06 Kunci Jawaban dan Pembahasan MAT IX

22. Jawaban: c
3 tahun cahaya = 3 × 9,46 × 1012 km
= 28,38 × 1012 km
= 2,838 × 1013 km
23. Jawaban: c
p = mv atau m = p
, sehingga diperoleh:
m= p
v =
5
5× 10
200
=
v
5× 10
2× 10
24. Jawaban: c
p= mv
= 8,5 × 10 –31 × 2,5 × 107 = 8,5 × 2,5 × 10 –24
= 21,25 × 10 –24
= 2,125 × 10 –23 kg m/s
25. Jawaban: c
Diketahui: A = 10 mm2 = 10 × 10 –6 m2 = 1 × 10 –5 m2 σ = F
A
66 Kunci Jawaban dan Pembahasan PR Matematika Kelas IX
5
2
= 2,5 × 10 3 = 2.500 kg
20
= 5
1 × 10 − = 20 × 105 = 2 × 106 N/m2 26. Jawaban: c
Diketahui: ρ = 1 g/cm3 = 10 –3 kg/10 –6 m3 = 103 kg/m3 P = P0 + ρgh
⇔ P – P0 = ρgh
P−P0 ⇔ h=
ρg
5 5
(4 × 10 ) − (1,01× 10 )
=
3
10 × 10
5
2,99 × 10
= 4
10
= 2,99 × 10
= 29,9 m
Jadi, penyelam berada pada kedalaman 29,9 m.
27. Jawaban: d
h = 1
2 gt2
⇔ 20 = 1
× 10 × t2
2
⇔ 20 = 5 × t2 ⇔ t2 = 4
⇔ t= 2
28. Jawaban: d
L = πr2 ⇔ 154 × 10 –6 = 22
× r2
7
⇔ 154 × 7 × 10 –6 = 22 × r2 ⇔ 7 × 7 × 10 –6 = r2 ⇔ r = 7 × 10 –3
29. Jawaban: a
Vbola = 4
× π × r3
3
⇔ 1,1304 × 10 –4 = 4
× 3,14 × r3
3
⇔ 3,3912 × 10 –4 = 4 × 3,14 × r3 ⇔ 3,3912 × 10 –4 = 12,56 × r3 ⇔ r3 = 3,3912
12,56
⇔ r3 = 339,12
12,56
⇔ r3 = 27 × 10 –6
⇔ r=
× 10–4
× 10–6
3 6
27 10 −
×
= 3 × 10 –2
30. Jawaban: d
Luas penampang logam = L = s 2
2,56 × 10 –4 = s 2
⇔ s 2 =
Panjang diagonal =
4
2,56 10 −
× = 1,6 × 10 –2 m = 1,6 cm
2 2
s + s =
2
2 1,6
×
= 2 m = 1,6 2 cm
B. Uraian
1. 275 – 5 × 313 – 312 = (33 ) 5 – 5 × 313 – 312 = 315 – 5 × 313 – 312 = 312 (33 – 5 × 3 – 1)
= 312 (27 – 15 – 1)
= 312 × 11
2. Vtangki = 30.800 liter = 30.800 dm3 = 30,8 m3 V = πr2t 30,8 = 22
7 × r2 × 5
⇔ r2 30,8 × 7
=
22× 5
1, 4 × 7
⇔ r= = 1, 4 × 1, 4 = 2
5
(1,4) = 1,4 m
Diameter = 2r = 2 × 1,4 = 2,8 m.
Jadi, diameter tangki 2,8 m.
3. Bola dimasukkan ke dalam kubus. Oleh karena
sisi-sisi kubus menyentuh bola maka panjang rusuk
kubus = diameter bola.
Vbola = 4
3 πr3 = 4 d
π(
3 2 )3
⇔ 4.851 = 4
3
× 22
7 ×
3
d
8
⇔ d3 4.851× 21
=
11
⇔ d3 = 441 × 21
⇔ d3 = 21 × 21 × 21
⇔ d= 3 21× 21× 21 = 21 cm
s
s
s