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2 years ago

Provide one example of a bad collision, and suggest an engineering solution to avoid the collision.

Engineering

1 answer:

JulsSmile [24]2 years ago

7 0

Answer:

1). Keep your distance. Drive far enough behind the car in front of you so you can stop safely. ...

Drive strategically. Avoid situations that could force you to suddenly use your brakes. ...

Don't get distracted. ...

Don't drive when drowsy or under the influence.

2). By far the deadliest accident type is the head-on collision. Head-on collisions consider both vehicle's speed at the time of the crash, which means even an accident at lower speeds can be catastrophic

Explanation:

first is how to avoid the collision and second is bad collision

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kow [346]

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3 years ago

Steam enters a turbine steadily at 7 MPa and 600°C with a velocity of 60 m/s and leaves at 25 kPa with a quality of 95 percent.

Rufina [12.5K]

Answer:

a) $\dot m = 16.168\,\frac{kg}{s}$ , b) $v_{out} = 680.590\,\frac{m}{s}$ , c) $\dot W_{out} = 18276.307\,kW$

Explanation:

A turbine is a steady-state devices which transforms fluid energy into mechanical energy and is modelled after the Principle of Mass Conservation and First Law of Thermodynamics, whose expressions are described hereafter:

Mass Balance

$\frac{v_{in}\cdot A_{in}}{\nu_{in}} - \frac{v_{out}\cdot A_{out}}{\nu_{out}} = 0$

Energy Balance

$-q_{loss} - w_{out} + h_{in} - h_{out} = 0$

Specific volumes and enthalpies are obtained from property tables for steam:

Inlet (Superheated Steam)

$\nu_{in} = 0.055665\,\frac{m^{3}}{kg}$

$h_{in} = 3650.6\,\frac{kJ}{kg}$

Outlet (Liquid-Vapor Mix)

$\nu_{out} = 5.89328\,\frac{m^{3}}{kg}$

$h_{out} = 2500.2\,\frac{kJ}{kg}$

a) The mass flow rate of the steam is:

$\dot m = \frac{v_{in}\cdot A_{in}}{\nu_{in}}$

$\dot m = \frac{\left(60\,\frac{m}{s} \right)\cdot (0.015\,m^{2})}{0.055665\,\frac{m^{3}}{kg} }$

$\dot m = 16.168\,\frac{kg}{s}$

b) The exit velocity of steam is:

$\dot m = \frac{v_{out}\cdot A_{out}}{\nu_{out}}$

$v_{out} = \frac{\dot m \cdot \nu_{out}}{A_{out}}$

$v_{out} = \frac{\left(16.168\,\frac{kg}{s} \right)\cdot \left(5.89328\,\frac{m^{3}}{kg} \right)}{0.14\,m^{2}}$

$v_{out} = 680.590\,\frac{m}{s}$

c) The power output of the steam turbine is:

$\dot W_{out} = \dot m \cdot (-q_{loss} + h_{in}-h_{out})$

$\dot W_{out} = \left(16.168\,\frac{kg}{s} \right)\cdot \left(-20\,\frac{kJ}{kg} + 3650.6\,\frac{kJ}{kg} - 2500.2\,\frac{kJ}{kg}\right)$

$\dot W_{out} = 18276.307\,kW$

6 0

2 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 training and test sets. The code goes thus :

import pandas as pd

#import the pandas dataframe and alias it as pd

from sklearn.model_selection import train_test_split

#import the train_test_split function

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

#read in the housing data

features_df = df.iloc[:,1:]

#seperate the features from the label ;

target_df = df.iloc[:,0]

#put the label into a seperate dataframe as well.

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

#uses tuple unpacking to randomly assign the data each of the 4 variables.

#Test size is test percent of the entire dataset

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

3 0

2 years ago

Anaircraft component is fabricated from an aluminum alloy that has a plane-strain fracture toughness of 40 MPa 1/2.It has been d

navik [9.2K]

Answer:

Yes, fracture will occur since toughness (42.4 MPa) is greater than the toughness of the material, 40MPa

Explanation:

Given

Toughness, k = 40Mpa

Stress, σ = 300Mpa

Length, l = 4mm = 4 * 10^-3m

Under which fracture occurred (i.e., σ= 300 MPa and 2a= 4.0 mm), first we solve for parameter Y (The dimensionless parameter)

Y = k/(σπ√a)

Where a = ½ of the length in metres

Y = 40/(300 * π * √(4/2 * 10^-3))

Y = 1.68 ---- Approximated

To check if fracture will occur of not; we apply the same formula.

Y = k/(σπ√a)

Then we solve for k, where

σ = 260Mpa and a = ½ * 6 * 10^-3

So,.we have

1.68 = k/(260 * π * √(6*10^-3)/2)

k = 1.68 * (260 * π * (6*10^-3)/2)

k = 42.4 MPa --- Approximately

Therefore, fracture will occur since toughness (42.4 MPa) is greater than the toughness of the material, 40 MPa

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3 years ago

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DedPeter [7]

Answer:

Explanation:

4 0

2 years ago

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