The thermal radiation from the earth’s surface peaks at a

Problem 1P The intensity of sunlight at the top of the earth’s atmosphere is approximately . Mars is about 1.5 times as far from the sun as the earth. What is the approximate intensity of sunlight at the top of Mars’s atmosphere?

Problem 2P Averaged over day, night, seasons, and weather conditions, a square meter of the earth’s surface receives an average of 240 W of radiant energy from the sun. The average power radiated back to space is A. Less than 240 W. B. More than 240 W. C. Approximately 240 W.

Problem 3P The thermal radiation from the earth’s surface peaks at a wavelength of approximately 10 ?m. If the surface of the earth warms, this peak will A. Shift to a longer wavelength. B. Stay the same. C. Shift to a shorter wavelength.

Problem 4P The thermal radiation from the earth’s surface peaks at a wavelength of approximately 10 ?m. What is the energy of a photon at this wavelength? A. 2.4 eV B. 1.2 eV C. 0.24 Ev D. 0.12 eV

Problem 5P Electromagnetic waves in certain wavelength ranges interact with water molecules because the molecules have a large electric dipole moment. The electric field of the wave A. Exerts a net force on the water molecules. B. Exerts a net torque on the water molecules. C. Exerts a net force and a net torque on the water molecules.

Problem 6P Taking an X Ray X rays are a very penetrating form of electromagnetic radiation. X rays pass through the soft tissue of the body but are largely stopped by bones and other more dense tissues. This makes x rays very useful for medical and dental purposes, as you know. A schematic view of an x-ray tube and a driver circuit is given in Figure VI.1. A filament warms the cathode, freeing electrons. These electrons are accelerated by the electric field established by a high-voltage power supply connected between the cathode and a metal target. The electrons accelerate in the direction of the target. The rapid deceleration when they strike the target generates x rays. Each electron will emit one or more x rays as it comes to rest. An x-ray image is essentially a shadow; x rays darken the film where they pass, but the film stays unexposed, and thus light, where bones or dense tissues block x rays. An x-ray technician adjusts the quality of an image by adjusting the energy and the intensity of the x-ray beam. This is done by adjusting two parameters: the accelerating voltage and the current through the tube. The accelerating voltage determines the energy of the x-ray photons, which can’t be greater than the energy of the electrons. The current through the tube determines the number of electrons per second and thus the number of photons emitted. In clinical practice, the exposure is characterized by two values: “kVp” and “mAs.” kVp is the peak voltage in kV. The value mAs is the product of the current (in mA) and the time (in s) to give a reading in mA . s. This is a measure of the total number of electrons that hit the target and thus the number of x rays emitted. Typical values for a dental x ray are a kVp of 70 (meaning a peak voltage of 70 kV) and mAs of 7.5 (which comes from a current of 10 mA for 0.75 s, for a total of 7.5 mAs). Assume these values in all of the problems that follow. In Figure VI.1, what is the direction of the electric field in the region between the cathode and the target electrode? A. To the left B. To the right C. Toward the top of the page D. Toward the bottom of the page

