Current and Resistance
95
17.63
The power the beam delivers to the target is
(
)
(
)(
)
6
3
4.0
10
V
25
10
A
1.0
10
W
V I
−
= ∆
=
×
×
=
×
P
5
The mass of cooling water that must flow through the tube each second if the rise in the
water temperature is not to exceed 50°C is found from
(
)
(
m
t c
T
)
= ∆
∆
∆
P
as
(
)
(
)
(
)
5
1.0
10 J s
0.48 kg s
4 186 J kg
C
50 C
m
t
c
T
×
∆
=
=
=
∆
∆
⋅°
°
√
17.64
The volume of the material is
3
6
3
3
6
3
50.0 g
1 m
6.36
10
m
7.86 g
cm
10
cm
mass
V
density
−
=
=
=
×
L
=
⋅
Since
V
A
, the crosssectional area of the wire is
A
V L
=
(a) From
2
L
L
L
R
A
V L
V
ρ
ρ
ρ
=
=
=
, the length of the wire is given by
(
)
(
)
6
3
8
1.5
6.36
10
m
9.3 m
11
10
m
R V
L
ρ
−
Ω
×
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 Fall '10
 STAFF
 Physics, Current, Resistance, Mass, Power, International System of Units, crosssectional area

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