SNMPTN2008kode302Fisika pembahasan31 38 jogjastudent.com
Pembahasan :
31. Jawab : D
EKfoton : EKelektron = h
= h
c
:
λ
1
2
m v2 = h
c 1 ( m v)
:
λ 2 m
2
c 1 p2
c 1 ( m v) 2
=h :
:
λ 2 m
λ 2 m
2
h
2
c λ
h
h 1
= h :
= c :
λ 2m
λ
λ 2m
h 1
= c:
= 2mcλ : h
λ 2m
EK foton
2mcλ
=
EK elektron
h
=
(
)(
2 x (9,1 x 10 -31 ) × 3 × 108 6,6 × 10 −11
6,63 × 10 −34
)
35. Jawab : C
I2 = Io cos2 θ
2
I2
1
1
= cos2 θ = cos2 60° = = = 25 %
Io
4
2
36. Jawab : B
M = 750 x
fob = 0,4 cm
d = 20 cm
so' k = ∞ ⇒ mata tak berakomodasi : sok = fok
pp = 25 cm
' + sok = s ' + fok
d = s ob
ob
' = d – fok = 20 – fok
s ob
' f
s ob
ob
sob =
= 54,35 ≈ 54
' −f
s ob
ob
=
(20 − f ok )× 0,4 = 0,4 (20 − f ok )
(20 − f ok ) − 0,4 19,6 − f ok
M = Mob x Mok
32. Jawab : B
750 =
'
s ob
20 − f ok
pp
25 25 (19,6 − fok )
x
=
(20 − f ok ) × f ok = 0,4 f ok
0
,
4
s ob f ok
19,6 − f ok
12 fok = 19,6 – fok ⇒ fok =
19,6
= 1, 48 ≈ 1,5 cm
13
37. Jawab : B
P=
w = m g = 3.000 x 10 = 30.000 N
lw = 3 – 2 = 1 m
Στ = 0
ND lD – w lw = 0
ND =
w l w 30.000 × 1
= 10.000 N = 10 kN
=
lD
3
33. Jawab : A
0,04 m
minyak
hm
3 3
2 2
nL
nL
PA : PU =
− 1 :
− 1 = − 1 : − 1
nA
nU
4 1
3
9 3 1 1
PA : 5 = − 1 : − 1 = : = 2 : 8 = 1 : 4
8 2 8 2
1
5
PA = × 5 = dioptri
4
4
38. Jawab : C
0,1 m
C1 =
r1 1×10 −2 1
= ×10 −11 F
=
k 9 × 109 9
C2 =
r2 2 ×10 −2 2
= × 10 −11 F
=
9
k
9
9 × 10
ha
air
hm = 0,1 – 0,04 – ha = 0,06 – ha
w = Fa + Fm
ρb Vb g = ρa Va g + ρm Vm g
ρb hb = ρa ha + ρm hm
0,5 x 0,1 = 1 x ha + 0,8 x (0,06 – ha)
0,05 = 0,2 ha + 0,048 ⇒ ha = 0,01 m
hm = 0,06 – 0,01 = 0,05 m
P = Po + Pa + Pm = 1 x 105 + ρa g ha + ρm g hm
= 105 + 1000 x 10 x 0,01 + 800 x 10 x 0,05
= 100.500 Pa = 100,5 kPa
34. Jawab : C
(Qlepas)1 = (Qterima)1
10 ct (125 – 23) = ma ca (23 – 20)
1020 ct (125 – 23) = ma 1 (23 – 20) ⇒ ma =
1020 c t
3
ma = 340 ct
(Qlepas)2 = (Qterima)2
m ct (125 – 25) = ma 1 (25 – 20)
100 m ct = 5 x 340 ct ⇒ m =
1
1 nL
1
=
− 1 ×
+
f n M
R
R
2
1
1700
100
= 17 gram
Mula-mula : q1 = 2 x 10–7 C dan q2 = 0
Setelah dihubungkan :
V1' = V2'
q1'
q'
= 2
C1 C 2
1
× 10 −11
C
1
9
1
q1' =
× q '2 =
× q '2 = q '2
2
C2
2
−11
×10
9
q1' + q 2' = q1 + q 2
1 '
–7
q 2 + q 2' = 2 x 10 + 0
2
3 '
4
–7
q 2 = 2 x 10 ⇒ q '2 = × 10 − 7 C ⇒ 2 benar
2
3
4
×10 − 7
q'
= 6 x 104 V ⇒ 4 benar
V2' = 2 = 3
C2 2
−11
×10
9
