Physics Solution Manual for 1100 and 2101

The power p is the energy heat in this case per unit

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Unformatted text preview: he product of the current I and the time ∆t, according to Equation 20.1, so that ∆q I ∆t (1.5 ×10−11 A ) ( 0.50 s ) = = = 4.7 ×107 Number of Na+ ions = −19 e e 1.60 × 10 C ____________________________________________________________________________________________ REASONING AND SOLUTION Ohm's law (Equation 20.2, V = IR ) gives the result directly V 240 V I= = = 22 A R 11 Ω ______________________________________________________________________________ 5. 6. REASONING Voltage is a measure of energy per unit charge (joules per coulomb). Therefore, when an amount ∆q of charge passes through the toaster and there is a potential difference across the toaster equal to the voltage V of the outlet, the energy that the charge delivers to the toaster is given by 1058 ELECTRIC CIRCUITS Energy = V ∆q (1) The charge ∆q that flows through the toaster in a time ∆t depends upon the magnitude I of ∆q the electric current according to I = (Equation 20.1). Therefore, we have that ∆t ∆q = I ∆t (2) V (Equation 20.2), to calculate the current in the toaster in R terms of the voltage V of the outlet and the resistance R of the toaster. We will employ Ohm’s law, I = SOLUTION Substituting Equation (2) into Equation (1) yields Energy = V ∆q = VI ∆t Substituting I = (3) V (Equation 20.2) into equation (3), we find that R (120 V ) V2 V ( 60.0 s ) = 6.2 ×104 J ∆t = Energy = VI ∆t = V ∆t = R R 14 Ω ______________________________________________________________________________ 2 7. REASONING According to Ohm’s law, the resistance is the voltage of the battery divided by the current that the battery delivers. The current is the charge divided by the time during which it flows, as stated in Equation 20.1. We know the time, but are not given the charge directly. However, we can determine the charge from the energy delivered to the resistor, because this energy comes from the battery, and the potential difference between the battery terminals is the difference in electric potential energy per unit charge, according to Equation 19.4. Thus, a 9.0-V battery supplies 9.0 J of energy to each coulomb of charge passing through it. To calculate the charge, then, we need only divide the energy from the battery by the 9.0 V potential difference. SOLUTION Ohm’s law indicates that the resistance R is the voltage V of the battery divided by the current I, or R = V/I. According to Equation 20.1, the current I is the amount of charge ∆q divided by the time ∆t , or I = ∆q/∆t . Using these two equations, we have R= V V V ∆t = = I ∆q / ∆t ∆q Chapter 20 Problems 1059 According to Equation 19.4, the potential difference ∆V is the difference ∆(EPE) in the ∆ ( EPE ) electric potential energy divided by the charge ∆q, or ∆V = . However, it is ∆q customary to denote the potential difference across a battery by V, rather than ∆V, so ∆ ( EPE ) ∆ ( EPE ) V= . Solving this expression for the charge gives ∆q = . Using this result ∆q V in the expression for the resistance, we find that ( 9.0 V ) ( 6 × 3600 s ) = 16 Ω V ∆t V ∆t V 2 ∆t R= = = = ∆q ∆ ( EPE ) /V ∆ ( EPE ) 1.1×105 J 2 8. REASONING AND SOLUTION a. The total charge that can be delivered is 3600 s 5 ∆q = (220 A ⋅ h) = 7.9 × 10 C 1h b. The maximum current is 220 A ⋅ h = 350 A 1h (38 min) 60 min ______________________________________________________________________________ I= 9. SSM REASONING The number N of protons that strike the target is equal to the amount of electric charge ∆q striking the target divided by the charge e of a proton, N = (∆q)/e. From Equation 20.1, the amount of charge is equal to the product of the current I and the time ∆t. We can combine these two relations to find the number of protons that strike the target in 15 seconds. The heat Q that must be supplied to change the temperature of the aluminum sample of mass m by an amount ∆ T is given by Equation 12.4 as Q = cm∆T , where c is the specific heat capacity of aluminum. The heat is provided by the kinetic energy of the protons and is equal to the number of protons that strike the target times the kinetic energy per proton. Using this reasoning, we can find the change in temperature of the block for the 15 secondtime interval. SOLUTION a. The number N of protons that strike the target is N= ∆q I ∆t (0.50 × 10−6 A)(15 s) = = = 4.7 × 1013 −19 e e 1.6 × 10 C 1060 ELECTRIC CIRCUITS b. The amount of heat Q provided by the kinetic energy of the protons is Q = (4.7 × 1013 protons)(4.9 × 10−12 J/proton) = 230 J Since Q = cm∆T and since Table 12.2 gives the specific heat of aluminum as c = 9.00 × 102 J/(kg.C°), the change in temperature of the block is Q 230 J = = 17 C° cm J 2 −3 9.00 × 10 kg ⋅ C° (15 × 10 kg) ______________________________________________________________________________ ∆T = 10. REASONING a. The resistance R of a piece of material is related to its length L and cross-sectional area A by Equation 20.3, R = ρ L / A , where ρ is the resistivity of the ma...
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This note was uploaded on 04/30/2013 for the course PHYS 1100 and 2 taught by Professor Chastain during the Spring '13 term at LSU.

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