Physics Exam 3

Hw6-What is the equivalent resistance of the 5 resistors in the circuit in the figure if R1 = 18 ohms

(R1R2)(R1R2)

Hw6-A certain substance has a dieletric constant of 2.8 and a dieletric strength of 18MV/m. If it is used as the dieletric material in a parallel pile capacitor, what minimum area showed the plates of the capacitor have to obtain a capacitence of 4*10^-8 F and to ensure that the capacitor will be able to withstand a potential difference of 3.7kV?

A=? C=[(k)(Eo)(A)/(d)] A/d = C/(k)(Eo) (dieletric strength)(A)=potential difference (dieletric strength)(A)/(A/d)=potetial differencce A = (potential difference)(A/d)/(dieletric strength)

Hw7-You are given a number of 10 Ω resistors, each capable of dissipating only 1.0 W without being destroyed. What is the minimum number of such resistors that you need to combine in series or parallel to make a 10 Ω resistance that is capable of dissipating at least 5.0 W?

Equivalent Resistance: 1/Req = E*(1/n*R) = m/n*R Ptotal = n^2*p ; P=l*w

Hw6-An unknown resistor is connected between the terminals of a 4.5V battery. Energy is dissipated in the resistor at the rate of .520W. The same resistor is then connected between the terminals of a 2.48V battery. At what rate is energy now dissipated?

P1/P2 = (V1/V2)^2 P2 = P1(V2/V1)^2

Hw6-A nichrome heater dissipates 480W when the applied potential difference is 110V and the wire temp is 750 deg C. WHat would be the dissipation rate if the wire temp were help at 200 deg C by immersing the wire in a bath of cooking oil? Applied potential difference stays the same and a for Nichrome at 750 deg C is 4*10^-4 K.

R = (v)^2/P R = R1*(1+a*(delta T)) P = (v)^2/R

Hw6-When a 30V emf device is placed across two resistors in series, a current of 14.0A is flowing in each of the resistors. When the same emf device is placed across the same 2 resistors in parallel, the current through the emf device is 62.0 A. What is the magnitude of the larger of the two resistances?

R = R1+R2 V = I*R R = (R1*R2)/(R1+R2) R1=?

Hw6-A common flash light bulb is rated at .300A and 2.90V. If the resistors of the bulb filament at room temp (20 deg C) is 2.50 ohms, what is the temp of the filament while the bulb is on? a=.004403

R2 = V/i R2 = R1(1+a(delta T)) solve for t

Hw6-2 conductors are made of the same material and have the same length. Conductor A is a solid wire of diameter 1.5mm. Conductor B is a hollow tube of outside diameter 7.1mm and inside diameter of 6.1mm. What is the resistance ratio Ra/Rb, measured between their ends?

Ra = (4*o*l)/(pi*d^2) Rb = (4*o*l)/(pi*d^2) Area of hollow tube: (pi*d^2)/4 = (pi/4)*(Douter^2-Dinner^2) Rb = (4*o*l)/(pi*(Douter^2-Dinner^2) Ra/Rb = ?

An electron is accelerated from rest by a potential difference of 380 V. It then enters a uniform magnetic field of magnitude 300 mT with its velocity perpendicular to the field. Calculate the speed of the electron.

given that V = 350v B = 200mT = 200 * 10-3T mass of electron = me = 9.11 *10-31kg electron charge = e = 1.6 *10-19C from the conservation of energy eV = 1/2 *mv2 v = √2eV/me plug in values v = 1.11 * 107 m/s radius r = mev/eB plug in values r = 3.16 *10-4m

Hw6-A 100W light bulb is plugged into a standard 120 V outlet. How much does it cost per month (30 days) to leave the light turned on continuously? Assume electrical cost of 6.5 cents/kWh b) Resistance of bulbs? c) current in bulb?

hours = (days)(hours) kWh = (.1 kw)(total hrs) cost = (kwh)(.065) b) P = V/i ; i = P/V V = Ri ; R = V/ c) i = P/V

Hw7-Assume that R1 = 1.0 Ω, R2 = 2.0 Ω, ε1 = 4.0 V, and ε2 = ε3 = 8.0 V. Take positive current as that flowing through the battery from - to +. (a) Calculate the current through each ideal battery in the figure. i1? b) i2? c) i3? d) calculate Va-Vb

i1 + i2 + i3 = 0 -R2i1 + E1 + R2i2 - E2 = 0 -R2i3+E3 + R2i2 - E2 = 0 i2 - d) Vb + E3 - R2i2 = Va

