cm as shown) and a width of 19.2 cm. What is the moment of inertia (in kg. m²) of the tire about an axis perpendicular to the page through its center? 33.0 cm 16.5 cm Sidewall Ο 30.5 cm Tread i Enter a number. Find the moment of inertia of the sidewall and the moment of inertia of the tread region. Each can be modeled as a cylinder of nonzero thickness. What is the inner and outer radius for each case? What is the formula for the moment of inertia for a thick-walled cylinder? How can you find the mass of a hollow cylinder?…arrow_forwardYou have just bought a new bicycle. On your first riding trip, it seems that the bike comes to rest relatively quickly after you stop pedaling and let the bicycle coast on flat ground. You call the bicycle shop from which you purchased the vehicle and describe the problem. The technician says that they will replace the bearings in the wheels or do whatever else is necessary if you can prove that the frictional torque in the axle of the wheels is worse than -0.02 N . m. At first, you are discouraged by the technical sound of what you have been told and by the absence of any tool to measure torque in your garage. But then you remember that you are taking a physics class! You take your bike into the garage, turn it upside down and start spinning the wheel while you think about how to determine the frictional torque. The driveway outside the garage had a small puddle, so you notice that droplets of water are flying off the edge of one point on the tire tangentially, including drops that…arrow_forward
- 2nd drop down is "up" or "down"arrow_forwardRomeo (79.0 kg) entertains Juliet (57.0 kg) by playing his guitar from the rear of their boat at rest in still water, 2.70 m away from Juliet, who is in the front of the boat. After the serenade, Juliet carefully moves to the rear of the boat (away from shore) to plant a kiss on Romeo's cheek. (a) How far (in m) does the 81.0 kg boat move toward the shore it is facing? m (b) What If? If the lovers both walk toward each other and meet at the center of the boat, how far (in m) and in what direction does the boat now move? magnitude m direction ---Select---arrow_forward2nd image is the same for all drop downsarrow_forward
- A mobile is constructed of light rods, light strings, and beach souvenirs as shown in the figure below. If m4 = 12.0 g, find values (in g) for the following. (Let d₁ = 3.20 cm, d₂ = 5.10 cm, d3 = 1.00 cm, d4 = 5.80 cm, d5 = 2.40 cm, and d6 = 3.20 cm.) d₁ d2 d3 d4 Mg d5 d6 mg MA mi (a) m₁ = g (b) m2 = (c) m3 = g g (d) What If? If m₁ accidentally falls off and shatters when it strikes the floor, the rod holding m will move to a vertical orientation so that m hangs directly below the end of the rod supporting m₂. To what values should m₂ equilibrium and be oriented horizontally? (Enter your answers in g.) m2 = m3 = and m3 be adjusted so that the other two rods will remain inarrow_forwardAn automobile tire is shown in the figure below. The tire is made of rubber with a uniform density of 1.10 × 103 kg/m³. The tire can be modeled as consisting of two flat sidewalls and a tread region. Each of the sidewalls has an inner radius of 16.5 cm and an outer radius of 30.5 cm as shown, and a uniform thickness of 0.600 cm. The tread region can be approximated as having a uniform thickness of 2.50 cm (that is, its inner radius is 30.5 cm and outer radius is 33.0 cm as shown) and a width of 19.2 cm. What is the moment of inertia (in kg . m²) of the tire about an axis perpendicular to the page through its center? 33.0 cm 30.5 cm kg. m² 16.5 cm Sidewall Treadarrow_forwardJohn is pushing his daughter Rachel in a wheelbarrow when it is stopped by a brick 8.00 cm high (see the figure below). The handles make an angle of 0 = 17.5° with the ground. Due to the weight of Rachel and the wheelbarrow, a downward force of 403 N is exerted at the center of the wheel, which has a radius of 16.0 cm. Assume the brick remains fixed and does not slide along the ground. Also assume the force applied by John is directed exactly toward the center of the wheel. (Choose the positive x-axis to be pointing to the right.) i (a) What force (in N) must John apply along the handles to just start the wheel over the brick? N (b) What is the force (magnitude in kN and direction in degrees clockwise from the -x-axis) that the brick exerts on the wheel just as the wheel begins to lift over the brick? magnitude direction kN ° clockwise from the -x-axisarrow_forward
- Your neighbor designs automobiles for a living. You are fascinated with her work. She is designing a new automobile and needs to determine how strong the front suspension should be. She knows of your fascination with her work and your expertise in physics, so she asks you to determine how large the normal force on the front wheels of her design automobile could become under a hard stop, when the wheels are locked and the automobile is skidding on the road. She gives you the following information. The mass of the automobile is m₂ = 1.10 × 103 kg and it can carry five passengers of average mass m = 80.0 kg. The front and rear wheels are separated by d = 4.45 m. The center of mass of the car carrying five passengers is dCM = 2.25 m behind the front wheels and hCM = 0.630 m above the roadway. A typical coefficient of kinetic friction between tires and roadway is μk = 0.840. (Caution: The braking automobile is not in an inertial reference frame. Enter the magnitude of the force in N.) Narrow_forwardThree solid, uniform boxes are aligned as in the figure below. Find the x- and y-coordinates (in m) of the center of mass of the three boxes, measured from the bottom left corner of box A. (Consider the three-box system.) HINT 0.200 m 0.280 m 0.120 m y A B C 0.350 m Origin 0.750 kg 1.00 kg 0.650 kg Х ст E m m Уст xarrow_forwardConsider the truss shown in the figure, built from three struts attached by three pins. The truss supports a downward force of F = 1,080 N applied at the point B. Assume the mass of the truss is negligible, the pins are frictionless, and the supports at A and C are also frictionless. 01 F B nc 02 C (a) Assuming 0₁ = 26.0° and 0 2 = 51.0°, what are n and n? (Enter the magnitudes in N.) ΠΑ пс = = N N (b) The force any strut applies on a pin must be directed along the length of the strut as a force of tension or compression. What are the directions of the forces that the struts exert on the pins joining them? strut AB on joint A: ---Select--- strut AB on joint B: strut BC on joint B: strut BC on joint C: strut AC on joint A: strut AC on joint C: |---Select--- --Select--- --Select--- --Select--- |---Select--- ✓ ✓ ✓ Find the force of tension or of compression (in N) in each of the three struts. bar AB N N bar BC bar AC Narrow_forward
- College PhysicsPhysicsISBN:9781305952300Author:Raymond A. Serway, Chris VuillePublisher:Cengage LearningUniversity Physics (14th Edition)PhysicsISBN:9780133969290Author:Hugh D. Young, Roger A. FreedmanPublisher:PEARSONIntroduction To Quantum MechanicsPhysicsISBN:9781107189638Author:Griffiths, David J., Schroeter, Darrell F.Publisher:Cambridge University Press
- Physics for Scientists and EngineersPhysicsISBN:9781337553278Author:Raymond A. Serway, John W. JewettPublisher:Cengage LearningLecture- Tutorials for Introductory AstronomyPhysicsISBN:9780321820464Author:Edward E. Prather, Tim P. Slater, Jeff P. Adams, Gina BrissendenPublisher:Addison-WesleyCollege Physics: A Strategic Approach (4th Editio...PhysicsISBN:9780134609034Author:Randall D. Knight (Professor Emeritus), Brian Jones, Stuart FieldPublisher:PEARSON





