... changes.
PRACTICE TEST 3
Practice Test 3
PHYSICS PHYSICS
PHYSICS PHYSICS
PHYSICS
TESTTEST
TESTTEST
TEST
Peterson’s: www.petersons.com 29 7
25 . 25 .
25 . 25 .
25 .
The corThe cor
The corThe cor
The cor
rr
rr
r
ect ... power
output.
II. The PE in equals the KE out.
III. The KE in equals the work out.
(A) I only
(B) II only
(C) I and III only
(D) II and III only
(E) I, II,...
... WS
1
O
A
O
B
O
C
O
D
O
E
2
O
A
O
B
O
C
O
D
O
E
3
O
A
O
B
O
C
O
D
O
E
4
O
A
O
B
O
C
O
D
O
E
5
O
A
O
B
O
C
O
D
O
E
6
O
A
O
B
O
C
O
D
O
E
7
O
A
O
B
O
C
O
D
O
E
8
O
A
O
B
O
C
O
D
O
E
9
O
A
O
B
O
C
O
D
O
E
10
O
A
O
B
O
C
O
D
O
E
11
O
A
O
B
O
C
O
D
O
E
12
O
A
O
B
O
C
O
D
O
E
13
O
A
O
B
O
C
O
D
O
E
14
O
A
O
B
O
C
O
D
O
E
15
O
A
O
B
O
C
O
D
O
E
16
O
A
O
B
O
C
O
D
O
E
17
O
A
O
B
O
C
O
D
O
E
18...
... 1. E
2. D
3. A
4. A
5. D
6. C
7. C
8. E
9. A
10. C
11. B
12. E
13. C
14. E
15. D
16. C
17. D
18. C
19. D
20 . B
21 . B
22 . A
23 . B
24 . A
25 . D
26 . C
27 . C
28 . E
29 . E
30. B
31. C
32. A
33. ... E
34. C
35. B
36. D
37. D
38. B
39. C
40. C
41. E
42. C
43. C
44. E
45. D
46. E
47. A
48. C
49. E
50 . C
51 . C
52 . A
53 . A
54 . A
55 . E
56 . B
57 . E
58 ....
... 1. B
2. E
3. D
4. D
5. B
6. A
7. C
8. E
9. A
10. D
11. C
12. A
13. A
14. B
15. D
16. E
17. D
18. E
19. A
20 . D
21 . B
22 . E
23 . A
24 . B
25 . A
26 . D
27 . B
28 . C
29 . A
30. C
31. C
32. A
33. ... D
34. C
35. E
36. C
37. E
38. C
39. C
40. E
41. B
42. B
43. B
44. A
45. B
46. A
47. B
48. E
49. E
50 . E
51 . C
52 . D
53 . E
54 . D
55 . A
56 . E
57 . C
58 ....
... 1. A
2. E
3. C
4. E
5. A
6. C
7. E
8. A
9. B
10. B
11. D
12. E
13. A
14. C
15. A
16. B
17. E
18. D
19. A
20 . E
21 . C
22 . D
23 . E
24 . D
25 . E
26 . A
27 . A
28 . D
29 . C
30. C
31. A
32. A
33. ... D
34. D
35. D
36. E
37. C
38. C
39. C
40. C
41. E
42. C
43. A
44. E
45. C
46. D
47. B
48. B
49. A
50 . D
51 . B
52 . C
53 . B
54 . A
55 . A
56 . E
57 . E
58 ....
... than 2. 4 m/s
2
.
(B) toward him at 2. 4 m/s
2
.
(C) downward at 2. 4 m/s
2
.
(D) greater than 2. 4 m/s
2
.
(E) none of these.
PHYSICS TEST
Peterson’s SAT II Success: Physics
25 0
PHYSICS PHYSICS
PHYSICS ... 1. B
2. E
3. B
4. D
5. E
6. C
7. A
8. A
9. E
10. C
11. E
12. B
13. B
14. B
15. D
16. D
17. E
18. D
19. B
20 . E
21 . E
22 . C
23 . D
24 . A
25...
... is 23 5. 043 923 1
The mass of is 139.9
23 5
140
Uu
Ba 11 059 95
The mass of is 91. 926 1 52 8
92
u
Kr u
1
0
23 5
92
140
56
92
36
4
1
0
1 0086 65 23 5 043 923 1 139 9
nU BaKrn
uu
+→ ++
+→ .11 059 9 91 926 1 52 8 ... to E = mc
2
.
Subtracting we have:
23 6 0 52 5 9
23 5 87140
18119
.
() .
.
u
u
u
−
RADIOACTIVITY
Peterson’s SAT II Success: Physics
23 2
This...
... SUMMARY
Peterson’s SAT II Success: Physics
98
The 25 0N boulder shown in the diagram above possesses PE,
which can be stated in equation form below.
PE = mgh = wt h
= ( 25 0N) (50 m)
= 12, 50 0 J
If the boulder ... PROPERTIES
Peterson’s: www.petersons.com 97
Work
Nm
PE
kg m/s m
Nm
KE
1
2
kg
m
2
=•
=•
=
=
=• •
=•
=
=
=•
Fs
Joule
mgh
Joule
mv
2
22
2
s
Nm=•
= Joule
It should be p...
... of
. 024 5m.
Stating the equations:
and
but
an
Vgd
Vad
VV
f
f
2
0
2
2
0
2
2
2
=
=−
=
dd
m/s m
2 . 024 5m
m/s
2
298 2
800
2
••
•−
=−
.
26 .26 .
26 .26 .
26 .
The corThe cor
The corThe cor
The cor
rr
rr
r
ect ... (D).
Coulombs Law is
Fk
qq
r
=
12
2
.
The solution is
F
m
C
ff
m
=
×
−× −×
=
−−
910
310 410
02
270
9
2
2
66
2
N
N
()()
(. )
52...