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yoon (jey283) – Ch17h1 – turner – (56545)
1
This printout should have 20 questions.
Multiplechoice questions may continue on
the next column or page – fnd all choices
beFore answering.
001
10.0 points
What is the kinetic energy oF a proton that is
traveling at a speed oF 3650 m
/
s? Take the
mass oF the proton to be 1
.
67
×
10

27
kg.
Correct answer: 1
.
11243
×
10

20
J.
Explanation:
Since this speed is much less than the speed
oF light, we just use the approximate Formula
For kinetic energy:
KE
=
1
2
mv
2
=
1
2
(1
.
67
×
10

27
kg)(3650 m
/
s)
2
=
1
.
11243
×
10

20
J
.
002
10.0 points
IF the kinetic energy oF an electron is
4
.
1
×
10

18
J, what is the speed oF the elec
tron?
You can use the approximate (non
relativistic) Formula here. Take the mass oF
the electron to be 9
.
11
×
10

31
kg.
Correct answer: 3
.
00018
×
10
6
m
/
s.
Explanation:
The nonrelativistic Formula For kinetic en
ergy is
KE
=
1
2
mv
2
.
Rearranging this expression to solve For the
speed, we obtain
v
=
r
2(
KE
)
m
=
R
2(4
.
1
×
10

18
J)
9
.
11
×
10

31
kg
=
3
.
00018
×
10
6
m
/
s
.
003 (part 1 of 4) 10.0 points
Locations
A
,
B
, and
C
are in a region oF
uniForm electric feld, as shown in the diagram
below.
b
b
b
A
B
C
V
E
Location
A
is at
Vr
A
=
a−
0
.
7
,
0
,
0
A
m
.
Location
B
is at
Vr
B
=
a
0
.
8
,
0
,
0
A
m
.
In the region the electric feld has the value
V
E
=
a
730
,
0
,
0
A
N
/
C
.
±or a path starting at
B
and ending at
A
,
the displacement vector Δ
V
ℓ
will be oF the Form
Δ
V
ℓ
=
a
Δ
x,
0
,
0
A
.
±ind Δ
x
.
Correct answer:
−
1
.
5 m.
Explanation:
This is pretty straightForward. To fnd the
vector pointing From
B
to
A
, we just subtract:
Δ
V
ℓ
=
Vr
A
−
Vr
B
=
a−
0
.
7
,
0
,
0
A
m
− a
0
.
8
,
0
,
0
A
m
=
a−
1
.
5
,
0
,
0
A
m
.
So Δ
x
is just
−
1
.
5 m
.
004 (part 2 of 4) 10.0 points
Now fnd the change in electric potential along
this path.
Correct answer: 1095 V.
Explanation:
We can just use the equation
Δ
±
=
−
V
E
·
Δ
V
ℓ.
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View Full Documentyoon (jey283) – Ch17h1 – turner – (56545)
2
Δ
V
=
−
E
x
Δ
x
=
−
(730)(
−
1
.
5)
=
1095 V
.
005 (part 3 of 4) 10.0 points
Find the change in potential energy when
a proton (
m
p
= 1
.
7
×
10

27
kg and
q
p
=
1
.
6
×
10

19
C) moves from
B
to
A
.
Correct answer: 1
.
752
×
10

16
J.
Explanation:
Here we can use the expression
Δ
U
=
q
Δ
V.
Δ
U
=
q
p
Δ
V
= (1
.
6
×
10

19
C)(1095 V)
=
1
.
752
×
10

16
J
.
006 (part 4 of 4) 10.0 points
Find the change in potential energy when
an electron (
m
e
= 9
.
1
×
10

31
kg and
q
e
=
−
1
.
6
×
10

19
C) moves from
B
to
A
.
Correct answer:
−
1
.
752
×
10

16
J.
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 Fall '11
 TURNER,J

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