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nignag [31]
1 year ago
5

The variation of the pressure of a fluid with density at constant temperature is known as the _____.

Engineering
1 answer:
qaws [65]1 year ago
7 0

The variation of the pressure of a fluid with density at constant temperature is known as the coefficient of compressibility.

<h3>What is a compressor?</h3>

A compressor can be defined as a mechanical device that is designed and developed to provide power to refrigerators, air conditioners, and other heating or cooling mechanical devices (engines), especially by increasing the pressure on air or other applicable gases.

In an isothermal process, the coefficient of compressibility is also known as isothermal compressibility or compressibility and it refers to a measure of the variation of the pressure and relative volume of a fluid with density at constant temperature.

Read more on coefficient of compressibility here: brainly.com/question/25237713

#SPJ1

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qwelly [4]

Answer:

1. Can you tell me something about yourself?

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7 0
3 years ago
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What should be given to a customer before doing a repair?
natima [27]
A. I believe, lmk if I’m right
7 0
2 years ago
A box-shaped aquarium has horizontal dimensions 0.5 m by 1 m, and depth 0.5 m, and is filled two-thirds of the way to the surfac
antoniya [11.8K]

Answer:

3270 N/m^2

Explanation:

we can calculate the pressure difference between the bottom and surface of the tank by applying the equation for the net vertical pressure

Py = - Ph ( g ± a )

for a downward movement

Py = - Ph ( g - a )  ------ ( 1 )

From the above data given will  be

p = 1000 kg/m^3, h = 2/3 * 0.5 = 0.33 m , a =2g , g = 9.81

input values into equation 1  becomes

Py =  -Ph ( g - 2g ) = Phg ------ ( 3 )

Py = 1000 * 0.33 * 9.81

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6 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.

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2 years ago
What is made in heaven?​
kramer

Answer:

Babies come from heaven didn't you know?

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2 years ago
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