Q2: The 300 N force as shown in figure acts on the point O. Resolve this force into two components one along AC and the other through point C? Calculate * R 300 N 4 A 4

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. Calculate current across the capacitor shown in Fig. P8.6 if the voltage input is v1(t)=30 cos (320t-50°) V . v(t) i(t) Figure P8.6 For a current i(t) = C cos(wt+P), enter the value of the phase angle (P). (be sure that your answer is between -180deg to 180deg.) Notes on entering solution: • Enter your solution in degrees Have your current in cosine format . Do not include units in your answer ex. 5° is entered as 5 C = 1 µF

A particle accelerator sends high-velocity charged particles down an evacuated pipe. Explain how a coil of wire wrapped around the pipe could detect the passage of individual particles. Sketch a graph of the voltage output of the coil as a single particle passes through it.

Explain how did they get this results table. What does 2V ac mean?

16.  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

The coil shown in the figure has 2 turns, a cross- sectional area of 0.20 m2, and a field (parallel to the axis of the coil) with a magnitude given by B = (4.0 + 3.0t2) T, where t is in s. What is the potential difference, VA - Vc, at t = 3.0 s? B A а. -7.2 V b. +7.2 V С. -12 V d. -4.8 V е. +4.8 V

A 1.13-mF capacitor is connected to a North American electrical outlet (AVms = 120 V, f = 60.0 Hz). Assuming the energy stored in the capacitor is zero at t = 0, determine the magnitude of the current in the wires at t = S. 174 x Your response differs from the correct answer by more than 10%. Double check your calculations. Al Need Help? Read It Submit Answer

Apply the loop rule to loop abcdefgha in the figure. E, = =D18 V d. 12 0.5 2 R2 2.5 Q 12 R, a 13 6.0 2 I3 R 1.5 2 0.5 2 E2 = 45 V %3D O -1,R, + E, -I1 + I;R3 + I3r2+ E2 = 0 O -I,R, + E, - I,r, + I,R3 + I,r2 - E2 = 0 O -I,R, + E, - I,r, - I,R3 - I,r2 + E2 = 0 O I,R, +E, + Ir, - 1,R, - Ir, - E, = 0 %3D %3D %3D

Problem 10: While preparing a laboratory experiment on capacitors, you connect an E-4.5 V DC power supply to fully charge an old capacitor that is marked C- 330 HF Your lab assistant hooks up the circuit as shown below. When the switch S is at point a, the capacitor is connected to the power supply. When the switch S is at point b, the capacitor is disconnected from the power supply and connected to a voltmeter, V. The voltmeter has an internal resistance R-10.0 MS2 al Otheexpertta.com Part (a) The capacitor is initially uncharged, when the switch is moved to position a, hooking the capacitor C to the power supply. What will happen to the charge on the capacitor? It will quickly (but not instantaneously) increase to a maximum value Correct! Part (b) Calculate the charge on the capacitor (in Coulombs) when it is fully charged by the power supply Part (c) After the capacitor is fully charged, you switch the switch to position b at time t-0 seconds. The voltmeter initially reads 4.5 V.…