Practice Final Exam PHY 2140 - Wayne State Universityalan/2140Website/Documents/PracticeExam… ·...

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Practice Final Exam PHY 2140 Spring/Summer 2006 1. A proton is sent at high speed toward a gold nucleus (charge +79e). What is the electrical force acting on the proton when it is 1.5 × 10 -14 m away from the gold nucleus? (e = 1.6 × 10 -19 C, k e = 8.99 × 10 9 N•m 2 /C 2 ) a. 29.1 N b. 81.0 N c. 90.8 N d. 396.6 N e. 7.6 × 10 6 N 2. A point charge of +3.0 μC is located at the origin of a coordinate system and a second point charge of -6.0 μC is at x = 1.00 m. What is the electric potential at the x = 0.50 m point? (k e = 8.99 × 10 9 Nm 2 /C 2 ) a. 16 × 10 4 V b. 11 × 10 4 V c. -11 × 10 4 V d. -5.4 × 10 4 V e. -3.9 × 10 4 V 3. Number 10 copper wire (radius = 1.3 mm) is commonly used for electrical installations in homes. What is the voltage drop in 40 m of #10 copper wire if it carries a current of 10 A? (The resistivity of copper is 1.7 × 10 8 Ω⋅m.)? a. 1.3 V b. 0.77 V c. 4.8 V d. 0.13 V e. 0.32 V

Transcript of Practice Final Exam PHY 2140 - Wayne State Universityalan/2140Website/Documents/PracticeExam… ·...

Page 1: Practice Final Exam PHY 2140 - Wayne State Universityalan/2140Website/Documents/PracticeExam… · A proton is sent at high speed toward a gold nucleus (charge +79e). What is the

Practice Final Exam PHY 2140

Spring/Summer 2006

1. A proton is sent at high speed toward a gold nucleus (charge +79e). What is the electrical force acting on the proton when it is 1.5 × 10-14 m away from the gold nucleus? (e = 1.6 × 10-19 C, ke = 8.99 × 109 N•m2/C2)

a. 29.1 N b. 81.0 N c. 90.8 N d. 396.6 N e. 7.6 × 106 N

2. A point charge of +3.0 μC is located at the origin of a coordinate system and a second point charge of -6.0 μC is at x = 1.00 m. What is the electric potential at the x = 0.50 m point? (ke = 8.99 × 109 N⋅m2/C2)

a. 16 × 104 V b. 11 × 104 V c. -11 × 104 V d. -5.4 × 104 V e. -3.9 × 104 V

3. Number 10 copper wire (radius = 1.3 mm) is commonly used for electrical installations in homes. What is the voltage drop in 40 m of #10 copper wire if it carries a current of 10 A? (The resistivity of copper is 1.7 × 10−8 Ω⋅m.)?

a. 1.3 V b. 0.77 V c. 4.8 V d. 0.13 V e. 0.32 V

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4. Three resistors connected in parallel have individual values of 4.0, 6.0 and 10.0 Ω,

respectively. If this combination is connected in series with a 12-V battery and a 2.0-Ω resistor, what is the current in the 10-Ω resistor?

a. 0.59 A 2.0 Ω

4.0 Ω 6.0 Ω 10.0 Ω12 V

b. 1.0 A c. 5.9 A d. 16 A e. 30 A

5. At the Fermilab accelerator in Weston, Illinois, singly-charged ions (+e) with momentum 4.8 × 10−16 kg⋅m/s are held in a circular orbit of radius 1 km by an upward magnetic field. What B-field must be used to maintain the ions in this orbit? (qion = 1.6 × 10−19 C)

a. 1.5 T b. 2.0 T c. 3.0 T d. 4.5 T e. 5.0 T

6. What is the self-inductance in a coil that experiences a 3.0 V induced emf when the current is changing at a rate of 110 A/s?

a. 83 mH b. 45 mH c. 37 mH d. 27 mH e. 13 mH

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7. A series RLC AC circuit contains a resistor of 30 Ω, a capacitor of 0.35 μF and an

inductor of 90 mH. If an effective (rms) voltage of 150 V is applied, what is the effective (rms) current when the circuit is in resonance?

a. 3.3 A b. 5.0 A c. 9.4 A d. 11.4 A e. 16.1 A

8. A spaceship of mass 106 kg is to be accelerated to 0.80 c. How much energy does this require?

a. 2.5 × 1023 J b. 1.2 × 1023 J c. 7.2 × 1022 J d. 6.0 × 1022 J e. 4.1 × 1021 J

9. In the Compton Effect, what is the greatest change in wavelength that can occur? (h = 6.63 × 10-34 J·s, melectron = 9.11 × 10-31 kg, c = 3.00 × 108 m/s)

a. 2.43 × 10-12 m b. 4.85 × 10-12 m c. 7.29 × 10-12 m d. equal to the incident wavelength e. infinite

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10. The ionization energy of the hydrogen atom is 13.6 eV. What is the energy of a photon emitted corresponding to a transition from the n = 5 to n = 2 state?

a. 2.9 eV b. 3.5 eV c. 4.0 eV d. 7.9 eV e. 13.1 eV

11. If a fossil bone is found to contain 1/8th as much Carbon-14 as the bone of a living animal, what is the approximate age of the fossil? (half-life of 14C = 5730 years)?

