SAT II Physics (Gary Graff) Episode 1 Part 1 pot
SAT II Physics (Gary Graff) Episode 2 Part 1 potx
... 10 .(.nm ×
−−−
××
××
714 19
14
39 10 26 10
63 8 10 32 1
mHzJ
nm Hz
). . /
photon
Orange 4 00
58 52 10 34 10
5
19
14 19
−
−
××
J
nm Hz J
/
/
.
photon
Yellow photon
Green 33 5 710 3 810
56 65 10 ... through a potential
difference of 1 volt. (Remember, 1 volt is equal to 1 Joule per
coulomb of charge.)
Work
=
=
×
()
()
=× = ×
−
−
QV
ev
.C
V
eV J
1
16 10
1
1...
SAT II Physics (Gary Graff) Episode 2 Part 7 pot
... ANSWERS
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. ... the
following?
I. Negative work
II. An isothermal process
III. Increasing energy
(A) I only
(B) II only
(C) I and III only
(D) II and III only
(E) I, II, and III
74...
SAT II Physics (Gary Graff) Episode 2 Part 6 potx
... faster.
II. The voltage decreases.
III. The current increases.
(A) I only
(B) II only
(C) I and III only
(D) II and III only
(E) I, II, and III
PRACTICE TEST 3
Peterson’s SAT II Success: Physics
332
is ... ANSWERS
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....
SAT II Physics (Gary Graff) Episode 2 Part 8 pptx
... 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...
SAT II Physics (Gary Graff) Episode 2 Part 5 pptx
... reciprocals.
That is to do the following
11 111
12 34
RRRRR
t
=+++
yielding the
decimal solution
1
05 025 016 7 012 4
R
t
=+ + + .
Which is
1
1042 9 6
R
t
== Ω
.
17 .17 .
17 .17 .
17 .
The corThe cor
The corThe ... (E)
235
92
4
2
234
90
0
1
234
91
UHe Th e Pa→+ →
−
+
13 . Identify the parent nucleus.
14 . Identify the alpha particle.
15 . Identify the beta particle.
PHYSIC...
SAT II Physics (Gary Graff) Episode 2 Part 4 ppsx
... ANSWERS
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. ... immediately
because:
I. Heat is melting the ice.
II. Condensation is occurring.
III. Phase change occurs.
(A) I only
(B) II only
(C) I and III only
(D) II and III on...
SAT II Physics (Gary Graff) Episode 2 Part 3 doc
... ANSWERS
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. E
26. E
27. C
28. D
29. C
30. B
31. C
32. ... object has no velocity.
II. The object has no acceleration.
III. The object does not move.
(A) I only
(B) II only
(C) I and III only
(D) II and III only
(E) I, II,...
SAT II Physics (Gary Graff) Episode 2 Part 2 pdf
... 235.04392 31
The mass of is 13 9.9
235
14 0
Uu
Ba 11 05995
The mass of is 91. 92 615 28
92
u
Kr u
1
0
235
92
14 0
56
92
36
4
1
0
1 008665 235 04392 31 139 9
nU BaKrn
uu
+→ ++
+→ .11 0599 91 92 615 28 4 03466
236 ... n
eV
=
−=
−
4
15 1 3
34
.
.
n
eV n
=
−=
2
13 6 1.
Solution
The energy emitted by the electron as it changes from n = 3 to n = 1
is:
∆
∆
∆
∆
EE E
EeV eV
EeV...
SAT II Physics (Gary Graff) Episode 1 Part 8 pps
... of a set of parallel resistors is found in the
following manner:
11 11
12
3
RRRR
t
=++
11
2
1
5
1
10
1
40
1
5 2 1 025
1
825
12 1
R
R
R
R
t
t
t
t
=++ +
=+++
=
=
ΩΩ Ω Ω
ΩΩΩ Ω
Ω
.
.
The value of ... R
3
– R
4
parallel set of resistors:
11 11
11
10
1
8
1
4
1
25
4
12
3
RRRR
R
R
R
t
t
t
t
=++
=++
=
=
ΩΩΩ
Ω
Ω
.
CHAPTER 5
Peterson’s SAT II Success: Physics
18 4...
SAT II Physics (Gary Graff) Episode 1 Part 7 pdf
... )(. )
(. )
12
2
9
2
2
19 19
2
910
16 10 16 10
01
N
=
ו
×
=×
−
−
−
F
m
m
F
2 304 10
23 10
28 2
2
24
.
.
N
11 0
4
That’s the force operating on the particles at a distance of 10
mm. Now ... )
−
−
°
++ = +
°
−
26 13 1 25
19 5 19 50 836 8
((, . )
,. . .
,
26 13 1 25
26 13 1 25 19 50 836 8 19 5
28 0 81
J
JJ
J
C
t
J
C
t
J
ff
+=+
°
+
°
==
°
=
°=
856...
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