Pengerjaan 23 Oktober 2019
Revision as of 17:53, 23 October 2019 by Alesdaniel (talk | contribs) (Created page with " ==1A== w4 = 4 t1 = 1 t2 = 2 t3 = 3 a = [[w4,-t2,-t3], \ [w4,-t1,-t3], \ [w4,-t1,-t2]] b = [[t1], \ [t2], \ [t3]] def gaussElimin(a,b): n = len...")
1A
w4 = 4
t1 = 1
t2 = 2
t3 = 3
a = [[w4,-t2,-t3], \
[w4,-t1,-t3], \ [w4,-t1,-t2]]
b = [[t1], \
[t2], \ [t3]]
def gaussElimin(a,b):
n = len(b) for i in range(0,n-1): for k in range(i+1,n,i+1): if a[i][k] != (0.0): a[i][k] = a[i][k]/a[i-1][k] lam = len(b) for i in range(n,k): b = a[i][k] - lam*(a[i][k]/a[i+1][k]) return b
print(gaussElimin(a,b))
Hasil yang didapatkan adalah none; kesalahan dikarenakan kesalahan pada algoritma yang sudah dilakukan.
2A
h = 0.1
v = 0
a = 2
Cd = 0.01
m = 100
def diff_t2(t2,v):
return((v2-v)*((m)/((Cd*(v2)**3)**0.5))
def rungeKutta(t2,v,v0,h):
v = v0 k1 = h*diff_t2(v0+h*t2) k2 = h*diff_t2(v0+0.5*k1*h+v2) k3 = h*diff_t2(v0+0.5*k2*h+v2) k4 = h*diff_t2(v0+k3*h+v2) t2 = t0 + (1/6)*(k1+2*k2+2*k3+k4)
return(t2) print('waktu yang diperlukan', rungeKutta(v2,v0,t2,h))
Hasilnya adalah syntax error, dikarenakan masih ada kesalahan input algoritma yang dikerjakan, terutama pada penulisan define persamaan rungeKutta(t2,v,v0,h)