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 TIME OF COMPLETION_______________   =          NAME____SOLUTION___= ______________________

 

 

<= span style=3D'mso-bidi-font-size:10.0pt'> &= nbsp;           &nbs= p;            &= nbsp;           &nbs= p;    DEPARTMENT OF NATURAL SCIENCES

 

PHYS 1112, Exam 3            =             &nb= sp;            =             &nb= sp;            =             &nb= sp;            =             &nb= sp;  Section 1

Version 1            &= nbsp;           &nbs= p;            &= nbsp;           &nbs= p;            &= nbsp;           &nbs= p;            &= nbsp;           &nbs= p;             = July 17, 2008

Total Weight: 100 points

 

1.     &nb= sp;   Check your examination for completeness prior to starting.  There are a total of ten (10) prob= lems on seven (7) pages.

 

2.   =       Authorized references include your calculator with calculator handbook, and the Refere= nce Data Pamphlet (provided by your instructor).

 

3.     &nb= sp;   You will have 75 minutes to complete the examination.

 

4.     &nb= sp;   The total weight of the examination is 100 points.

 

5.     &nb= sp;   There are six (6) multiple choice and four (4) calculation problems. Work five (5= ) multiple choice problems and four (4) calculation problems. Show all work; partial credit will be given for correct work shown.

 

6.     &nb= sp;   If you have any questions during the examination, see your instructor who will=

be located in the classroom.

 

7.     &nb= sp;   Start:        &= nbsp;      1:30 p.m.

Stop:&n= bsp;            = ;   2:45 p.m.

 

 

 

 = ;             <= /span>PROBLEM

 

 = ;            &n= bsp;   POINTS

 

 = ;            &n= bsp;  CREDIT

 

     &nb= sp;            =    1-6

 

     &nb= sp;            =    20

 

 = ;

 

     &nb= sp;            =     7

 

     &nb= sp;            =    20

 

 = ;

 

8

 

     &nb= sp;            =    20

 

 = ;

 

     &nb= sp;            =     9

 

     &nb= sp;            =    20

 

 = ;

 

 = ;            &n= bsp;       10

 

 = ;            &n= bsp;       20

 

 

 

 = ;            &n= bsp;   TOTAL

 

 = ;            &n= bsp;      100

 

 

 

 

 

 = ;          PERCENTAGE

 

 

 

CIRCLE THE SINGLE BEST ANSWER FOR ALL MULTIPLE CHOICE QUESTIONS. IN MULTIPLE CHOICE QUESTIONS WHICH REQUIRE A CALCULATION SHOW WORK FOR PARTIAL CREDIT.

 

 

 

 

1. The inductive reactance of a 20.0-mH inductor at a certain frequency is 120 W. What is the angular frequency in rad/s?

            = (A) 2000 rad/s.

            = (B) 4000 rad/s.

(4)       (C) 500 rad/s.

            = (D) 6000 rad/s.<= /o:p>

            = (E) 7000 rad/s.

 

2.  In a RLC circuit, the values of the inductance and capacitance are both doubled.  In comparison with the resonance fr= equency of the original circuit, the new frequency will be  

 

            = (A) The same as before.

            = (B)  Reduced to one-half the original va= lue.

(4)       (C) Reduced to one-quarter the original value.

            = (D) Increased by a factor of two.

        &= nbsp;   (E) Increased by a fac= tor of four.

 

3.  The angle of incidence is the angle between

            = (A) The incident ray and the reflected ray.

            = (B) The incident ray = and the normal drawn at the point where this ray meets the mirror. <= /span>

(4)       (C) The reflected ray and the normal drawn at the point where this ray meets the mirror .

            = (D) The incident ray and the surface of the mirror.

            = (E) The reflected ray and the surface of the mirror.

 

 

 

4. The = index of refraction, n is given by

 

(A) n =3D c v .

 

(B) n =3D cv2.

(4)

(C) n =3D=   v/c.=

 

(D) n =3D c/v.

&n= bsp;

(E) Non= e .of the other answers is correct

 

 

5.  A ray of light strikes the mirror perpendicular to the mirror. What is the angle of reflection?

 

(A) 0o.

 

(B)  90o.

(4)

(C) 180= o.

 

(D) 360= o.

&n= bsp;

(E) 270= o.

 

 

 

 

6. Lucite has an index of refra= ction of 1.50. What is its critical angle of incidence?

            = (A) 1.16o.

            = (B) 15.0o.

