W
L
2
cos
θ
015
10.0points
A student weighing 700 N climbs at constant
speed to the top of an 8 m vertical rope in 10
s.
The average power expended by the student
to overcome gravity is most nearly
1.
P
= 87
.
5 W.
2.
P
= 1
.
1 W.
3.
P
= 5
,
600 W.
4.
P
= 560 W.
correct
5.
P
= 875 W.
Explanation:
P
=
W
t
=
F d
t
=
(700 N)(8 m)
10 s
=
560 W
.
016
10.0points
A 15 kg
piece
of metal (specific
heat of
0
.
05 cal
/
g
·
◦
C) having a temperature of 65
◦
C
is added to 0
.
8 kg of water having a tempera
ture of 7
◦
C.
What is the final equilibrium temperature
of the system?
Assume there is no loss of
heat to the surroundings. The specific heat of
water is 1 cal
/
g
·
◦
C.
1. 46.8235
2. 64.9333
3. 37.4848
4. 35.0645
5. 39.4545
6. 33.6667
7. 54.5
8. 41.4
9. 43.8913
10. 23.0
Correct answer: 35
.
0645
◦
C.
1.
There is no change in the internal energy
of the gas.
correct
2.
There is no volume change undergone by
the gas.
3.
None of these is true.
4.
There is no work done by the gas.
Explanation:
Let :
m
1
= 15 kg
,
c
1
= 0
.
05 cal
/
g
·
◦
C
,
T
1
= 65
◦
C
,
m
2
= 0
.
8 kg
,
c
2
= 1 cal
/
g
·
◦
C
,
and
T
2
= 7
◦
C
.
There is no loss of heat to the surroundings,
so
m
1
c
1
(
T
1

T
f
) =
m
2
c
2
(
T
f

T
2
)
m
1
c
1
T
1

m
1
c
1
T
f
=
m
2
c
2
T
f

m
2
c
2
T
2
m
1
c
1
T
1
+
m
2
c
2
T
2
= (
m
2
c
2
+
m
1
c
1
)
T
f
T
f
=
m
1
c
1
T
1
+
m
2
c
2
T
2
m
1
c
1
+
m
2
c
2
Since
m
1
c
1
T
1
+
m
2
c
2
T
2
= (15 kg) (0
.
05 cal
/
g
·
◦
C) (65
◦
C)
+ (0
.
8 kg) (1 cal
/
g
·
◦
C) (7
◦
C)
= 54
.
35 kg
·
cal
/
g
and
m
1
c
1
+
m
2
c
2
= (15 kg) (0
.
05 cal
/
g
·
◦
C)
+ (0
.
8 kg) (1 cal
/
g
·
◦
C)
= 1
.
55 kg
·
cal
/
g
·
◦
C
,
then the final temperature will be
T
f
=
54
.
35 kg
·
cal
/
g
1
.
55 kg
·
cal
/
g
·
◦
C
=
35
.
0645
◦
C
.
017
10.0points
In an isothermal process working on an ideal
gas,
Version 025 – Final – yao – (92695)
8
5.
There is no heat exchange during the
process.
Explanation:
The temperature is constant in an isother
mal process. The internal energy of an ideal
gas consists only of the kinetic energy of its
molecules, which, in turn, depends only on
the temperature. Thus, there is no change in
the internal energy of an ideal gas.
018
10.0points
A
circular
disk
with
moment
of
inertia
1
2
m R
2
, mass
m
and radius
R
is mounted
at its center, about which it can rotate freely.
A light cord wrapped around it supports a
weight
m g
.
R
I
ω
m
T
g
Find the total kinetic energy of the system
when the weight is moving at a speed
v
.
8.
K
=
3
4
m v
2
correct
9.
K
=
5
2
m v
2
Explanation:
v
=
r ω ,
so the total kinetic energy is
K
tot
=
K
m
+
K
rot
=
1
2
m v
2
+
1
2
I ω
2
=
1
2
m v
2
+
1
2
parenleftbigg
1
2
m R
2
parenrightbigg
parenleftBig
v
R
parenrightBig
2
=
3
4
m v
2
.
019
10.0points