Problem 7P Taking an X Ray X rays are a very penetrating form of electromagnetic radiation. X rays pass through the soft tissue of the body but are largely stopped by bones and other more dense tissues. This makes x rays very useful for medical and dental purposes, as you know. A schematic view of an x-ray tube and a driver circuit is given in Figure VI.1. A filament warms the cathode, freeing electrons. These electrons are accelerated by the electric field established by a high-voltage power supply connected between the cathode and a metal target. The electrons accelerate in the direction of the target. The rapid deceleration when they strike the target generates x rays. Each electron will emit one or more x rays as it comes to rest. An x-ray image is essentially a shadow; x rays darken the film where they pass, but the film stays unexposed, and thus light, where bones or dense tissues block x rays. An x-ray technician adjusts the quality of an image by adjusting the energy and the intensity of the x-ray beam. This is done by adjusting two parameters: the accelerating voltage and the current through the tube. The accelerating voltage determines the energy of the x-ray photons, which can’t be greater than the energy of the electrons. The current through the tube determines the number of electrons per second and thus the number of photons emitted. In clinical practice, the exposure is characterized by two values: “kVp” and “mAs.” kVp is the peak voltage in kV. The value mAs is the product of the current (in mA) and the time (in s) to give a reading in mA . s. This is a measure of the total number of electrons that hit the target and thus the number of x rays emitted. Typical values for a dental x ray are a kVp of 70 (meaning a peak voltage of 70 kV) and mAs of 7.5 (which comes from a current of 10 mA for 0.75 s, for a total of 7.5 mAs). Assume these values in all of the problems that follow. If the distance between the cathode and the target electrode is approximately 1.0 cm, what will be the maximum acceleration of the free electrons? Assume that the electric field is uniform. Step-by-step solution

Problem 8P Taking an X Ray X rays are a very penetrating form of electromagnetic radiation. X rays pass through the soft tissue of the body but are largely stopped by bones and other more dense tissues. This makes x rays very useful for medical and dental purposes, as you know. A schematic view of an x-ray tube and a driver circuit is given in Figure VI.1. A filament warms the cathode, freeing electrons. These electrons are accelerated by the electric field established by a high-voltage power supply connected between the cathode and a metal target. The electrons accelerate in the direction of the target. The rapid deceleration when they strike the target generates x rays. Each electron will emit one or more x rays as it comes to rest. An x-ray image is essentially a shadow; x rays darken the film where they pass, but the film stays unexposed, and thus light, where bones or dense tissues block x rays. An x-ray technician adjusts the quality of an image by adjusting the energy and the intensity of the x-ray beam. This is done by adjusting two parameters: the accelerating voltage and the current through the tube. The accelerating voltage determines the energy of the x-ray photons, which can’t be greater than the energy of the electrons. The current through the tube determines the number of electrons per second and thus the number of photons emitted. In clinical practice, the exposure is characterized by two values: “kVp” and “mAs.” kVp is the peak voltage in kV. The value mAs is the product of the current (in mA) and the time (in s) to give a reading in mA . s. This is a measure of the total number of electrons that hit the target and thus the number of x rays emitted. Typical values for a dental x ray are a kVp of 70 (meaning a peak voltage of 70 kV) and mAs of 7.5 (which comes from a current of 10 mA for 0.75 s, for a total of 7.5 mAs). Assume these values in all of the problems that follow. What, physically, does the product of a current (in mA) and a time (in s) represent? A. Energy in mJ B. Potential difference in mV C. Charge in mC D. Resistance in m?

Problem 9P Taking an X Ray X rays are a very penetrating form of electromagnetic radiation. X rays pass through the soft tissue of the body but are largely stopped by bones and other more dense tissues. This makes x rays very useful for medical and dental purposes, as you know. A schematic view of an x-ray tube and a driver circuit is given in Figure VI.1. A filament warms the cathode, freeing electrons. These electrons are accelerated by the electric field established by a high-voltage power supply connected between the cathode and a metal target. The electrons accelerate in the direction of the target. The rapid deceleration when they strike the target generates x rays. Each electron will emit one or more x rays as it comes to rest. An x-ray image is essentially a shadow; x rays darken the film where they pass, but the film stays unexposed, and thus light, where bones or dense tissues block x rays. An x-ray technician adjusts the quality of an image by adjusting the energy and the intensity of the x-ray beam. This is done by adjusting two parameters: the accelerating voltage and the current through the tube. The accelerating voltage determines the energy of the x-ray photons, which can’t be greater than the energy of the electrons. The current through the tube determines the number of electrons per second and thus the number of photons emitted. In clinical practice, the exposure is characterized by two values: “kVp” and “mAs.” kVp is the peak voltage in kV. The value mAs is the product of the current (in mA) and the time (in s) to give a reading in mA . s. This is a measure of the total number of electrons that hit the target and thus the number of x rays emitted. Typical values for a dental x ray are a kVp of 70 (meaning a peak voltage of 70 kV) and mAs of 7.5 (which comes from a current of 10 mA for 0.75 s, for a total of 7.5 mAs). Assume these values in all of the problems that follow. During the 0.75 s that the tube is running, what is the electric power? A. 7.0 kW B. 700 W C. 70 W D. 7.0 W

