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MakcuM [25]
3 years ago
9

Which option identifies the step skipped in the following scenario?

Engineering
2 answers:
Irina-Kira [14]3 years ago
5 0

Answer:

He did not pay attention to the cost analysis of the project.

Explanation:

- Massmaster34

olga nikolaevna [1]3 years ago
4 0

Answer:

he didn't use the proper material

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Explanation:

3 0
3 years ago
Someone claims that in fully developed turbulent flow in a tube, the shear stress is a maximum at the tube surface. Is this clai
Alika [10]

Answer:

Yes this claim is correct.

Explanation:

The shear stress at any point is proportional to the velocity gradient at any that point. Since the fluid that is in contact with the pipe wall shall have zero velocity due to no flow boundary condition and if we move small distance away from the wall the velocity will have a non zero value thus a maximum gradient will exist at the surface of the pipe hence correspondingly the shear stresses will also be maximum.

5 0
4 years ago
Consider a system whose temperature is 18°C. Express this temperature in R, K, and °F.
zvonat [6]

Answer:

In Rankine 524.07°R

In kelvin 291 K

In Fahrenheit 64.4°F  

Explanation:

We have given temperature 18°C

We have to convert this into Rankine R

From Celsius to Rankine we know that  T(R)=(T_{C}+273.15)\frac{9}{5}

We have to convert 18°C

So T(R)=(18+273.15)\frac{9}{5}=524.07^{\circ}R

Conversion from Celsius to kelvin

T(K)=(T_{C}+273)

We have to convert 18°C

T(K)=(18+273)=291K

Conversion of Celsius to Fahrenheit

T(F)=T_{C}\times \frac{9}{5}+32=64.4^{\circ}F

7 0
4 years ago
1. Implement the k-means clustering algorithm either in Java or Python. • The program should be executable with at least 3 param
givi [52]

Answer:

The code for this Question in Python is as follows:

matplotlib inline

from copy import deepcopy

import numpy as np

import pandas as pd

from matplotlib import pyplot as plt

plt.rcParams['figure.figsize'] = (16, 9)

plt.style.use('ggplot')

# Importing the dataset

data = pd.read_csv('xclara.csv')

print(data.shape)

data.head()

# Getting the values and plotting it

f1 = data['V1'].values

f2 = data['V2'].values

X = np.array(list(zip(f1, f2)))

plt.scatter(f1, f2, c='black', s=7)

# Number of clusters

k = 3

# X coordinates of random centroids

C_x = np.random.randint(0, np.max(X)-20, size=k)

# Y coordinates of random centroids

C_y = np.random.randint(0, np.max(X)-20, size=k)

C = np.array(list(zip(C_x, C_y)), dtype=np.float32)

print(C)

# To store the value of centroids when it updates

C_old = np.zeros(C.shape)

# Cluster Lables(0, 1, 2)

clusters = np.zeros(len(X))

# Error func. - Distance between new centroids and old centroids

error = dist(C, C_old, None)

# Loop will run till the error becomes zero

while error != 0:

   # Assigning each value to its closest cluster

   for i in range(len(X)):

       distances = dist(X[i], C)

       cluster = np.argmin(distances)

       clusters[i] = cluster

   # Storing the old centroid values

   C_old = deepcopy(C)

   # Finding the new centroids by taking the average value

   for i in range(k):

       points = [X[j] for j in range(len(X)) if clusters[j] == i]

       C[i] = np.mean(points, axis=0)

   error = dist(C, C_old, None)

# Initializing KMeans

kmeans = KMeans(n_clusters=4)

# Fitting with inputs

kmeans = kmeans.fit(X)

# Predicting the clusters

labels = kmeans.predict(X)

# Getting the cluster centers

C = kmeans.cluster_centers_

fig = plt.figure()

ax = Axes3D(fig)

ax.scatter(X[:, 0], X[:, 1], X[:, 2], c=y)

ax.scatter(C[:, 0], C[:, 1], C[:, 2], marker='*', c='#050505', s=1000)

4 0
4 years ago
Description: An apartment building has a pumped ring main that circulates hot water at 75 C through insulated 40 mm pipe with he
Ratling [72]

Answer:

hfhhfhff

Explanation:

7 0
3 years ago
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