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masya89 [10]
4 years ago
7

Describe how this instrument works. Operation:

Engineering
1 answer:
LenKa [72]4 years ago
5 0

Answer:

An instrument takes the image that you are looking at and reflects it off the aluminum coating. The image then bounces off the coating to the glass. The glass is angle toward the eyepiece so that you have a full visual of the image you are trying to see. The telescope allows us to see further; they are able to collect and focus more light from distant objects than our eyes can alone. This is achieved by refracting or reflecting the light using lenses or mirrors.

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Which is/are not a mechanism commonly associated with tool wear (mark all that apply)?a. Adhesion b. Attrition c. Abrasion d. Co
butalik [34]

Answer: d)Coercion

Explanation:Tool wear is defined as the situation when the cutting tool is subjected to the regular process of cutting metal then they tend to wear because of the continuous action of cutting and facing stresses and pressure . The mechanism that does not happen during this process are coercion that means the process of exerting forces on any material forcefully against the will or need. Therefore, adhesion,attrition and abrasion are the process of tool wear .So the correct option is (d)

5 0
3 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
Please help i give brainliest​
Mazyrski [523]

Answer:

A mock-up

Explanation:

It is made of cheap and easy to access parts.

5 0
3 years ago
Why do engineers play a variety of roles in the engineering process? They are part of a team. They need to act as both scientist
Kamila [148]

Answer: They showed that many people worked on the project and demonstrate how hard they worked.

Explanation:

4 0
3 years ago
Read 2 more answers
A ball bearing has been selected with the bore size specified in the catolog as 35.000 mm to 35.020 mm. Specify appropriate mini
Fofino [41]

Answer:

the minimum shaft diameter is 35.026 mm

the maximum shaft diameter is 35.042mm

Explanation:

Given data;

D-maximum = 35.020mm and d-minimum = 35.000mm

we have to go through Tables "Descriptions of preferred Fits using the Basic Hole System" so from the table, locational interference fits H7/p6

so From table, Selection of International Trade Grades metric series

the grade tolerance are;

ΔD = IT7(0.025 mm)

Δd = IT6(0.016 mm)

Also from Table "Fundamental Deviations for Shafts" metric series

Sf = 0.026

so  

D-maximum

Dmax = d + Sf + Δd

we substitute

Dmax = 35 + 0.026 + 0.016

Dmax = 35.042 mm

therefore the maximum diameter of shaft is 35.042mm

d-minimum

Dmin = d + Sf

Dmin = 35 + 0.026

Dmin = 35.026 mm

therefore the minimum diameter of shaft is 35.026 mm

8 0
3 years ago
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