di dalam bola : E = 0 ⇒ 1 salah, maka 3 juga salah
31. Jawab : D
EKfoton : EKelektron = h
= h
c
:
λ
1
2
m v2 = h
c 1 ( m v)
:
λ 2 m
2
c 1 p2
c 1 ( m v) 2
=h :
:
λ 2 m
λ 2 m
2
h
2
c λ
h
h 1
= h :
= c :
λ 2m
λ
λ 2m
h 1
= c:
= 2mcλ : h
λ 2m
EK foton
2mcλ
=
EK elektron
h
=
(
)(
2 x (9,1 x 10 -31 ) × 3 × 108 6,6 × 10 −11
6,63 × 10 −34
)
35. Jawab : C
I2 = Io cos2 θ
2
I2
1
1
= cos2 θ = cos2 60° = = = 25 %
Io
4
2
36. Jawab : B
M = 750 x
fob = 0,4 cm
d = 20 cm
so' k = ∞ ⇒ mata tak berakomodasi : sok = fok
pp = 25 cm
' + sok = s ' + fok
d = s ob
ob
' = d – fok = 20 – fok
s ob
' f
s ob
ob
sob =
= 54,35 ≈ 54
' −f
s ob
ob
=
(20 − f ok )× 0,4 = 0,4 (20 − f ok )
(20 − f ok ) − 0,4 19,6 − f ok
M = Mob x Mok
32. Jawab : B
750 =
'
s ob
20 − f ok
pp
25 25 (19,6 − fok )
x
=
(20 − f ok ) × f ok = 0,4 f ok
0
,
4
s ob f ok
19,6 − f ok
12 fok = 19,6 – fok ⇒ fok =
19,6
= 1, 48 ≈ 1,5 cm
13
37. Jawab : B
P=
w = m g = 3.000 x 10 = 30.000 N
lw = 3 – 2 = 1 m
Στ = 0
ND lD – w lw = 0
ND =
w l w 30.000 × 1
= 10.000 N = 10 kN
=
lD
3
33. Jawab : A
0,04 m
minyak
hm
3 3
2 2
nL
nL
PA : PU =
− 1 :
− 1 = − 1 : − 1
nA
nU
4 1
3
9 3 1 1
PA : 5 = − 1 : − 1 = : = 2 : 8 = 1 : 4
8 2 8 2
1
5
PA = × 5 = dioptri
4
4
38. Jawab : C
0,1 m
C1 =
r1 1×10 −2 1
= ×10 −11 F
=
k 9 × 109 9
C2 =
r2 2 ×10 −2 2
= × 10 −11 F
=
9
k
9
9 × 10
ha
air
hm = 0,1 – 0,04 – ha = 0,06 – ha
w = Fa + Fm
ρb Vb g = ρa Va g + ρm Vm g
ρb hb = ρa ha + ρm hm
0,5 x 0,1 = 1 x ha + 0,8 x (0,06 – ha)
0,05 = 0,2 ha + 0,048 ⇒ ha = 0,01 m
hm = 0,06 – 0,01 = 0,05 m
P = Po + Pa + Pm = 1 x 105 + ρa g ha + ρm g hm
= 105 + 1000 x 10 x 0,01 + 800 x 10 x 0,05
= 100.500 Pa = 100,5 kPa
34. Jawab : C
(Qlepas)1 = (Qterima)1
10 ct (125 – 23) = ma ca (23 – 20)
1020 ct (125 – 23) = ma 1 (23 – 20) ⇒ ma =
1020 c t
3
ma = 340 ct
(Qlepas)2 = (Qterima)2
m ct (125 – 25) = ma 1 (25 – 20)
100 m ct = 5 x 340 ct ⇒ m =
1
1 nL
1
=
− 1 ×
+
f n M
R
R
2
1
1700
100
= 17 gram
Mula-mula : q1 = 2 x 10–7 C dan q2 = 0
Setelah dihubungkan :
V1' = V2'
q1'
q'
= 2
C1 C 2
1
× 10 −11
C
1
9
1
q1' =
× q '2 =
× q '2 = q '2
2
C2
2
−11
×10
9
q1' + q 2' = q1 + q 2
1 '
–7
q 2 + q 2' = 2 x 10 + 0
2
3 '
4
–7
q 2 = 2 x 10 ⇒ q '2 = × 10 − 7 C ⇒ 2 benar
2
3
4
×10 − 7
q'
= 6 x 104 V ⇒ 4 benar
V2' = 2 = 3
C2 2
−11
×10
9
di dalam bola : E = 0 ⇒ 1 salah, maka 3 juga salah