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Crazy boy [7]
3 years ago
5

Which of the following is a stealth aircraft designed to be invisible to enemy radar?

Engineering
1 answer:
BartSMP [9]3 years ago
4 0
Which of the following is a stealth aircraft designed to be invisible to enemy radar? It is A)B-2
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Identify the true statements about the relationship between the distillate and the sample mixture in a simple distillation.
fenix001 [56]

Answer:

The answer is "Choice b,c, and d".

Explanation:

The distillation would be the separation phase of working liquids that boil close to each other. Whenever the formulation is long-term, a much more fraction of a compound collected will have a lower boiling point than it would otherwise be. In other phrases, it'll be pulverized with both the lower level, also with the boiling point. Its structure of both the diluted volatile component may therefore include another cell's contaminants.

8 0
3 years ago
In a tensile test on a steel specimen, true strain = 0.12 at a stress of 250 MPa. When true stress = 350 MPa, true strain = 0.26
scZoUnD [109]

Answer:

The strength coefficient is 625 and the strain-hardening exponent is 0.435

Explanation:

Given the true strain is 0.12 at 250 MPa stress.

Also, at 350 MPa the strain is 0.26.

We need to find  (K) and the (n).

\sigma =K\epsilon^n

We will plug the values in the formula.

250=K\times (0.12)^n\\350=K\times (0.26)^n

We will solve these equation.

K=\frac{250}{(0.12)^n} plug this value in 350=K\times (0.26)^n

350=\frac{250}{(0.12)^n}\times (0.26)^n\\ \\\frac{350}{250}=\frac{(0.26)^n}{(0.12)^n}\\  \\1.4=(2.17)^n

Taking a natural log both sides we get.

ln(1.4)=ln(2.17)^n\\ln(1.4)=n\times ln(2.17)\\n=\frac{ln(1.4)}{ln(2.17)}\\ n=0.435

Now, we will find value of K

K=\frac{250}{(0.12)^n}

K=\frac{250}{(0.12)^{0.435}}\\ \\K=\frac{250}{0.40}\\\\K=625

So, the strength coefficient is 625 and the strain-hardening exponent is 0.435.

5 0
3 years ago
Under normal conditions, gauge pressure is ______ absolute pressure.
levacccp [35]

Answer:

Option B  Lower than

Explanation:

Gauge pressure is a relative measurement based on atmospheric pressure. Gauge pressure can be positive if it is above atmospheric pressure or it can also be negative it is below.  On another hand, absolute pressure is an actual pressure in a space and its value has always to be zero or above. Basically absolute pressure is zero if it is in a perfect vacuum. So the measurement of absolute pressure is gauge pressure + atmospheric pressure.  This is the reason in normal condition the gauge pressure = absolute pressure - atmospheric pressure and therefore is lower than absolute pressure

3 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
NEED HELP this is more than 1 Answer .
bearhunter [10]
A.Ensure all circuits are de-energized before beginning work
7 0
3 years ago
Read 2 more answers
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