Problem 10P Taking an X Ray X rays are a very penetrating form of electromagnetic radiation. X rays pass through the soft tissue of the body but are largely stopped by bones and other more dense tissues. This makes x rays very useful for medical and dental purposes, as you know. A schematic view of an x-ray tube and a driver circuit is given in Figure VI.1. A filament warms the cathode, freeing electrons. These electrons are accelerated by the electric field established by a high-voltage power supply connected between the cathode and a metal target. The electrons accelerate in the direction of the target. The rapid deceleration when they strike the target generates x rays. Each electron will emit one or more x rays as it comes to rest. An x-ray image is essentially a shadow; x rays darken the film where they pass, but the film stays unexposed, and thus light, where bones or dense tissues block x rays. An x-ray technician adjusts the quality of an image by adjusting the energy and the intensity of the x-ray beam. This is done by adjusting two parameters: the accelerating voltage and the current through the tube. The accelerating voltage determines the energy of the x-ray photons, which can’t be greater than the energy of the electrons. The current through the tube determines the number of electrons per second and thus the number of photons emitted. In clinical practice, the exposure is characterized by two values: “kVp” and “mAs.” kVp is the peak voltage in kV. The value mAs is the product of the current (in mA) and the time (in s) to give a reading in mA . s. This is a measure of the total number of electrons that hit the target and thus the number of x rays emitted. Typical values for a dental x ray are a kVp of 70 (meaning a peak voltage of 70 kV) and mAs of 7.5 (which comes from a current of 10 mA for 0.75 s, for a total of 7.5 mAs). Assume these values in all of the problems that follow. If approximately 1% of the electric energy ends up in the x-ray beam (a typical value), what is the approximate total energy of the x rays emitted? A. 500 J B. 50 J C. 5 J D. 0.5 J

Problem 11P Taking an X Ray X rays are a very penetrating form of electromagnetic radiation. X rays pass through the soft tissue of the body but are largely stopped by bones and other more dense tissues. This makes x rays very useful for medical and dental purposes, as you know. A schematic view of an x-ray tube and a driver circuit is given in Figure VI.1. A filament warms the cathode, freeing electrons. These electrons are accelerated by the electric field established by a high-voltage power supply connected between the cathode and a metal target. The electrons accelerate in the direction of the target. The rapid deceleration when they strike the target generates x rays. Each electron will emit one or more x rays as it comes to rest. An x-ray image is essentially a shadow; x rays darken the film where they pass, but the film stays unexposed, and thus light, where bones or dense tissues block x rays. An x-ray technician adjusts the quality of an image by adjusting the energy and the intensity of the x-ray beam. This is done by adjusting two parameters: the accelerating voltage and the current through the tube. The accelerating voltage determines the energy of the x-ray photons, which can’t be greater than the energy of the electrons. The current through the tube determines the number of electrons per second and thus the number of photons emitted. In clinical practice, the exposure is characterized by two values: “kVp” and “mAs.” kVp is the peak voltage in kV. The value mAs is the product of the current (in mA) and the time (in s) to give a reading in mA . s. This is a measure of the total number of electrons that hit the target and thus the number of x rays emitted. Typical values for a dental x ray are a kVp of 70 (meaning a peak voltage of 70 kV) and mAs of 7.5 (which comes from a current of 10 mA for 0.75 s, for a total of 7.5 mAs). Assume these values in all of the problems that follow. What is the maximum energy of the emitted x-ray photons?

