In response to the market revolution:the legal system worked with local governments to find better ways to regulate entrepreneurs.Chief Justice John Marshall ruled that legislatures could not alter or rescind charters and contracts that previous legislatures had created.local judges protected businessmen from paying property damages associated with factory construction and from workers seeking to unionize.Massachusetts Chief Justice Lemuel Shaw held in Commonwealth v. Hunt that workers had no right to organize.corporations proved less able to raise capital than chartered companies did.

Answers

Answer 1

Answer:

Answer:

Local judges protected businessmen from paying property damages associated with factory construction and from workers seeking to unionize.

Explanation:

The Market Revolution is the name given to change in the economy that occurred in the 19th century. This drastic change led to various important changes in the United States and across the world. During this period, capitalism became more entrenched and society became, for the first time, predominantly capitalist. This gave businessmen great power, as they played an increasingly important role when it came to economic growth. The power that they had influenced society deeply, including legislation. Judges often protected businessmen from paying property damages that were associated with their business enterprises. Moreover, workers had few rights and protections, and judges prevented them from unionizing.

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A wind turbine system has the following specifications: Diameter:45 m Rated power 700 kW at the wind speed of 12 m/s Turbine speed 1500 rpm Determine the swept area of the wind turbine. a)- 1640 m^2 B)- 1690 m^2 c)- 1590 m^2 d)- 1540 m^2
A 4-pole, 3-phase induction motor operates from a supply whose frequency is 60 Hz. calculate: 1- the speed at which the magnetic field of the stator is rotating
A conical funnel of half-angle θ = 30 drains through a small hole of diameter d = 6:25 mm at the vertex. The speed of the liquid leaving the funnel is V= √ 2gy where y is the height of the liquid free surface above the hole. The funnel initially is filled to height y0 = 300 mm. Obtain an expression for the time, t, for the funnel to completely drain, and evaluate. Find the time to drain from 300 mm to 150 mm (a change in depth of 150 mm), and from 150 mm to completely empty (also a change in depth of 150 mm). Can you explain the discrepancy in these times?

A 1-m3 tank containing air at 10°C and 350 kPa is connected through a valve to another tank containing 3 kg of air at 35°C and 150 kPa. Now the valve is opened, and the entire system is allowed to reach thermal equilibrium with the surroundings, which are at 19.5°C. Determine the volume of the second tank and the final equilibrium pressure of air. The gas constant of air is R = 0.287 kPa·m3/kg·K.

Answers

Answer:

V₂=1.76 m³

P=222.03 KPa

Explanation:

Given that

For tank 1

V₁=1 m³

T₁= 10°C = 283 K

P₁=350 KPa

For tank 2

m₂=3 kg

T₂=35°C = 308 K

P₂=150 KPa

We know that for air

P V = m R T

P=pressure ,V= Volume,R= gas constant ,T= temperature ,m =mass

for tank 2

P₂ V₂ = m₂ R T₂

By putting the values

150 x V₂ = 3 x 0.287 x 308

V₂=1.76 m³

Final mass = m₁+m₂

m =m₁+m₂

The final volume V= V₂+V₁

V= 1.76 + 1 m³

V= 2.76 m³

The final temperature T= 19.5°C

T= 292.5 K

$m=(PV)/(RT)$

$m_1=(P_1V_1)/(RT_1)$

$m_1=(350* 1)/(0.287* 283)$

$m_1=4.3\ kg$

m =m₁+m₂

m =4.3 + 3 = 7.3 kg

Now at final state

P V = m R T

P x 2.76 = 7.3 x 0.287 x 292.5

P=222.03 KPa

A 2-m3insulated rigid tank contains 3.2 kg of carbon dioxide at 120 kPa.Paddle-wheel work is done on the system until the pressure in the tank rises to 180kPa. Determine the entropy change in the carbon dioxide during this process. Assumeconstant specific heat and room temperature at 300 K

Answers

Answer:

The change in entropy is found to be 0.85244 KJ/k

Explanation:

In order to solve this question, we first need to find the ration of temperature for both state 1 and state 2. For that, we can use Charles' law. Because the volume of the tank is constant.

