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4 years ago

10

Suppose that the voltage is reduced by 10 percent (to 90 VV). By what percentage is the power reduced? Assume that the resistanc

e remains constant.

1 answer:

pickupchik [31]4 years ago

5 0

Answer:

The power is reduced by 19 percent.

Explanation:

The formula of power is given by:

$P = \frac{V^{2}}{R}$

In which V is the voltage, and R is the resistance.

I am going to use R = 1 in both cases.

With the original voltage, V = 1, we have

$P = \frac{V^{2}}{R} = \frac{1}{1} = 1$

With the modified voltage, V = 0.9, we have:

$P = \frac{V^{2}}{R} = \frac{0.9^{2}}{1} = 0.81$

So the power is reduced by 1-0.81 = 0.19 = 19 percent.

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A mass of air occupying a volume of 0.15m^3 at 3.5 bar and 150 °C is allowed [13] to expand isentropically to 1.05 bar. Its enth

e-lub [12.9K]

Answer:

Total work: -5.25 kJ

Total Heat: 52 kJ

Explanation:

V0 = 0.15

P0 = 350 kPa

t0 = 150 C = 423 K

P1 = 105 kPa (isentropical transformation)

Δh1-2 = 52 kJ (at constant pressure)

Ideal gas equation:

P * V = m * R * T

m = (R * T) / (P * V)

R is 0.287 kJ/kg for air

m = (0.287 * 423) / (350 * 0.15) = 2.25 kg

The specifiv volume is

v0 = V0/m = 0.15 / 2.25 = 0.067 m^3/kg

Now we calculate the parameters at point 1

T1/T0 = (P1/P0)^((k-1)/k)

k for air is 1.4

T1 = T0 * (P1/P0)^((k-1)/k)

T1 = 423 * (105/350)^((1.4-1)/1.4) = 300 K

The ideal gas equation:

P0 * v0 / T0 = P1 * v1 / T1

v1 = P0 * v0 * T1 / (T0 * P1)

v1 = 350 * 0.067 * 300 / (423 * 105) = 0.16 m^3/kg

V1 = m * v1 = 2.25 * 0.16 = 0.36 m^3

The work of this transformation is:

L1 = P1*V1 - P0*V0

L1 = 105*0.36 - 350*0.15 = -14.7 kJ/kg

Q1 = 0 because it is an isentropic process.

Then the second transformation. It is at constant pressure.

P2 = P1 = 105 kPa

The enthalpy is raised in 52 kJ

Cv * T1 + P1*v1 = Cv * T2 + P2*v2 + Δh

And the idal gas equation is:

P1 * v1 / T1 = P2 * v2 / T2

T2 = T1 * P2 * v2 / (P1 * v1)

Replacing:

Cv * T1 + P1*v1 + Δh = Cv * T1 * P2 * v2 / (P1 * v1) + P2*v2

Cv * T1 + P1*v1 + Δh = v2 * (Cv * T1 * P2 / (P1 * v1) + P2)

v2 = (Cv * T1 + P1*v1 + Δh) / (Cv * T1 * P2 / (P1 * v1) + P2)

The Cv of air is 0.7 kJ/kg

v2 = (0.7 * 300 + 105*0.16 + 52) / (0.7 * 300 * 105 / (105 * 0.16) + 105) = 0.2 m^3/kg

V2 = 2.25 * 0.2 = 0.45 m^3

T2 = 300 * 105 * 0.2 / (105 * 0.16) = 375 K

The heat exchanged is Q = Δh = 52 kJ

The work is:

L2 = P2*V2 - P1*V1

L2 = 105 * 0.45 - 105 * 0.36 = 9.45 kJ

The total work is

L = L1 + L2

L = -14.7 + 9.45 = -5.25 kJ

8 0

3 years ago

Which of the following wouldn't be pictured on a fan motor's ladder logic diagram? A. Auxiliary contacts B. two Push-button cont

pychu [463]

Answer:

c

Explanation:

:)

7 0

3 years ago

You want to plate a steel part having a surface area of 160 with a 0.002--thick layer of lead. The atomic mass of lead is 207.19

Pepsi [2]

Answer:

<u><em>To answer this question we assumed that the area units and the thickness units are given in inches.</em></u>

The number of atoms of lead required is 1.73x10²³.

Explanation:

To find the number of atoms of lead we need to find first the volume of the plate:

$V = A*t$

<u>Where</u>:

A: is the surface area = 160

t: is the thickness = 0.002

<u><em>Assuming that the units given above are in inches we proceed to calculate the volume: </em></u>

$V = A*t = 160 in^{2}*0.002 in = 0.32 in^{3}*(\frac{2.54 cm}{1 in})^{3} = 5.24 cm^{3}$

Now, using the density we can find the mass:

$m = d*V = 11.36 g/cm^{3}*5.24 cm^{3} = 59.5 g$

Finally, with the Avogadros number ( $N_{A}$ ) and with the atomic mass (A) we can find the number of atoms (N):

$N = \frac{m*N_{A}}{A} = \frac{59.5 g*6.022 \cdot 10^{23} atoms/mol}{207.19 g/mol} = 1.73 \cdot 10^{23} atoms$

Hence, the number of atoms of lead required is 1.73x10²³.

I hope it helps you!

3 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

A unidirectional E-Glass fiber-epoxy composite material contains 61% by volume E-Glass fibers stressed under isostrain condition

zalisa [80]

Answer:

The total load carried by the fiber will be "98%".

Explanation:

The given values are:

$V_{f}=0.61$

$V_{m}=1-V_{f}$

$=1-0.61$

$=0.39$

$E_{f}=10 \ Mpa$

$\sigma_{f}=0.35 \ Mpa$

$E_{m}=0.45 \ Mpa$ , $\sigma_{m}=9\times 10^{-3} \ Mpa$

As we know,

⇒ $E_{e}=fE_{f}+mE_{m}$

On putting the estimated values, we get

⇒ $=0.61\times 10+0.39\times 0.95$

⇒ $=6.27 \ Mpa$

Now,

⇒ $\sigma_{c}=f\sigma_{f}+m\sigma_{m}$

On putting the estimated values, we get

⇒ $=0.61\times 0.35+0.39\times 0.009$

⇒ $=0.217 \ Mpa$

Therefore,

The load carried by fiber,

$=\frac{f\sigma_{f}}{\sigma_{c}}$

$=\frac{0.35\times 0.61}{0.217}$

$=0.98$ i.e., 98%

4 0

4 years ago