homework 01 – BOBBITT, JAMES – Due: Jan 21 2008, 4:00 am
1
Question 1, chap 22, sect 1.
part 1 of 1
10 points
You have 1
.
4 kg of water. One mole of water
has a mass of 18
.
1 g
/
mol and each molecule
of water contains 10 electrons since water is
H
2
O.
What is the total electron charge contained
in this volume of water?
Correct answer:
−
7
.
45281
×
10
7
C (tolerance
±
1 %).
Explanation:
Let :
N
A
= 6
.
02214
×
10
23
molec
/
mol
,
q
e
=
−
1
.
6
×
10
−
19
C
/
electron
,
M
= 18
.
1 g
/
mol = 0
.
0181 kg
/
mol
,
m
= 1
.
4 kg
,
and
Z
= 10 electrons
/
molec
.
The mass is proportional to the number of
molecules, so for
m
grams in
n
molecules and
M
grams in
N
A
molecules,
m
M
=
n
N
A
n
=
m
M
N
A
.
Since 10 electrons are in each molecule of
water, then the total number of electrons
n
e
in the coin is
n
e
=
Z n
=
Z
m
M
N
A
and the total charge
q
for the
n
e
electrons is
q
=
n
e
q
e
=
Z
m N
A
q
e
M
= (10 electrons
/
molec)
1
.
4 kg
0
.
0181 kg
/
mol
×
(6
.
02214
×
10
23
molec
/
mol)
×
(
−
1
.
6
×
10
−
19
C
/
electron)
=
−
7
.
45281
×
10
7
C
.
Question 2, chap 22, sect 1.
part 1 of 1
10 points
What happens to the mass of an object
when it acquires a positive net charge by the
transfer of electrons?
1.
Decreases
correct
2.
Cannot be determined
3.
Increases
4.
Doesn’t change
Explanation:
When an object acquires a positive charge
it loses electrons, so its mass decreases.
Question 3, chap 22, sect 2.
part 1 of 1
10 points
A particle of mass 79 g and charge 14
μ
C
is released from rest when it is 72 cm from a
second particle of charge
−
14
μ
C.
Determine the magnitude of the initial ac
celeration of the 79 g particle.
Correct answer: 43
.
0133 m
/
s
2
(tolerance
±
1
%).
Explanation:
Let :
m
= 79 g = 0
.
079 kg
,
q
= 14
μ
C = 1
.
4
×
10
−
5
C
,
r
= 72 cm = 0
.
72 m
,
Q
=
−
14
μ
C =
−
1
.
4
×
10
−
5
C
,
and
k
e
= 8
.
9875
×
10
9
N
·
m
2
/
C
2
.
The force exerted on the particle is
F
=
k
e

q
 
Q

r
2
=
m

a


a

=
k
e

q
 
Q

m r
2
= (8
.
9875
×
10
9
N
·
m
2
/
C
2
)
×
vextendsingle
vextendsingle
1
.
4
×
10
−
5
C
vextendsingle
vextendsingle
vextendsingle
vextendsingle
−
1
.
4
×
10
−
5
C
vextendsingle
vextendsingle
(0
.
079 kg) (0
.
72 m)
2
=
43
.
0133 m
/
s
2
.
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homework 01 – BOBBITT, JAMES – Due: Jan 21 2008, 4:00 am
2
Question 4, chap 22, sect 2.
part 1 of 2
10 points
What can we conclude from the attrac
tive nature of the force between a positively
charged rod and an object?
+
+
+
F
F
?
1.
The object is a conductor.
2.
Cannot be determined
correct
3.
The object is an insulator.
Explanation:
The attractive nature of the force will be
the same whether the object is a conductor or
insulator.
Question 5, chap 22, sect 2.
part 2 of 2
10 points
What can we conclude from the attrac
tive nature of the force between a positively
charged rod and an object?
1.
The object has no net charge.
2.
The object has an appreciable positive
charge.
3.
The object does not have an appreciable
negative charge.
4.
Cannot be determined.
5.
The object has an appreciable negative
charge.
6.
The object does not have an appreciable
positive charge.
correct
Explanation:
Because of the electric induction caused
by the (positively or negatively) charged rod,
there is always attractive force between the
rod and the object unless the object has some
net charge with the same sign as the charge of
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