E=1, S=1 , R=0, Q = 0, Q t+1* R. Qr 0 E- Or S.
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Q: The ammeter shown in the figure reads 4.4 A, R1 =7.0 2, R2=5.0 2, and R3 = 2.0 Q RI 10.0 V R2 A R3…
A: Given: R1=7.0 ΩR2=5.0 ΩR3=2.0 ΩAmmeter reading= 4.4 A
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Q: Compute the following if Voltage supply is 143 Volts. A. total capacitance B. total charge 8 µF
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- SOLVE FOR I, Iz amd E Using KIRCHHO F'S RULES E 14V 巧 12= 0-97A a 1052 102Original Object Image on Screen Object Distance Irmage Distance Which equation could you use to calculate the focal length of this lens?Answer provided. answer should be A.)22.95J,B.)4.28m/s Please show your complete solution and write clearly. Thank you.
- Given: Use the following label: R, = 3.00 Q, R, = 10.0 O, R3 = 5.00 Q, R, = 4.00 Q, and Rs = 6.00 Q. 3.00 N 10.0 N 4.00 N 5.00 N 6.00 N 15.0 V Find: (a) R, (b) I, (c) I,, l2, l3, l4, ls (d) V,, V2, V3, V4, V5 (e) P2, PsHow much work does the battery do in moving 5 μ C from its low terminal to its high terminal? Answer in μJ. Diagram is attached.In 10 minute please
- Six uncharged capacitors with equal capacitances are combined in parallel. The combination is connected to a 7.29 V battery, which charges the capacitors. The charging process involves 0.000313 C of charge moving through the battery. Find the capacitance C of each capacitor. С — FEn la figura = 12V, R = 62 R2 = 122 R3 = 42 R4 = 32 R5 = 5 2 ¿Cuál es la diferencia de potencial en R5? ww ww Ra www wwDistance from E, E velocity ET centre of earth R. -6.23*107 J 5*10' m/s 1*10“ J -1.25*10'J 2R. -3.13*107 J 1.88*107 m/s 0.613*10“ J -1.25*107 J 3R -2.08*107 J 0.83*10' m/s 0.41*10ª J -1.25*107 J 4R. -1.58*107 J 0.33*107 m/s 0.26*10ª J -1.25*107 J 6R. -1.04*10' J -0.21*107 m/s -1.25*10' J (Not possible) 8R. -7.82*106 J -047*10'm/s -1.25*107 J (Not possible) Not possible Really, really far O J -1.25* 2. Plot on one graph E, E and E, as a function of distance. Use smooth curves and straight lines to highlight the patterns. Indicate on the graph the maximum distance the rock will travel.
- Use the following label: R, = 3.00 Q, R̟ = 10.0 Q, R3 = 5.00 Q, R, = 4.00 Q, and R$ = 6.00 Q. %3D %3D 3.00 n 10.0 n 4.00 N 5.00 N 6.00 N + 15.0 V Find: (a) Rr (b) I (c) l,, I2, ls, la, Is (d) V,, V2, Va, V, Vs (e) P2, Ps: 60 74.901 V What Have I Learned So Far? olay oli siisq a llad snil to 90 ai vibolov ori llite ei modets soos ali tud os 8 fristeno V=- V Fig. 4.11 A ball is thrown upward near the edge of a cliff. eti stugmo 7 s v=-39.2 m/s Solve the following problems. Show your solutions. disq gru 2601 910-1) Diswauoit d 56 General Physics 1(Second Edition) & slammd tovarl Joy seoqque to qoi srli no busw a\mes to b990 thay Ot.p.pi ser 1. Shelley threw a tennis ball upward near the edge of a cliff that is about 100 meters above sea level. Suppose the ball was thrown with a speed of 30.1 m/s. Determine the vertical positions of the ball for every interval of 1 s until 10.0 seconds. ed er 2. To test the durability of an sample of it from the top of a 50-m high building. ova 119116 viipolsy ari bad n engineered plastic container, its inventor drops a a. How long will it take for the container to hit the ground? b. Determine the speed of the container as it hits the ground. diod (ol jon er i Isaihoz…v can be determined from: morbol Combining like terms gives: e where AV- accelerating voltage. Rearranging gives: 2eAV m Substitution of (equation 2) into (equation 1) gives: Data N R V T 1 2 m I 22 v² = mv² = eAV, 150 2V nottom of B²r² e 1 2eV m Br mm, qque aswog ogslov rigid lino sifodinilladut m Procedure: C xt)s (equation 2) Sad m 99781 0.140[m] 150 V 2.15CM 10.0215 M 2.04 (equation 3) 1. Set the accelerating voltage and current as given by the instructor.non W 2. Measure the diameter of the circle formed by the electron beam with the centimeter scale next to the vacuum tube. Record the radius of the beam in the data table (not the diameter). :noftaups 901 maoil bur N R V T 101 I Insanal del inobra woman de 190hef de (012) 150 turns 0.140[m] I, SA M 200 V 123.3 cm 0.033m