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german
3 years ago
7

A body weighs 50 N and hangs from a spring with spring constant of 50 N/m. A dashpot is attached to the body. If the body is rai

sed 2 m from its equilibrium position and released from rest, determine the amplitude of vibration if (a) c = 17.7 kg/sec and (b) c = 40 kg/sec. What is the maximum amplitude of vibration in both of these cases?

Engineering
1 answer:
lbvjy [14]3 years ago
3 0

Answer:

a) 3.607 m

b) 1.5963 m

Explanation:

See that attached pictures for explanation.

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A 50-lbm iron casting, initially at 700o F, is quenched in a tank filled with 2121 lbm of oil, initially at 80o F. The iron cast
insens350 [35]

Answer:

a) The final equilibrium temperature is 83.23°F

b) The entropy production within the system is 1.9 Btu/°R

Explanation:

See attached workings

8 0
3 years ago
Let's model this housing price data! Before we can do this, however, we need to split the data into training and test sets. Reme
Lilit [14]

The program reads in a dataset into a pandas dataframe, and uses the train_test_split function in the sklearn library to split the data into <em>training and test sets</em>. The code goes thus :

import pandas as pd

<em>#import</em><em> </em><em>the</em><em> </em><em>pandas</em><em> </em><em>dataframe</em><em> </em><em>and</em><em> </em><em>alias</em><em> </em><em>it</em><em> </em><em>as</em><em> </em><em>pd</em>

from sklearn.model_selection import train_test_split

<em>#import</em><em> </em><em>the</em><em> </em><em>train_test_split</em><em> </em><em>function</em><em> </em>

housing_df = pd.read_csv('housing price.csv')

<em>#read</em><em> </em><em>in</em><em> </em><em>the</em><em> </em><em>housing</em><em> </em><em>data</em><em> </em>

features_df = df.iloc[:,1:]

<em>#seperate</em><em> </em><em>the</em><em> </em><em>features</em><em> </em><em>from</em><em> </em><em>the</em><em> </em><em>label</em><em> </em><em>;</em>

target_df = df.iloc[:,0]

<em>#put</em><em> </em><em>the</em><em> </em><em>label</em><em> </em><em>into</em><em> </em><em>a</em><em> </em><em>seperate</em><em> </em><em>dataframe</em><em> </em><em>as</em><em> </em><em>well</em><em>.</em><em> </em>

X_train, X_test, Y_train, Y_test = train_test_split(features_df, target_df, test_size = 0.1, random_state = 1)

<em>#uses</em><em> </em><em>tuple</em><em> </em><em>unpacking</em><em> </em><em>to</em><em> </em><em>randomly</em><em> </em><em>assign</em><em> </em><em>the</em><em> </em><em>data</em><em> </em><em>each</em><em> </em><em>of</em><em> </em><em>the</em><em> </em><em>4</em><em> </em><em>variables</em><em>.</em><em> </em>

<em>#</em><em>Test</em><em> </em><em>size</em><em> </em><em>is</em><em> </em><em>test</em><em> </em><em>percent</em><em> </em><em>of</em><em> </em><em>the</em><em> </em><em>entire</em><em> </em><em>dataset</em><em> </em>

Learn more :brainly.com/question/4257657?referrer=searchResults

3 0
3 years ago
A fire hose nozzle has a diameter of 1.125 in. According to some fire codes, the nozzle must be capable of delivering at least 2
Furkat [3]

Answer:

P_{1} = 403,708\,kPa\,(58.553\,psi)

Explanation:

Let assume that changes in gravitational potential energy can be neglected. The fire hose nozzle is modelled by the Bernoulli's Principle:

\frac{P_{1}}{\rho\cdot g} = \frac{P_{2}}{\rho \cdot g} + \frac{v^{2}}{2\cdot g}

The initial pressure is:

P_{1} = P_{2}+ \frac{1}{2}\cdot \rho v^{2}

The speed at outlet is:

v=\frac{\dot Q}{\frac{\pi}{4}\cdot D^{2}}

v=\frac{(250\,\frac{gal}{min} )\cdot (\frac{3.785\times 10^{-3}\,m^{3}}{1\,gal} )\cdot(\frac{1\,min}{60\,s} )}{\frac{\pi}{4}\cdot [(1.125\,in)\cdot(\frac{0.0254\,m}{1\,in} )]^{2} }

v\approx 24.592\,\frac{m}{s}\,(80.682\,\frac{ft}{s} )

The initial pressure is:

P_{1} = 101.325\times 10^{3}\,Pa+\frac{1}{2}\cdot (1000\,\frac{kg}{m^{3}} )\cdot (24.592\,\frac{m}{s} )^{2}

P_{1} = 403,708\,kPa\,(58.553\,psi)

7 0
3 years ago
Read 2 more answers
A 0.40-m3 insulated piston-cylinder device initially contains 1.3 kg of air at 30°C. At this state, the piston is free to move.
Setler79 [48]

Answer:

(a) The Final Temperature is 315.25 K.

(b) The amount of mass that has entered  0.5742 Kg.

(c) The work done is 56.52 kJ.

(d) The entrophy generation is 0.0398 kJ/kgK.

Explanation:

Explanation is in the following attachments.

6 0
3 years ago
A 4-L pressure cooker has an operating pressure of 175 kPa. Initially, one-half of the volume is filled with liquid and the othe
vodomira [7]

Answer:

the highest rate of heat transfer allowed is 0.9306 kW

Explanation:

Given the data in the question;

Volume = 4L = 0.004 m³

V_f = V_g = 0.002 m³

Using Table ( saturated water - pressure table);

at pressure p = 175 kPa;

v_f = 0.001057 m³/kg

v_g = 1.0037 m³/kg

u_f = 486.82 kJ/kg

u_g 2524.5 kJ/kg

h_g = 2700.2 kJ/kg

So the initial mass of the water;

m₁ = V_f/v_f + V_g/v_g

we substitute

m₁ = 0.002/0.001057  + 0.002/1.0037

m₁ = 1.89414 kg

Now, the final mass will be;

m₂ = V/v_g

m₂ = 0.004 / 1.0037

m₂ = 0.003985 kg

Now, mass leaving the pressure cooker is;

m_{out = m₁ - m₂

m_{out = 1.89414  - 0.003985

m_{out = 1.890155 kg

so, Initial internal energy will be;

U₁ = m_fu_f + m_gu_g

U₁ = (V_f/v_f)u_f  + (V_g/v_g)u_g

we substitute

U₁ = (0.002/0.001057)(486.82)  + (0.002/1.0037)(2524.5)

U₁ = 921.135288 + 5.030387

U₁ = 926.165675 kJ

Now, using Energy balance;

E_{in -  E_{out = ΔE_{sys

QΔt - m_{outh_{out = m₂u₂ - U₁

QΔt - m_{outh_g = m₂u_g - U₁

given that time = 75 min = 75 × 60s = 4500 sec

so we substitute

Q(4500) - ( 1.890155 × 2700.2 ) = ( 0.003985 × 2524.5 ) - 926.165675

Q(4500) - 5103.7965 = 10.06013 - 926.165675

Q(4500) = 10.06013 - 926.165675 + 5103.7965

Q(4500) = 4187.690955

Q = 4187.690955 / 4500

Q = 0.9306 kW

Therefore, the highest rate of heat transfer allowed is 0.9306 kW

5 0
3 years ago
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