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3 years ago

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(a) Please show that the heat capacity at constant volume and constant pressure for air are given by,Cv=macvmwherecvm= (1 + 0.92

rv)cvdandCp=macpmwherecpm= (1 + 0.84rv)cpdIn the above equations, the mass of air isma=md+mv, wheremdandmvare the masses of dry air andvapor, respectively. The specific heats at constant volume and pressure for moist air (the combination ofdry air and vapor) arecvmandcpm, respectively. (b) The vapor pressure varies greatly in the atmosphere.At a temperature of 0oC, the vapor pressure can reach 611 Pa, while at 30oC, it can reach 4350 Pa. Howmuch do the specific heats at constant volume and pressure for moist air,cvmandcpm, vary at Earth’ssurface? Assume a surface pressure of 100,000 Pa.

1 answer:

forsale [732]3 years ago

7 0

Answer:

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dmitriy555 [2]

Answer:

age = int(input())

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Explanation:

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6 0

2 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

Two vertical parallel plates are spaced 0.01 ft apart. If the pressure decreases at a rate of 60 psf/ft in the vertical z-direct

Whitepunk [10]

Answer:

umax = 0.1259ft/s

Explanation:

Given:

•Distance between plates, B = 0.01ft

•Pressure difference decrease, $\frac{dp}{dz}=60ps/ft$

•Fluid viscosity, u = 10^-³lbf-s/ft²

Specific gravity, S = 0.80

Max velocity in the z-direction will be:

$u_max= [\frac{B^2y}{8u}]\frac{dh}{ds}$

$But h = \frac{P}{y}+z$

Substituting for h in the first equation, we have:

$\frac{d}{dz}[\frac{p}{y}+z]$

$\frac{dh}{dz}=\frac{1}{y}\frac{dp}{ds}+\frac{dz}{dz}$

$= \frac{1}{0.8*62.4}(-60)+1$

= -0.20192

Substituting dh/dz value in the first equation (umax), we have:

$umax = \frac{0.01^2(0.8*62.4)}{8*10^-^3}(-0.20192)$

umax = 0.1259ft/s

4 0

3 years ago

The shaft is made of A992 steel. It has a diameter of 1 in. and is supported by bearings at A and D, which allows free rotation.

zysi [14]

Answer:

the angle of twist of B with respect to D is -1.15°

the angle of twist of C with respect to D is 1.15°

Explanation:

The missing diagram that is supposed to be added to this image is attached in the file below.

From the given information:

The shaft is made of A992 steel. It has a diameter of 1 in and is supported by bearing at A and D.

For the Modulus of Rigidity G = 11 × 10³ Ksi = 11 × 10⁶ lb/in²

The objective are :

1) To determine the angle of twist of B with respect to D

Considering the Polar moment of Inertia at the shaft $J\tau$

shaft $J\tau$ = $\dfrac{\pi}{2}r^4$

where ;

r = 1 in /2

r = 0.5 in

shaft $J \tau$ = $\dfrac{\pi}{2} \times 0.5^4$

shaft $J\tau$ = 0.098218

Now; the angle of twist at B with respect to D is calculated by using the expression

$\phi_{B/D} = \sum \dfrac{TL}{JG}$

$\phi_{B/D} = \dfrac{T_{CD}L_{CD}}{JG}+\dfrac{T_{BC}L_{BC}}{JG}$

where;

$T_{CD} \ \ and \ \ L_{CD}$ are the torques at segments CD and length at segments CD

${T_{BC} \ \ and \ \ L_{BC}}$ are the torques at segments BC and length at segments BC

Also ; from the diagram; the following values where obtained:

$L_{BC}}$ = 2.5 in

$J\tau$ = 0.098218

G = 11 × 10⁶ lb/in²

$T_{BC$ = -60 lb.ft

$T_{CD$ = 0 lb.ft

$L_{CD$ = 5.5 in

$\phi_{B/D} = 0+ \dfrac{[(-60 \times 12 )] (2.5 \times 12 )}{ (0.9818)(11 \times 10^6)}$

$\phi_{B/D} = \dfrac{[(-720 )] (30 )}{1079980}$

$\phi_{B/D} = \dfrac{-21600}{1079980}$

$\phi_{B/D} =$ − 0.02 rad

To degree; we have

$\phi_{B/D} = -0.02 \times \dfrac{180}{\pi}$

$\mathbf{\phi_{B/D} = -1.15^0}$

Since we have a negative sign; that typically illustrates that the angle of twist is in an anti- clockwise direction

Thus; the angle of twist of B with respect to D is 1.15°

(2) Determine the angle of twist of C with respect to D.Answer unit: degree or radians, two decimal places

For the angle of twist of C with respect to D; we have:

$\phi_{C/D} = \dfrac{T_{CD}L_{CD}}{JG}+\dfrac{T_{BC}L_{BC}}{JG}$

$\phi_{C/D} = 0+\dfrac{T_{BC}L_{BC}}{JG}$

$\phi_{B/D} = 0+ \dfrac{[(60 \times 12 )] (2.5 \times 12 )}{ (0.9818)(11 \times 10^6)}$

$\phi_{C/D} = \dfrac{21600}{1079980}$

$\phi_{C/D} =$ 0.02 rad

To degree; we have

$\phi_{C/D} = 0.02 \times \dfrac{180}{\pi}$

$\mathbf{\phi_{C/D} = 1.15^0}$

3 0

3 years ago

What is the meaning of the measurement met?

san4es73 [151]

Answer: MET(Metabolic equivalent of task) is the measurement technique that is based on the ratio of the energy that is being spend by an individual by any physical activity to the energy spent by individual sitting in quiet position.

This ration has some reference in which is ratio is measured.The energy being measured in the activity time of the person considers his/her relative mass at that time. This relative mass is compared in accordance with the oxygen amount of 3.5 ml per kg/min that is almost equal to energy of the person in the sitting position.

6 0

3 years ago