Hw8- An electron that is moving through a uniformmagnetic field has a velocity v = (30km/s)i + 21km/s)j when it experiences a force F =(-3.100×10-15 N)i +(4.429×10-15 N)j due tothe magnetic field. If Bx = 0, calculate the magnitude of themagnetic field B. (Tesla = T)

we know force F = q ( v X B ) given F = (-3.100×10-15 N)i + (4.429×10-15N)j v = (30km/s)i + 21km/s)j = (30000 m / s ) i + ( 21000 m / s ) j q =charge of electron = 1.6 * 10 -19 C B = By j + B z k Cross product of v and B is = i [ 21000 B z ] - j [ 30000 B z] + k [ 30000 B y] So, F = q ( v X B ) (-3.100×10-15 N)i + (4.429×10-15N)j = q[ i [ 21000 B z ] - j [ 30000 B z] + k [30000 B y ] ] from this 21000 q B z =(-3.100×10-15 N) from this B z= (-3.100×10-15 N) / 21000 q = 0.922 T -30000q B z = (4.429×10 -15N) 30000 q B y = 0 So, B y = 0 T

hw7- A solar cell generates a potential difference of .1269 V when a 500 ohm resistor is connected across it, and a potential difference of .1833 V when a 1000 ohm resistor is substituted. a) What is the internal resistance of the solar cell? b) What is the emf of the solar cell? c) The area of the cell is 5.0 cm2, and the rate per unit area at which it receives energy from light is 2.0m W/ cm2. What is the efficiency of the cell for converting the light energy to thermal energy in the 1000 Ω external resistor?

V = E - i*R =E-(V/R)*r Pr = (p)(a) n (efficiency)= (pd2)/r*(pr)

hw8- In a nuclear experiment a proton with kinetic energy 3.5 MeV moves in a circular path in a uniform magnetic field. If the magnetic field is B = 0.6 T what is the radius of the orbit? B) what energy must an alpha particle (q =+2e, m=4.0u) and a deuteron (q = +e, m=2.0u) have if they are to circulate in the same orbit? energy of the alpha? c) energy of the deuteron?

a) K = 3.5e6*1.6e-19 = 5.6e-13 J v = √2K/m = √(2*5.6e-13/1.673e-27) = 2.5874e7 m/s m v^2/R = q v B R = mv/qB = 1.673e-27*2.5874e7/1.6e-19/1.1 = 0.24595 ≈ 0.246 m b) m v^2/R = q v B v = R q B/m = 0.24595*2*1.6e-19*1.1/4/1.661e-27 = 1.30305e7 m/s K = 0.5 m v^2 = 0.5*(4*1.661e-27)*1.30305e7*1.30305e7 = 5.6406e-13 J K = 5.6406e-13/1.6e-19 = 3.53 MeV deuteron: m v^2/R = q v B v = R q B/m = 0.24595*1.6e-19*1.1/2/1.661e-27 = 1.30305e7 m/s K = 0.5 m v^2 = 0.5*(2*1.661e-27)*1.30305e7*1.30305e7 = 5.6406e-13/2 J K = 5.6406e-13/2/1.6e-19 = 1.76 MeV

Hw6-A cylindrical resistor of radius 5.1mm and length 1.8cm is made of material that has a resistivity of 3.50*10^-5 ohms. a) What is the current density when the energy dissipation rate in the resistor is 1.0W. b) What is the potential difference?

a) Current density= J = E/o = (V/l)/o = (i)(R)/(l)(o) P=i^2(R) ; i = (P/R)^(1/2) J = (P/R)^(1/2) * (R)/(l)(o) = (P)/(A)(o)(l)^(1/2) b) V = (P*R)^(1/2) = (P*o*(l/A))^(1/2) R = o * (l/A)

Hw8- An electron in a TV camera tube is moving at 6.60×106 m/s in a magnetic field of strength 77 mT. Without knowing the direction of the field, what can you say about the greatest and least magnitude of the force acting on the electron due to the field? Maximum force? b) minimum force? c) at one point the acceleration of the electron is 8.387*10^16; What is the angle between the electron velocity and the magnetic field?

a) F=q(v*B)=qvBsin(theta) b) F=q(v*B)sin(theta) sin(theta)=m*a/q*v*b theta =

Hw8- For the vectors in the figure, a = 5.3, b = 3.2 and c = 6.19. Calculate ab. A positive value is for the z-component coming out of the page. The units are m. b) calculate ac c) calculate bc

a+b+c=0 c=-a-b ab = 5.3i * 3.2j = ac = 5.3i * (-5.3i-3.2j) = bc = 3.2j * (-5.3i-3.2j) =

Hw7- A battery of e=2.10V and internal resistance R=.450 ohms is driving a motor. The motor is lifting a 2.0N mass at constant speed v=.50m/s. Assuming no energy loss, find the current i in the circuit. a)Enter lower current b)Enter higher current c)FInd the potential difference V across the terminals of the motor for the lower current d) Find the potential difference V across the terminals of the motor for the highest current.

a/b) Power = F*V P = r*i^2 ; r = P/i^2 ; r = 1/i^2 V = i*(R+r) = E c) V = i*r d) V = i*r