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2 years ago

15

In a team, a person’s efforts are less identifiable than when that person works independently. Because the person’s efforts are

less identifiable, in extreme circumstances this can lead to __________.

1 answer:

Eduardwww [97]2 years ago

3 0

The answer is deindividuation - a psychological state in which a person does not feel individual responsibility.

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H2O enters a conical nozzle, operates at a steady state, at 2 MPa, 300 oC, with the inlet velocity 30 m/s and the mass flow rate

Colt1911 [192]

Answer:

The flow velocity at outlet is approximately 37.823 meters per second.

The inlet radius of the nozzle is approximately 0.258 meters.

Explanation:

A conical nozzle is a steady state device used to increase the velocity of a fluid at the expense of pressure. By First Law of Thermodynamics, we have the energy balance of the nozzle:

Energy Balance

$\dot m \cdot \left[\left(h_{in}+\frac{v_{in}^{2}}{2} \right)-\left(h_{out}+\frac{v_{out}^{2}}{2} \right)\right]= 0$ (1)

Where:

$\dot m$ - Mass flow, in kilograms per second.

$h_{in}$ , $h_{out}$ - Specific enthalpies at inlet and outlet, in kilojoules per second.

$v_{in}, v_{out}$ - Flow speed at inlet and outlet, in meters per second.

It is recommended to use water in the form of superheated steam to avoid the appearing of corrosion issues on the nozzle. From Property Charts of water we find the missing specific enthalpies:

Inlet (Superheated steam)

$p = 2000\,kPa$

$T = 300\,^{\circ}C$

$h_{in} = 3024.2\,\frac{kJ}{kg}$

$\nu_{in} = 0.12551\,\frac{m^{3}}{kg}$

Where $\nu_{in}$ is the specific volume of water at inlet, in cubic meters per kilogram.

Outlet (Superheated steam)

$p = 600\,kPa$

$T = 160\,^{\circ}C$

$h_{out} = 2758.9\,\frac{kJ}{kg}$

If we know that $\dot m = 50\,\frac{kJ}{kg}$ , $h_{in} = 3024.2\,\frac{kJ}{kg}$ , $h_{out} = 2758.9\,\frac{kJ}{kg}$ and $v_{in} = 30\,\frac{m}{s}$ , then the flow speed at outlet is:

$35765-25\cdot v_{out}^{2} = 0$ (2)

$v_{out} \approx 37.823\,\frac{m}{s}$

The flow velocity at outlet is approximately 37.823 meters per second.

The mass flow is related to the inlet radius ( $r_{in}$ ), in meters, by this expression:

$\dot m = \frac{\pi \cdot v_{in}\cdot r_{in}^{2} }{\nu_{in}}$ (3)

If we know that $\dot m = 50\,\frac{kJ}{kg}$ , $v_{in} = 30\,\frac{m}{s}$ and $\nu_{in} = 0.12551\,\frac{m^{3}}{kg}$ , then the inlet radius is:

$r_{in} = \sqrt{\frac{\dot m\cdot \nu_{in}}{\pi\cdot v_{in}}}$

$r_{in}\approx 0.258\,m$

The inlet radius of the nozzle is approximately 0.258 meters.

7 0

2 years ago

What do we need to travel farther distances in space

kakasveta [241]

Answer

Definitely a

Explanation:

Because OK yes the space suit is necessary and everything but u can't get to space without the shuttle

6 0

1 year ago

Read 2 more answers

For a bronze alloy, the stress at which plastic deformation begins is 275 MPa, and the modulus of elasticity is 115 GPa. (a) Wha

Anton [14]

Answer:

89375 N

Explanation:

Rearrange the formula for normal stress for F:

$\sigma=\frac{F}{A}$

$F=\sigma*A$

Convert given values to base units:

275 MPa = $275*10^{6}$ Pa

325 $mm^{2}$ = 0.000325 $m^{2}$

Substituting in given values:

F = $(275*10^{6})*(0.000325)=89375$ N

3 0

2 years ago

High speed increases the risk of collision because of ALL of these things EXCEPT:

Fofino [41]

Answer:

visibility is increased

3 0

2 years ago

9.21 LAB: Sorting TV Shows (dictionaries and lists) Write a program that first reads in the name of an input file and then reads

natka813 [3]

The following code or the program will be used

<u>Explanation:</u>

def readFile(filename):

dict = {}

with open(filename, 'r') as infile:

lines = infile.readlines()

for index in range(0, len(lines) - 1, 2):

if lines[index].strip()=='':continue

count = int(lines[index].strip())

name = lines[index + 1].strip()

if count in dict.keys():

name_list = dict.get(count)

name_list.append(name)

name_list.sort()

else:

dict[count] = [name]

print(count,name)

return dict

def output_keys(dict, filename):

with open(filename,'w+') as outfile:

for key in sorted(dict.keys()):

outfile.write('{}: {}\n'.format(key,';'.join(dict.get(key))))

print('{}: {}\n'.format(key,';'.join(dict.get(key))))

def output_titles(dict, filename):

titles = []

for title in dict.values():

titles.extend(title)

with open(filename,'w+') as outfile:

for title in sorted(titles):

outfile.write('{}\n'.format(title))

print(title)

def main():

filename = input('Enter input file name: ')

dict = readFile(filename)

if dict is None:

print('Error: Invalid file name provided: {}'.format(filename))

return

print(dict)

output_filename_1 ='output_keys.txt'

output_filename_2 ='output_titles.txt'

output_keys(dict,output_filename_1)

output_titles(dict,output_filename_2)

main()

8 0

2 years ago