a. 5730 years b. 7640 years c. 17200 years d. 22900 years e. 45800 years

12. Determine the amount of energy released in the fusion reaction: 2 3 4 1

1 1 2 0H H He+ → + n ?

a. 3.27 MeV b. 7.21 MeV c. 17.6 MeV d. 21.5 MeV e. 43.7 MeV

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mmvKE

22p1 2

2 == = maF r

a =v 2

IRV =Δ

VIP Δ= ( )RVRIP

22 Δ

== …++= 21 RRReq …++=21

111RRReq

θsinqvBF = θsinBIF = θτ sinF= θτ sinNBIA=

qBmvr =

rI

μ2

0= NI

B 0μ= θcosBAAB ==Φ ⊥

max max

2 2rms rmsI VI V= =

,R rms rmsV IΔ = RX

,C rms rms CV IΔ =

,L rms LV rms I XΔ = Cf

X C π21

= fLX L π2=

( )22CL XXRZ −+= IZV =Δ

RXX CL −

=φtan cosav rms rmsP I V φ= Δ

Nt

ΔΦ= −

ΔE B v=E ( )sinNAB tω ω=E max NABω=E

fπω 2= ILt

Δ= −

ΔE I

NL Φ=

ANL2

0μ=

2

21 LIPE = RCτ = Q C V= Δ

LR

τ =

EvB

= E cB

= λfc = LCf

π21

=

2 2max max max max

0 02 2 2E B E cB

c 0μ μ μ= = =P Ptt Δ=Δ γ γPLL = 221

1cv−

2mcE γ= mvp γ= 22 mcmcKE −= γ 42222 cmcpE +=

nhfEn = φ−= hfKEmax 2max 0.2898 10T m Kλ −= × ⋅

21c

vvvvv

CBAC

CBACAB

+

+=

h hp mv

λ = = 4hx pπ

Δ Δ ≥ 4hE tπ

Δ Δ ≥ minhc

e Vλ =

Δ

0 (1 cos )e

hm c

λ λ λ θΔ = − = − hcEλ

= Compton wavelength:

0.00243 nme

hm c

= 2 sin( 1, 2,3,

d mm )

θ λ== …

2

2

0 mkea = hfEEE fi ==− γ ⎟

⎟⎠

⎞⎜⎜⎝

⎛−= 22

111

ifH nn

Rλ eV)6.13(2

2

nZEn −=

⎟⎠⎞

⎜⎝⎛−= 22

422 12 n

eZkmE een

3/10 Arr = N

tNR λ=

ΔΔ

= teNN λ−= 0

λλ693.0)2ln(

2/1 ==T 2mcE Δ= nmvr = Zanr /02=

fπω 2= λfc =

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if

if

ttxx

v−−

= if

if

ttvv

a−−

= θcosAAx = θsinAAy =

22yx AAA +=

x

y

AA

=θtan atvv += 0 20 2

1 attvx +=

axvv 220

2 += tvvx ⎟⎠⎞

⎜⎝⎛ +

=2

0 ΣFx = m ax ΣFy = m ay

sFW )cos( θ= 2

21 mvKE = Wnet = KEf - KEi KEf +PEf = KEi + PEi

cosE EA θΦ = F qE= PEW Δ−=

221

rqq

kF e= 0q

FE =

2rq

kE e= ( )AB VVqPE −=Δ

EdVV AB −=− rqkV e=

rqqkPE e

21= VCQ Δ=

dAC 0κε= …++=

21

111CCCeq

…++= 21 CCCeq ( )2VC Δ=21energy

21energy2 2

QQ VC

= Δ = tQI

ΔΔ

= IRV =Δ

AR ρ= ( )[ ]00 1 TT −+= αρρ ( )[ ]00 1 TTRR −+= α VIP Δ=

( )RVRIP

22 Δ

== 04

1πε

=ek circle, cylinder:

lrVrA 22 , ππ ==AnqvI d=

g = 9.8 m/s2 ke = 8.99 x 109 Nm2/C2 μ0 = 4π × 10−7 Tm/A 1 eV = 1.6 x 10-19 J ε0 = 8.85 x 10-12 C2/Nm2 electron charge = 1.60 x 10-19 C 1 kWh = 3.60 x 106 J 1 hour = 3600 sec For an electron: mec2=511 keV c = 3.00 x 108 m/s me= 9.11 x 10−31 kg mp = 1.67 x 10-27 Kg h = 6.63 x 10-34 J s asJ 1005.1 34 −×= 0 = 0.0529 × 10−9 m 1 u = 931.5 MeV/c2 1 u = 1.66 × 10−27 kg r0 = 1.2 × 10−15 m RH =1.097 × 107 m-1

Mass (u)

H11 1.007825

10 n 1.008665

H21 2.014102

H31 3.016049

32 He 3.016029

He42 4.002602

U23892 238.050784

Power Prefix Symbol10-12 pico- p10-9 nano- n10-6 micro- μ 10-3 milli- m10-2 centi- c103 kilo- k106 mega- M109 giga- G