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 3575 m
/
s? Take the
mass oF the proton to be 1
.
67
×
10

27
kg.
Correct answer: 1
.
06718
×
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)(3575 m
/
s)
2
=
1
.
06718
×
10

20
J
.
002
10.0 points
IF the kinetic energy oF an electron is
5
.
3
×
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
.
41109
×
10
6
m
/
s.
Explanation:
The nonrelativistic Formula For kinetic en
ergy is
=
1
2
2
.
Rearranging this expression to solve For the
speed, we obtain
v
=
±
2(
)
m
=
²
2(5
.
3
×
10

18
J)
9
.
11
×
10

31
kg
=
3
.
41109
×
10
6
m
/
s
.
003 (part 1 of 4) 10.0 points
Locations
A
,
B
,a
n
d
C
are in a region oF
uniForm electric feld, as shown in the diagram
below.
AB
C
%
E
Location
A
is at
%r
A
=
±
0
.
7
,
0
,
0
²
m
.
Location
B
is at
B
=
±
0
.
7
,
0
,
0
²
m
.
In the region the electric feld has the value
%
E
=
±
740
,
0
,
0
²
N
/
C
.
±or a path starting at
B
and ending at
A
,
the displacement vector Δ
%
$
will be oF the Form
Δ
%
$
=
±
Δ
x,
0
,
0
²
.
±ind Δ
x
.
Correct answer:

1
.
4m.
Explanation:
This is pretty straightForward. To fnd the
vector pointing From
B
to
A
,wejustsubtract:
Δ
%
$
=
A

B
=
±
0
.
7
,
0
,
0
²
m
±
0
.
7
,
0
,
0
²
m
=
±
1
.
4
,
0
,
0
²
m
.
So Δ
x
is just

1
.
4m
.
004 (part 2 of 4) 10.0 points
Now fnd the change in electric potential along
this path.
Correct answer: 1036 V.
Explanation:
We can just use the equation
Δ
V
=

%
E
·
Δ
%
$.
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View Full DocumentΔ
V
=

E
x
Δ
x
=

(740)(

1
.
4)
=
1036 V
.
005 (part 3 of 4) 10.0 points
Find the change in potential energy when
ap
r
o
t
on(
m
p
=1
.
7
×
10

27
kg and
q
p
=
1
.
6
×
10

19
C) moves from
B
to
A
.
Correct answer: 1
.
6576
×
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)(1036 V)
=
1
.
6576
×
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
.
6576
×
10

16
J.
Explanation:
We can again use the expression
Δ
U
=
q
Δ
Δ
U
=
q
e
Δ
V
=(

1
.
6
×
10

19
C)(1036 V)
=

1
.
6576
×
10

16
J
.
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 Spring '08
 Turner
 Physics, Electron, Energy, Kinetic Energy, Potential Energy, Correct Answer, Electric charge

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