(4)       (C) 41.8o.=

        &= nbsp;   (D) 65.2o.

        &= nbsp;   (E) 87.4o.<= /span>

 

 

 

 

 

 

  1. An ac generator with a maximum emf of 20.0 V is connected in series with a 20.0-mF capacitor and an 80= .0-W resistor, There is no inductance in the circuit. Find

 

 

a. the power factor,

XL =3D 0

 

XC =3D 1/(w C) =3D 1/((400 rad/s) (20.0 x 10= -6 F)) =3D 125 W

 = ;

tan f =3D (X<= sub>L – XC)/R =3D (0 - 125 W)/ (80.0 W)

 

f =3D -57.4o

 

cos f =3D 0.539

 

b. the rms cur= rent,

Z =3D (R2 + (X<= sub>L – XC)2)1/2  =3D 148 = W

 

Irms =3D DVrms/Z =3D [(20 V)/sqrt(2)]= /(148 W) =3D 0.0955= A

 

 

c. the average power if the angular frequency of the gener= ator is 400 rad/s.

P =3D Irms DVrms cos f =3D (0.0955 A)(14.1 V) cos(-57.4o)  =3D 0.72= 8 W

 

 

8.  A slab of glass with an index of refraction of 1.50 is submerged in water with an index of refraction of 1.3= 3. Light in the water is incident on the glass. Find the angle of refraction if the angle of incidence is

 

a. 60.0o.

 

n1 sin(q= 1) =3D n2 sin(q2) 

 

sin(q2<= /sub>) =3D n1 sin(q1)/= n2  

 

sin(q2<= /sub>) =3D (1.33) sin(60.0o)/(1.50) =3D 0.768 

 

q1 =3D 50.2o

 

b. 45.0o.

 

n1 sin(q= 1) =3D n2 sin(q2) 

 

sin(q2<= /sub>) =3D n1 sin(q1)/= n2  

 

sin(q2<= /sub>) =3D (1.33) sin(45.0o)/(1.50) =3D 0.627 

 

q1 =3D 38.8o

 

c. 30.0o

 

 

n1 sin(q= 1) =3D n2 sin(q2) 

 

sin(q2<= /sub>) =3D n1 sin(q1)/= n2  

 

sin(q2<= /sub>) =3D (1.33) sin(30.0o)/(1.50) =3D 0.443 

 

q1 =3D 26.3o

 

 

  1. A serie= s ac circuit contains a 100 -W resistor, a 30.0-mF capacitor, a 0.100-H inductor, and a 50.0-Hz generator that has an rms output of 45.0 V.   =

 

    1. Find = the impedance of the circuit.

 

XL =3D 2 p f L =3D 2 p (50.0 Hz) (0.100 H) =3D 31.4 W

 

XC =3D 1/(2 p f C) =3D 1/(2 p (50.0 Hz) (30.0 x= 10-6 F)) =3D 106 W

 

Z =3D (R2 + (X<= sub>L – XC)2)1/2  =3D 125 = W

 

    1. Find = the maximum current that flows in the resistor.

 

 

Imax =3D DVmax/Z =3D (45.0 V) sqrt(2)= /(125 W) =3D 0.510 = A

 

 

    1. Find = the maximum potential difference across the capacitor.<= /li>

 

 

DVmax, C =3D XC I= max 

 

DVmax, C =3D (106 W)(0.510 A) =3D 54.1 V

 

 

    1. Does = the voltage lead the current in the circuit or the current leads the volt= age? EXPLAIN.

 

tan f =3D (XL – XC)/R =3D (31.4 W - 106 W)/ (100 W)

 

f =3D -36.7o

 

Current leads the voltage since the phase angle is negative.

 

 

 

10.  The current in a 0.800= -H solenoid increases from 20.0 mA to 160 mA in 7.00 s.

a.      What is the average emf induced in the solenoid during= that time interval?

e =3D -L DI/Dt

 

|e| =3D (0.80= 0 H)(0.160 A – 0.020 A)/(7.00 s) =3D  0.0160 V

 

b.     How much energy is sto= red in the solenoid at the moment the current reaches 160-mA value?<= /span>


U =3D ½ L I2 =3D ½ (0.800 H)(0.160 A)2 = =3D  0.0102 J<= o:p>

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

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PHYS 1112 Exam 3, Version 1

Summer 2008        =             &nb= sp;            =             &nb= sp;            =             &nb= sp;    9

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