Difference between revisions of "UTS Senno"
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Soal A | Soal A | ||
+ | import math | ||
+ | print('massa balok 1') | ||
+ | a =eval(input()) | ||
+ | print('massa balok 2') | ||
+ | b =eval(input()) | ||
+ | print('massa balok 3') | ||
+ | c =eval(input()) | ||
+ | print('massa balok 4') | ||
+ | d =eval(input()) | ||
+ | |||
+ | print( ) | ||
+ | |||
+ | alfa=d/(a+b+c) | ||
+ | print('maka didapat sinus alfa sebesar',alfa) | ||
+ | |||
+ | print( ) | ||
+ | |||
+ | sudutalfa=math.degrees(math.asin(alfa)) | ||
+ | print('Jadi besar sudut alfa adalah','/n',sudutalfa,'derajat') | ||
Revision as of 20:27, 27 October 2019
Soal A
import math
print('massa balok 1') a =eval(input()) print('massa balok 2') b =eval(input()) print('massa balok 3') c =eval(input()) print('massa balok 4') d =eval(input())
print( )
alfa=d/(a+b+c) print('maka didapat sinus alfa sebesar',alfa)
print( )
sudutalfa=math.degrees(math.asin(alfa)) print('Jadi besar sudut alfa adalah','/n',sudutalfa,'derajat')
Soal B
import numpy as np a = 5 b = 1 c = 2
def diff_v (t,v):
fungsi = a - b -c return (fungsi)
v = 0 h = 1 step_size = np.arange (0,20,h)
for t in step_size:
k1 = diff_v (t,v) k2 = diff_v ((t+0.5*h), (v+0.5*k1*h)) k3 = diff_v ((t+0.5*h), (v+0.5*k2*h)) k4 = diff_v ((t+h), (v+k1*h))
v = v + 1/6*(k1+2*k2+2*k3+k4)*h
print ('maka v setelah 20 detik adalah', v)