Problem 12P Electric Cars In recent years, practical hybrid cars have hit the road—cars in which the gasoline engine runs a generator that charges batteries that run an electric motor. These cars offer increased efficiency, but significantly greater efficiency could be provided by a purely electric car run by batteries that you charge by plugging into an electric outlet in your house. But there’s a practical problem with such vehicles: the time necessary to recharge the batteries. If you refuel your car with gas at the pump, you add 130 MJ of energy per gallon. If you add 20 gallons, you add a total of 2.6 GJ in about 5 minutes. That’s a lot of energy in a short time; the electric system of your house simply can’t provide power at this rate. There’s another snag as well. Suppose there were electric filling stations that could provide very high currents to recharge your electric car. Conventional batteries can’t recharge very quickly; it would still take longer for a recharge than to refill with gas. One possible solution is to use capacitors instead of batteries to store energy. Capacitors can be charged much more quickly, and as an added benefit, they can provide energy at a much greater rate— allowing for peppier acceleration. Today’s capacitors can’t store enough energy to be practical, but future generations will. A typical home’s electric system can provide 100 A at a voltage of 220 V. If you had a charger that ran at this full power, approximately how long would it take to charge a battery with the equivalent of the energy in one gallon of gas? A. 100 min B. 50 min C. 20 min D. 5 min

Problem 13P Electric Cars In recent years, practical hybrid cars have hit the road—cars in which the gasoline engine runs a generator that charges batteries that run an electric motor. These cars offer increased efficiency, but significantly greater efficiency could be provided by a purely electric car run by batteries that you charge by plugging into an electric outlet in your house. But there’s a practical problem with such vehicles: the time necessary to recharge the batteries. If you refuel your car with gas at the pump, you add 130 MJ of energy per gallon. If you add 20 gallons, you add a total of 2.6 GJ in about 5 minutes. That’s a lot of energy in a short time; the electric system of your house simply can’t provide power at this rate. There’s another snag as well. Suppose there were electric filling stations that could provide very high currents to recharge your electric car. Conventional batteries can’t recharge very quickly; it would still take longer for a recharge than to refill with gas. One possible solution is to use capacitors instead of batteries to store energy. Capacitors can be charged much more quickly, and as an added benefit, they can provide energy at a much greater rate— allowing for peppier acceleration. Today’s capacitors can’t store enough energy to be practical, but future generations will. The Tesla Roadster, a production electric car, has a 375 V battery system that can provide a power of 200 kW. At this peak power, what is the current supplied by the batteries? A. 75 kA B. 1900 A C. 530 A D. 75 A

Problem 14P Electric Cars In recent years, practical hybrid cars have hit the road—cars in which the gasoline engine runs a generator that charges batteries that run an electric motor. These cars offer increased efficiency, but significantly greater efficiency could be provided by a purely electric car run by batteries that you charge by plugging into an electric outlet in your house. But there’s a practical problem with such vehicles: the time necessary to recharge the batteries. If you refuel your car with gas at the pump, you add 130 MJ of energy per gallon. If you add 20 gallons, you add a total of 2.6 GJ in about 5 minutes. That’s a lot of energy in a short time; the electric system of your house simply can’t provide power at this rate. There’s another snag as well. Suppose there were electric filling stations that could provide very high currents to recharge your electric car. Conventional batteries can’t recharge very quickly; it would still take longer for a recharge than to refill with gas. One possible solution is to use capacitors instead of batteries to store energy. Capacitors can be charged much more quickly, and as an added benefit, they can provide energy at a much greater rate— allowing for peppier acceleration. Today’s capacitors can’t store enough energy to be practical, but future generations will. To charge the batteries in a Tesla Roadster, a transformer is used to step up the voltage of the household supply. If you step a 220 V, 100 A system up to 400 V, what is the maximum current you can draw at this voltage? A. 180 A B. 100 A C. 55 A D. 45 A

