Reclung.cA onri gleFigure 12. The circuit shown above is constructed with two 6.0 V batteries and three resistors with the values shown.The currents 1₁, 12, and 13 in each branch of the circuit are indicated.2GEOMETRY AND TRIGONOMETRYCircleA = ²²C = 2πrCy.inderbh- reRectangular SolidVlhSphere1. Using Kirchhoff's rules, write, but DO NOT SOLVE, equations that can be used to solve for thecurrent in each resistor.ii. Calculate the current in the 200 9 resistor.iii. Calculate the power dissipated by the 200 2 resistor.V = ar²lS..Figure 2The two 6.0 V batteries are replaced with a battery with voltage & and a resistor of resistance 50 2, as shownabove. The voltmeter V shows that the voltage across the 200 £2 resistor is 4.4 V.(b) Calculate the current through the 50 2 resistor.(c) Calculate the voltage & of the battery.V=²S = 4; 7²Rig Triangle2 +6² = c²sin =COS==tan 8 =OJAbC2m.ab6.0 V7,4&A - areaC=circumferenceh-hei£= le LW = IULr=fuu.0V-volumeS=sur. ce arca1. FC5arc le tung.e50150 Ωwww1₂200 ΩΣADVANCED PLACEMENT PHYSICS C EQUATIONSb150 92www100 £2www90%Ta6.0 V100 92a200 ΩΣ (V) 50 ΩΣCALCULUSdf df dudu dx(*²)=(e) = ae(In cx) ==[sin(x)] = acos(x)cs(x)] = x√x²dx=¹¹_n-1Joth dx = 10mx-1dx√= In|x + al[cos(ax)dx = =sin(x)[sin (ax) dx = cos(x)VECTOR PRODUCTSA-B= AB cose|Ã x B= ABsine v=x0+ axtx = xo + x² + ₂a2=²+2a₂ (-0)FΣF F-m m|J = √F 1 = App = mvFSHFAE=W=F-drP =dtP = F.jAU = mg h4 = 2²2 = 6³²7T=FxPà-4-441-Sr dm = 2mr¹Σm.x,m₂Xcm-V=Y@= F x1 = loK=@=000 + at6 = 6 + + ½a²²UNITSYMBOLS₁-2BMECHANICSFactor Prefix10⁹ giga1010³PREFIXESmegakilocentimillimicro161.9nano10-12 picoJaE- ene y= _orceF=СADVANCED PLACEMENT PHYSICS C EQUATIONSItI1mnP7Jm, ulse--.cy enc= .ght- kinetic e e yk spring constantl = len ulaccelerationLangu. ar...mentu....7h = massP =0::crp mo...ent mSymbolGMkotatio ali e tia=1 = ....ea5.ampere,kelvin,TJotential ener tyOCHevioa=elocity or s, ceaW = work done on a > stemXProton mass, m,1.67 x 10 27 kgNet ron... ss, m, = 1.67 x 10 kgElectron mass, m, = 9.11 10-31 kgA og du's numbe, N = 5.02 x 1023 mol-1Universal gas constant, R = 8.31 J/(mol-K)Boltzmann's constant, kg = 1.38 x 10-23 J/K1 unified atomic mass unit,Planck's constant,e-angleosition= coefficient of frictiond.sta ceR = -KAXtorquea gulur speeda guluf accelerationplase a glemmeter,kilogram, kgsecond,SAKU. - * (4x)²J = max cos(col + 0)1-²-3-7=T₁ = 2₁1₂ =2= √//|Fal - GimmUg =Gm,m₂r||FR|- 1992AneoF₂mole, molhertz, Hz.newton, Npascal, Pajoule, Joinecosetane$8-dÀ=&E-di-dVXAV = -√E-dirE₁ =-ELECTRICITY AND MAGNETISMA = area-B magnetic fieldC = capacitancedistanceE electric fieldemfE-forceV = 159ArzoC-U₂ -97=AV = 2C=ECG-241-d=R-PL=لم = F1 = Nev AΔΙΑR.-ERΣΕADVANCED PLACEMENT PHYSICS C TABLE OF INFORMATIONNLANIS AND CONVERN FIORSΕ-ΣΕRPP = 1AV1 9192Απερ TElectron charge na gn..ude,Uc-AV=C(AV)² = He Idex4Sd of light,Universal gravitationalconstant,Acceleratio i due to gravityat Earth su..ace,1 = currentJ= current density-L-inductance<= lengthn = number of loops of wireper unit lengthNnumber of charge carriersper unit volumePRwatt,coulomb,volt,ohm,henry,P = powerQ-chargeVac ium permittiv t,,Coulomb's law constant, k = 1/(4n) = 9.0x10° (Nm²)/c²Vacuum permeability.Ho = 4 x 10 (T-m)/AMagnetic constant, k = Ho/(4x) = 1x 10-7 (T-m)/A1 atmosphere ressure,R- resistancer = radius or distancet-timeWCVΩH= point c' argeJ= potential or stored energy✓= electric potentialv = velocity or speedp = resistivity- fluxx = dielectric constantFy=qvxBb-dl=Hol.70 1/2 3/5 √2/210√3/2 4/5√3/3 3/4F-S1 dixBB₂ = onl•=fBui8150x1¹9e cron ol, 1 V-1.60 x 10 'Jc = ..00 x 10 m/sG=7 x 10-¹1 (N-m²j/kg²8 = 9.8 m/s²=$E-di-dl8 = -1U₁--1-4²1 u = 1.66 x 10-27 kg = 931 MeV/e²h = 6.63 x 10-34 J-s = 4.14 x 10-15 eV-sh: 1.99 x 10-25 J-m = 1.24 x 10³ eV nm8 = 8.85 x 10-¹2 c²/(N-m²)1 atm = 1.0 x 10' N/m² = 1.0 x 10³ Pafarad,tesla.degree Celsius,electron volt,60°4/5 √3/23/5√2/21 4/3WVALUES OF TRIGONOMETRIC FUNCTIONS FOR COMMON ANGLES0. 0°30°90°101/2√√dtFTCeV80The following assumptions are used in this exam.I. The frame of reference of any problem is inertial unless otherwisestated.II. The direction of current is the direction in which positive chargeswould drift.V. Edge effects for the electric field of a parallel plate capacitor arenegligible unless otherwise stated.III. The electric potential is zero at an infinite distance from an isolatedpoint charge.IV. All batteries and meters are ideal unless otherwise stated.

