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- A charged belt, 50 cm wide, travels at 30 m/s between a source of charge and a sphere.The belt carries charge into the sphere at a rate corresponding to 100 mA. Compute the surface charge density on the beltA car battery is rated at 1,020 AH (i.e., amps x hours). Calculate the number of coulombs this is equivalent to. Your Answer:mobile battery during a charge cycle are shown in 1.26 The voltage and current at the terminals of an auto- Fig. P1.26. a) Calculate the total charge transferred to the battery. b) Calculate the total energy transferred to the battery. Figure P1.26 v (V) bnohostib 24 Loiwollol sd lo 16 8- 8. 12 16 20 t (ks) i (A) maxi 24 16 8- the che 4 8. 12 16 20 t (ks)
- The amount of charge flowing through a particular conducting wire is represented by the equation q (t) = at³ + bt? +c, where a = 5 A/s?, b = 3 A/s, and c is an unknown constant. When t = 0.9s, the voltage across the wire is measured to be 12 V. What is the resistance of the wire in Ohms?A car battery is rated at 600 AH (i.e., amps x hours). Calculate the number of coulombs this is equivalent to.Consider the circuit shown in the figure below, where C, = 8.00 µF, C, = 6.00 µF, and AV = 18.0 V. Capacitor C, is first charged by closing switch S,. Switch S, is then opened, and the charged capacitor is connected to the uncharged capacitor by closing S,. AV (a) Calculate the initial charge (in µC) acquired by C,. (Round your answer to at least one decimal place.) 144 (b) Calculate the final charge (in µC) on each capacitor. (Round your answers to at least the nearest integer.) Q, = 61.7 X µC Q2 = 82.3 X µC (c) What If? After a very long time, switch S, is also closed. By what amount does the charge on the second capacitor change after S, has been closed for a very long time? (Give your answer in µC.) 46.3 V µC
- Consider the circuit shown in the figure below, where C, = 8.00 µF, C, = 6.00 µF, and AV = 18.0 V. Capacitor C, is first charged by closing switch S,. Switch s, is then opened, and the charged capacitor is connected to the uncharged capacitor by closing S,. AV (a) Calculate the initial charge (in µC) acquired by C,. (Round your answer to at least one decimal place.) (b) Calculate the final charge (in µC) on each capacitor. (Round your answers to at least the nearest integer.) Q1 = %3D (c) What If? After a very long time, switch S, is also closed. By what amount does the charge on the second capacitor change after S, has been closed for a very long time? (Give your answer in µC.)An electron is fired at a speed vi = 4.3 × 106 m/s and at an angle θi = 39.7° between two parallel conducting plates as shown in the figure. If s = 1.7 mm and the voltage difference between the plates is ΔV = 99.8 V, determine how close, w, the electron will get to the bottom plate. Put your answer in meters and include at 6 decimal places in your answer. Do not include units. The x-axis of the coordinate system is in the middle of the parallel plate capacitor. Round your answer to 6 decimal places.During 33 seconds of use, 340 C of charge flow through a microwave oven. Compute the size of the electric current.
- A metallic wire with resistance 8 0, crosS-sectional area 8.2x10 m², and conductivity o= 5.81×10 0 m mV/m) inside the wire is; -1 is connected to a battery of & = 14 V. The electric field E (in 3.sq 4. Maximum surface resistance. Consider a square sheet of side L, thickness d, and electrical resistivity p. The resistance measured between opposite edges of the sheet is called the surface resistance: R = pL/ Ld = p/d, which is independent of the area of the sheet. (R is called the resistance per square and is expected in ohms per square, because p/d has the dimensions of ohms.) If we express p by p=m/ne²t, then Rq=m/ndet. Suppose now that the minimum value of the collision time is determined by scattering from surfaces of the sheet, so that Td/v, where v is the Fermi velocity. Thus the maximum surface resistivity is Rmvp/nd²e². Show for a monatomic metal sheet one atom thickness that sq Rħ/e² = 4.1k, where lkn is 10³ ohms. sqAn electron is fired at a speed vi = 3.1 × 106 m/s and at an angle θi = 36.8° between two parallel conducting plates as shown in the figure. If s = 1.8 mm and the voltage difference between the plates is ΔV = 98.8 V, determine how close, w, the electron will get to the bottom plate. Put your answer in meters and include at 6 decimal places in your answer. Do not include units. The x-axis of the coordinate system is in the middle of the parallel plate capacitor