A plant has ten machines and currently operates two 8-hr shift per day, 5 days per week, 50 weeks per year. the ten machines pro
1 answer:
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Answer:
import numpy as np
import time
def matrixMul(m1,m2):
if m1.shape[1] == m2.shape[0]:
t1 = time.time()
r1 = np.zeros((m1.shape[0],m2.shape[1]))
for i in range(m1.shape[0]):
for j in range(m2.shape[1]):
r1[i,j] = (m1[i]*m2.transpose()[j]).sum()
t2 = time.time()
print("Native implementation: ",r1)
print("Time: ",t2-t1)
t1 = time.time()
r2 = m1.dot(m2)
t2 = time.time()
print("\nEfficient implementation: ",r2)
print("Time: ",t2-t1)
else:
print("Wrong dimensions!")
Explanation:
We define a function (matrixMul) that receive two arrays representing the two matrices to be multiplied, then we verify is the dimensions are appropriated for matrix multiplication if so we proceed with the native implementation consisting of two for-loops and prints the result of the operation and the execution time, then we proceed with the efficient implementation using .dot method then we return the result with the operation time. As you can see from the image the execution time is appreciable just for large matrices, in such a case the execution time of the efficient implementation can be 1000 times faster than the native implementation.
Answer:
The inductance of the inductor is 0.051H
Explanation:
From Ohm's law;
V = IR .................. 1
The inductor has its internal resistance referred to as the inductive reactance, X, which is the resistance to the flow of current through the inductor.
From equation 1;
V = IX
X =
................ 2
Given that; V = 240V, f = 50Hz, =
, I = 15A, so that;
From equation 2,
X=
= 16Ω
To determine the inductance of the inductor,
X = 2
fL
L =
=
= 0.05091
The inductance of the inductor is 0.051H.
Answer:
Manometric difference x=142.85 mm.
Explanation:
Given :
Pipe diameter
venturi meter
We can know that discharge through venturi meter is given as
h=1.8 m
We know that
Where x is the manometric deflection
⇒
So x=14.28 mm
Manometric difference x=142.85 mm.