Problem 15P Electric Cars In recent years, practical hybrid cars have hit the road—cars in which the gasoline engine runs a generator that charges batteries that run an electric motor. These cars offer increased efficiency, but significantly greater efficiency could be provided by a purely electric car run by batteries that you charge by plugging into an electric outlet in your house. But there’s a practical problem with such vehicles: the time necessary to recharge the batteries. If you refuel your car with gas at the pump, you add 130 MJ of energy per gallon. If you add 20 gallons, you add a total of 2.6 GJ in about 5 minutes. That’s a lot of energy in a short time; the electric system of your house simply can’t provide power at this rate. There’s another snag as well. Suppose there were electric filling stations that could provide very high currents to recharge your electric car. Conventional batteries can’t recharge very quickly; it would still take longer for a recharge than to refill with gas. One possible solution is to use capacitors instead of batteries to store energy. Capacitors can be charged much more quickly, and as an added benefit, they can provide energy at a much greater rate— allowing for peppier acceleration. Today’s capacitors can’t store enough energy to be practical, but future generations will. One design challenge for a capacitor-powered electric car is that the voltage would change with time as the capacitors discharged. If the capacitors in a car were discharged to half their initial voltage, what fraction of energy would still be left? A. 75% B. 67% C. 50% D. 25%

Problem 16P The wireless power transfer system is outlined in Figure VI.2. An AC supply generates a current through the primary coil, creating a varying magnetic field. This field induces a current in the secondary coil, which is connected to a resistance (the lightbulb) and a capacitor that sets the resonance frequency of the secondary circuit to match the frequency of the primary circuit. At a particular moment, the current in the primary coil is clockwise, as viewed from the secondary coil. At the center of the secondary coil, the field from the primary coil is A. To the right. B. To the left. C. Zero.

Problem 17P The wireless power transfer system is outlined in Figure VI.2. An AC supply generates a current through the primary coil, creating a varying magnetic field. This field induces a current in the secondary coil, which is connected to a resistance (the lightbulb) and a capacitor that sets the resonance frequency of the secondary circuit to match the frequency of the primary circuit. At a particular moment, the magnetic field from the primary coil points to the right and is increasing in strength. The field due to the induced current in the secondary coil is A. To the right. B. To the left. C. Zero.

Problem 18P The wireless power transfer system is outlined in Figure VI.2. An AC supply generates a current through the primary coil, creating a varying magnetic field. This field induces a current in the secondary coil, which is connected to a resistance (the lightbulb) and a capacitor that sets the resonance frequency of the secondary circuit to match the frequency of the primary circuit. The power supply drives the primary coil at 9.9 MHz. If this frequency is doubled, how must the capacitor in the secondary circuit be changed?

Problem 19P The wireless power transfer system is outlined in Figure VI.2. An AC supply generates a current through the primary coil, creating a varying magnetic field. This field induces a current in the secondary coil, which is connected to a resistance (the lightbulb) and a capacitor that sets the resonance frequency of the secondary circuit to match the frequency of the primary circuit. What are the rms and peak currents for a 60 W bulb? (The rms voltage is the usual 120 V.) A. 0.71 A, 0.71 A B. 0.71 A, 0.50 A C. 0.50 A, 0.71 A D. 0.50 A, 0.50 A

Problem 20AIP A 20 ? resistor is connected across a 120 V source. The resistor is then lowered into an insulated beaker, containing 1.0 L of water at 20° C, for 60 s. What is the final temperature of the water?

Problem 21AIP As shown in Figure VI.3, a square loop of wire, with a mass of 200 g, is free to pivot about a horizontal axis through one of its sides. A 0.50 T horizontal magnetic field is directed as shown. What current I in the loop, and in what direction, is needed to hold the loop steady in a horizontal plane?