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- Q1/By, using IC {7408 ,7432,7404 , 7400 ,7402 , and 7486} draw the following equation and find the truth table for each one? 1- Y=Ā.B+ (A.B + B.C) 2- Y= (A+B+C).Ā 3- Y= (A+B). (A + B). (B.C)There is one part to this question. I need to know the nm. Thank you!3. Let E = (-2)x+ây +5₂ V/m and calculate (a) VPQ given P(-7,2,1) and Q(4,1,2) (b) Vp if V = 0 at Q (c) Vp if V = 0 at (2,0,-1)
- CV below shows the electrochemical behaviour of Species A that undergoes EC reaction at the electrode-electrolyte interface. Explain the electrochemical behaviour. The scan rates are A= 500mV/s, B= 250 mV/S and C= 100 mV/S. -0.80 -0.60 -0.40 -0.20 0.1 0.2 0.3 0.4 0.5 0.00 B 0.20 Potential, V 0.40 Current, µAHW/2 0, If the switch in Fig. 10 opens at 1 , find v(7) for 7 = 0 and we(0). Answer: 8¢V, 533 1. =0 UV PEquation1: Q(t)=CVcap(t) Equation 2: Qcharging(t)=CV(1−e^(−t/RC)) 1. Combine equations 1 and 2 to create an equation capable of finding the time-dependent voltage across a charging capacitor. Equation 3: I(t)≡(dQ(t))/(dt) 2. Combine equations 2 and 3 to create an equation capable of finding the time-dependent current across a charging capacitor.
- 1. Find: RT = ? IT =? 65 0 84 2 25 V 73 2 10 2In a parallel-plate diode, the anode is at 100 V with respect to the cathode which is 10 mm away from it. An electron is emitted from cathode with an initial velocity towards cathode of 1x10 m/s. The arrival velocity of the electron at anode is ..3V 1 U = 2n² v? w?dw (Н. W.) %3D hw eKBT-1
- 9 V Hol 2 R₁ 10 Q 3 www R₂ 20 8 7 6 11. What is the value of V₁? b. 6 V b. 8 V c. 9V 12. What is the value of I₁? b. 11.4 A b. 13.4 A d. 015.4A 13. What is the values of V₁? b. 1.5 V b. 5 V d. 9V 14. What is the value of V₂? b. 1.5 V b. 5 V d. 9V 15. What is the value of V3 ? b. 1.5 V b. 5 V d. 9V 4 www R3 10 5 d. 10V c. 14.4A c. 2.5 V c. 2.5 V c. 2.5 VHow was it solved. need g-k Answers: a. 98.9949 Vb. 197.9899 Vc. 70 Vd. 100 mse. 10 Hzf. 20π rad/secg. 125.0876°h. v(t) = 98.9949sin(20πt + 125.0876°)i. -29.0880 Vj. 0.99959°k. 63.643 msThe electrons in a cathode ray tube are accelerated from cathode(negative terminal) to anode(positive terminal) by a potential difference of 2000 V. If this potential difference is increased to 8000 V the electrons will arrive at the screen with a.twice the kinetic energy and four times the velocity. b.twice the kinetic energy and twice the velocity. c.four times the kinetic energy and twice the velocity. d.four times the kinetic energy and four times the velocity. e.twice the kinetic energy and the same velocity.