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Answered: 1 σου. 150 Ω ΑΛΛ www 200 ΩΣ 100 Ω…

Science

Physics

1 σου. 150 Ω ΑΛΛ www 200 ΩΣ 100 Ω Figure 1 1 601-

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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In one of the classic nuclear physics experiments at the beginning of the 20th century, an alpha particle was accelerated toward a gold nucleus, and its path was substantially deflected by the Coulomb interaction. If the energy of the doubly charged alpha nucleus was 7.80 MeV, how close (in m) to the gold nucleus (79 protons) could it come before being deflected?
Use the following constants if necessary. Coulomb constant, k = 8.987×10^9 N⋅ m^2/C^2 . Vacuum permittivity, ϵ0 = 8.854×10^−12 F/m . Magnitude of the Charge of one electron, e = −1.60217662×10^−19 C . Mass of one electron, me = 9.10938356×10^−31 kg . Unless specified otherwise, each symbol carries their usual meaning. For example, μC means microcoulomb .
Use the following constants if necessary. Coulomb constant, k=8.987×109N⋅m2/C2. Vacuum permittivity, ϵ0=8.854×10−12F/m. The magnitude of the Charge of one electron, e=−1.60217662×10−19C. Mass of one electron, me=9.10938356×10−31kg. Unless specified otherwise, each symbol carries its usual meaning. For example, μC means microcoulomb. Consider two charges q1=32e and q2=34e at positions (−20,50,44) and (36/√3, 36/√2,−31) respectively where all the coordinates are measured on the scale of 10−9m or nanometers. If the position vector of the charge q1 is r1 and charge q2 is r2. Note: the second set of coordinates will read as thirty-six by root 3, thirty-six by root 2 and negative thirty-one. Now consider another charge q3=−12e is in the xyz system positioned at (−37/√3, 41/√2, −22). [Note: The coordinates will be read as negative thirty-seven by root 3, forty-one by root 2 and negative 22. a) Calculate the x, y and z components of the net force acting on q1 and q2. b) Calculate the x, y…
Use the following constants if necessary. Coulomb constant, k=8.987×109N⋅m2/C2. Vacuum permittivity, ϵ0=8.854×10−12F/m. Magnitude of the Charge of one electron, e=−1.60217662×10−19C. Mass of one electron, me=9.10938356×10−31kg. Unless specified otherwise, each symbol carries their usual meaning. For example, μC means microcoulomb . A planet with earth like magnetic field can be considered as a bar magnet. Suppose, we have such a bar magnet. We will treat it a magnetic dipole. Its dipole moment is given by the μ⃗ =μyj^+μzk^ which remains unchanged without the application of external magnetic field. Suppose, a uniform magnetic field given by, B⃗ =((39.0)j^+(61.0)k^)×109Tesla. Some neutron stars can generate such intense magnetic field. When this magnetic field is applied, the bar magnet starts to rotate. At some instant during rotation, consider it as "position 1" of the dipole, it's torque is τ⃗ =(2892.875181053008)×1029.0i^N⋅m and potential energy is U=−3190.214598870666×1029.0J.…
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been having trouble with this, how would you solve this? An electron is injected between two oppositely charged parallel plates connected to a DC voltage supply. The distance between the plates is 6.50 mm and the electron, initially at rest, accelerates towards the positive plate gaining 8.25 ´ 10-18 J of kinetic energy. Based on the above information, what will the mption of electrons be? (uniform motion towards the neg/pos plate, accelerated motion towards neg/pos plate) please explain step by step
For the figure below, obtain the actual value of Io . 6Ω 14 2 V I 3 I, = 15 A 7Ω www 2Ω 1 V www 4Ω www Μ 3 Ω 5Ω www
Use the following constants if necessary. Coulomb constant, k=8.987×10^9N⋅m2/C2. Vacuum permittivity, ϵ0=8.854×10^−12F/m. Magnitude of the Charge of one electron, e=−1.60217662×10^−19C. Mass of one electron, me=9.10938356×10^−31kg. Unless specified otherwise, each symbol carries their usual meaning. For example, μC means microcoulomb . Step 1 Three point charge q1,q2 and q3 are shown in figure (This figure is not drawn to scale). Does it form a dipole? If it does then find the dipole moment. Use the following data: Value of q1=25nC and coordinate is (27,−31), Value of q2=25nC and coordinate is (40,−34), Value of q3=−50nC and coordinate is (−60,25). These coordinates are given in nano meters and 1nC means 1.0 nano coulomb charge. a) The dipole moment of the given system x component of the dipole moment y component of the dipole moment Now consider three infinite sheets A,B and C as shown in figure 2. These sheets lie on the xz plane and separates four different regions. Charge…
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Calculate IL, using the Thevinin’s Theorem. the following are the values: V1 = 36 V, R1 = R3 = 22 Ω, R2 = R4 = 33 Ω, R5 = RL = 5 Ω REQUIRED VALLUES: RTHI, VTH1, RTH2, VTH2, IL