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3 years ago

Write a single statement that prints outsideTemperature with 2 digits in the fraction

Engineering

1 answer:

lord [1]3 years ago

3 0

Answer:

outsideTemperature=108.90

Explanation:

#include <iostream>

#include <ios>

#include <iomanip>

using namespace std;

int main()

{

double outsideTemperature = 108.90;

/* print outside temperature */

cout << setprecision(2) << fixed << outsideTemperature << endl;

return 0;

}

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Before cutting coarse screw threads, the operator should lubricate: A. The leadscrew and gearbox B. The ways and cross slide C.

Maksim231197 [3]

Answer:

(d) all of the above

Explanation:

before cutting the screw threads the operator should lubricate all of the machine parts given in the option that is lead screw and gearbox , the ways and the cross slide and the carriage and half-nuts. we should use lubrication because it reduces the overall system friction and if friction is reduced then heat generated due to friction is also decreases which is beneficial

so option (D) will be correct because we need lubricate in all the given parts

8 0

3 years ago

Indicate the correct statement about the effect of Reynolds number on the character of the flow over an object.

sergeinik [125]

Answer:

If Reynolds number increases the extent of the region around the object that is affected by viscosity decreases.

Explanation:

Reynolds number is an important dimensionless parameter in fluid mechanics.

It is calculated as;

$R_e__N} = \frac{\rho vd}{\mu}$

where;

ρ is density

v is velocity

d is diameter

μ is viscosity

All these parameters are important in calculating Reynolds number and understanding of fluid flow over an object.

In aerodynamics, the higher the Reynolds number, the lesser the viscosity plays a role in the flow around the airfoil. As Reynolds number increases, the boundary layer gets thinner, which results in a lower drag. Or simply put, if Reynolds number increases the extent of the region around the object that is affected by viscosity decreases.

5 0

3 years ago

Write a naive implementation (i.e. non-vectorized) of matrix multiplication, and then write an efficient implementation that uti

erik [133]

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.