practice final exam solutions

7/22/2019 Practice Final Exam Solutions

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1070 Practice Final Exam SolutionsW08

1. Travelling sound wave in +x directionForms for travelling wave are:

y= y0 sin (2 )

= y0sin [(t )], = 2f=

Can only be (C) or (D)

Angular frequency = 2f, f=

f = = 2000 Hz

= 4000

= = where v= 340

y = y0sin [4000 (t - ) ] ANS (C)

2. Intensity level, IL at 15m

0

IL = 10 og I

I

Given as 85 db

0

I

I = og -1(8.5) I =I0(10

8.5)

The standard reference intensityI0= 10-12w/m2


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I= 10-12 108.5= 10-3.5= 10+0.5 10-4

= x 10-4

= 3.16 x 10-4w/m2

Power =IAwhere (A=area)

A= 2.0 1.0 m2= 2.0 m2

P= (3.16 10-4) 2.0 = 6.32 10-4w

ANS (A)

(note, the distance 15.0 m is irrelevant)

3. Beat FrequenciesIf two waves of frequency f1and f2are added, the beat frequency fBis

fB= | f1f2|

(Abs. value is not in text, but isin lab manual)

5 = | fstring- 440| = | fstring450 |

The only common solution is fstring= 445, hence the 2ndharmonic is at

2(445) = 890 Hz,

ANS (B)

4. Clearly, only (D) is false5. Lens

r1 r2 oil (n2)

light water (n1)

By conventions, both r1and r2are positive.


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Second case = , narrow pupil a2 = 2.0mm

= = = =

2= (6.0m) = 3.0 m ANS A)

8. 8-Carbon atoms with bondlength b = 0.15 nm, n = 2. Find the probability

density at the 2ndcarbon,x= 0.15nm,

P =

2

= sin

2

( ), where = 7 0.15nm= 1.05 nm

P = sin2( )

= sin2( ) = 1.16

1.2 nm-1 ANS (E)

9. Hexatriene

6 carbons, 6 es, linear

Ground State 3

2 Main transition is from n = 3 to n = 4

n=1

=2 2 2

2

(4 3 )

8e

h

m , =5b


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5

=2

em b c

h=

2

em b c

h

= (9.1 10-31)

= 2.647 10-7m

= 265 nm ANS (E)

10. 20 bonds = 20 C atoms in ring (20 electrons).

Energy-level ground-state diagram is

_ _ n=6 (empty)

n=5

n=4

n=3

n=2

n=1

n=0

The lowest-energy transition in this case is from n=4 to n=5.

ANS (B)

11. % Transmittances

Relation between absorbance A and %T:

A= log = log


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%T = 100 (10-A)

A1= log ( ) = log (0.4) = 0.39794

From Beers law, A2= 3A1 = 3 0.39794

= 1.1938

%T2= 100 (10-A2) = 100 (10-1.1938)

= 100 (0.064) = 6.4 ANS (A)

12. Time-line for rat tissue

Ingest Killed Analysis

--------------------------------->|-------------------------------->|

21 days (t1) 64 days ( = 85-21) (t2)

e= p+ b p

C0=? C1= ? C2= 1600 c/s

The question really asks for C0, the initial count rate in rat tissue sample.

First relate C2to C1: C2= C1e- p t2

or C1= C2ep t2

where p=1/2

0.693

pT= = 0.00693 day-1

C1= 1600 e+(0.00693) (64)= 2493 c/s

Now C1= C0e- e t1 or C0= C1e

e t1


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e= p+ b= p+1/2

0.693

bT= p+ = 0.0300 day

-1

Co= 2493 e+(0.0300) (21)= 4684 c/s ANS(B)

13.

C= C0ex

n C= n C0x

= -

=

Using first and last points in data set,

=

= = = 0.501 cm-1 ANS (B)

14. CircuitFirst find the equivalent resistance of the parallel resistors;

Req= (-1= 2.0

Then the total resistance of the circuit is

R = Req+ R3= 2.0 + R3

Now V = IR, where V = 12V

R =

R3= 6.02.0 = 4.0 ANS (A)


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