Nguyen, Thanh – Homework 19 – Due: Apr 11 2008, 7:00 pm – Inst: Weathers
1
This
printout
should
have
10
questions.
Multiplechoice questions may continue on
the next column or page – find all choices
before answering.
The due time is Central
time.
001
(part 1 of 2) 10 points
A plane electromagnetic wave has an energy
flux of 331 W
/
m
2
. A flat, rectangular surface
of dimensions 35 cm
×
29 cm is placed per
pendicular to the direction of the wave. The
surface absorbs half of the energy and reflects
half.
The speed of light is 2
.
99792
×
10
8
m
/
s.
Calculate the total energy absorbed by the
surface in 0
.
8 min.
Correct answer: 806
.
316 J.
Explanation:
Let :
I
= 331 W
/
m
2
,
a
= 35 cm = 0
.
35 m
,
b
= 29 cm = 0
.
29 m
,
and
t
= 0
.
8 min
.
The total power incident on the surface is
equal to the intensity times the area.
Since
this is energy per unit time, the total energy
hitting the surface is
U
tot
=
I a b t .
From this, only half the energy is absorbed,
so
U
abs
=
U
tot
2
=
1
2
(331 W
/
m
2
) (0
.
35 m)
×
(0
.
29 m) (0
.
8 min)
=
806
.
316 J
.
002
(part 2 of 2) 10 points
Calculate the momentum absorbed in this
time.
Correct answer: 8
.
06874
×
10

6
kg
·
m
/
s.
Explanation:
The momentum of the incident wave is
p
inc
=
U
abs
c
= 2
U
abs
c
.
Since the surface reflects only 50% of the
wave, the momentum upon reflection is
p
refl
=

1
2
U
tot
c
=

U
abs
c
.
By conservation of momentum
p
inc
=
p
abs
+
p
refl
p
abs
=
p
inc

p
refl
= 2
U
abs
c


U
abs
c
¶
= 3
U
abs
c
=
3 (806
.
316 J)
2
.
99792
×
10
8
m
/
s
=
8
.
06874
×
10

6
kg
·
m
/
s
.
keywords:
003
(part 1 of 2) 10 points
Consider a light bulb
S
emitting light isotrop
ically, (
i.e.
uniformly in all directions), with
a power of 42 W. The paper is 4 m away and
has an area of 0
.
04 m
2
,
with the coefficient
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 Spring '00
 Littler
 Work, Light, Correct Answer, Thanh

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