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Q: The generator of a car idling at 1.582E+3 rpm produces 15.3 V. What will the output be at a rotation…
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- Ɛz = 18 V 0.5 Q R, 2.50- R a R 1.5 0.5 N E2 = 45 V What equation do you get when you apply the loop rule to the loop abcdefgha, in terms of the variables in the figure? 0 = If the current through the top branch is Iɔ = 0.57 A, what is the current through the bottom, I3, in amps?1) The correct unit for current, I, are which of the following answers: A) 1 Amp = | C/s B) 1 Amp = ! C?/53 C) 1 Amp = |J/C•S D) 1 Amp = | N•m/C %3DThe severity of an electrical shock depends on all of the following except: O The duration of the contact with the source of the shock. O Whether the shock is AC (alternating current) or DC (direct current). Whether the source of the electricity is a battery or a DC power supply. O The path the current takes through the body.
- The generator of a car idling at 1.638E+3 rpm produces 11.6 V. What will the output be at a rotation speed of 2.762E+3 rpm, assuming nothing else changes? Answer in units of V.What is the current on the path 1-2 in the circuit below? 3 A 6.92e-4 A 1.4e-2 A 0 A16. A parallel plate capacitor used in a flash for a camera must be able to store 31.0 J of energy when connected to 312 V. (Most electronic flashes actually use a 1.50- to 6.00-V battery, but increase the effective voltage using a dc–dc inverter.) Assuming the capacitor completely discharges to produce a flash in 7.40×10−3 s, what average power is dissipated in the flashbulb during this time? _____kW