Jawaban:
18.
Lt = Lo + ∆L
= 20 + ( ɑ . Lo . ∆T)
= 20 + (1,2 . 10^-5 . 20. 80)
= 20 + 0,0192
= 20,0192 m
19.
Lt = 10 + (1,2 . 10^-5 . 10 . 50)
= 10 + 0,006
= 10,006 m
20.
50 cm = 0,5 m
0,5011 = 0,5 + (1,1 . 10^-5 . 0,5 . ∆T)
0,5011 - 0,5 = 1,1 . 10^-5 . 0,5 . ∆T
0,0011/ 1,1 . 10^-5 . 0,5 = ∆T
∆T = 200
To = 256 - 200 = 56° C
ket:
ɑ = koefisien muai panjang
Lo = panjang mula² (m)
Lt = panjang akhir (m)
∆T = perubahan suhu
To = suhu mula²
∆L = perubahan panjang
" Life is not a problem to be solved but a reality to be experienced! "
© Copyright 2013 - 2024 KUDO.TIPS - All rights reserved.
Verified answer
Jawaban:
18.
Lt = Lo + ∆L
= 20 + ( ɑ . Lo . ∆T)
= 20 + (1,2 . 10^-5 . 20. 80)
= 20 + 0,0192
= 20,0192 m
19.
Lt = 10 + (1,2 . 10^-5 . 10 . 50)
= 10 + 0,006
= 10,006 m
20.
50 cm = 0,5 m
0,5011 = 0,5 + (1,1 . 10^-5 . 0,5 . ∆T)
0,5011 - 0,5 = 1,1 . 10^-5 . 0,5 . ∆T
0,0011/ 1,1 . 10^-5 . 0,5 = ∆T
∆T = 200
To = 256 - 200 = 56° C
ket:
ɑ = koefisien muai panjang
Lo = panjang mula² (m)
Lt = panjang akhir (m)
∆T = perubahan suhu
To = suhu mula²
∆L = perubahan panjang