P1/T1 = P2/T2

T2/T1 = P2/P1

T2/T1 = 180 KPa/120KPa

T2/T1 = 1.5

Now, the change in entropy is given as:

ΔS = m(s2 - s1)

where,

s2 = Cv ln(T2/T1)

s1 = R ln(V2/V1)

ΔS = change in entropy

m = mass of CO2 = 3.2 kg

Therefore,

ΔS = m[Cv ln(T2/T1) - R ln(V2/V1)]

Since, V1 = V2, therefore,

ΔS = mCv ln(T2/T1)

Cv at 300 k for carbondioxide is 0.657 KJ/Kg.K

Therefore,

ΔS = (3.2 kg)(0.657 KJ/kg.k) ln(1.5)

ΔS = 0.85244 KJ/k

The following program includes fictional sets of the top 10 male and female baby names for the current year. Write a program that creates: A set all_names that contains all of the top 10 male and all of the top 10 female names. A set neutral_names that contains only names found in both male_names and female_names. A set specific_names that contains only gender specific names. Sample output for all_names: {'Michael', 'Henry', 'Jayden', 'Bailey', 'Lucas', 'Chuck', 'Aiden', 'Khloe', 'Elizabeth', 'Maria', 'Veronica', 'Meghan', 'John', 'Samuel', 'Britney', 'Charlie', 'Kim'}

Answers

The program analyses a set of male and female names and displays, the combined set of names, specific names and neutral names. The program written in python 3 goes thus :

male_names = {'John', 'Bailey', 'Charlie', 'Chuck', 'Michael', 'Samuel', 'Jayden', 'Aiden', 'Henry', 'Lucas'}

#setofmalenames

female_names = {'Elizabeth', 'Meghan', 'Kim', 'khloe','Bailey', 'Jayden' , 'Aiden', 'Britney', 'Veronica', 'Maria'}

#setoffemalenames

neutral_names = male_names.intersection(female_names)

#names common to both males and females

all_names = male_names.union(female_names)

#set of all baby names ; both male and female

specific_names = male_names.symmetric_difference(female_names)

#name in one set and not in the other

print(all_names)

print(' ')

#leavesaspaceinbetweenthelines

print(specific_names)

print(' ')

print(neutral_names)

Asamplerunoftheprogramisattached.

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Answer:

Please see attachment

Explanation:

Please see attachment

The information on a can of pop indicates that the can contains 360 mL. The mass of a full can of pop is 0.369 kg, while an empty can weighs 0.153 N. Determine the specific weight, density, and specific gravity of the pop and compare your results with the corresponding values for water at Express your results in SI units.

Answers

Answer:

Specific weight of the pop, $w_(s) = 8619.45 N/m^(3)$

Density of the pop, $\rho_(p) = 8790.76 kg/m^(3)$

$g_(s) = 8.79076$

$w_(w) = 9782.36 N/m^(3)$

Given:

Volume of pop, V = 360 mL = 0.36 L = $0.36* 10^(-3) m^(3)$

Mass of a can of pop , m = 0.369 kg

Weight of an empty can, W = 0.153 N

Solution:

Now, weight of a full can pop, W

W' = mg = $0.369* 9.8 = 3.616 N$

Now weight of the pop in can is given by:

w = W' - W = 3.616 - 0.513 = 3.103 N

Now,

The specific weight of the pop, $w_(s) = (weight of pop)/(volume of pop)$

$w_(s) = (3.103)/(0.36* 10^(- 3)) = 8619.45 N/m^(3)$

Now, density of the pop:

$\rho_(p) = (w_(s))/(g)$

$\rho_(p) = (86149.45)/(9.8) = 8790.76 kg/m^(3)$

Now,

Specific gravity, $g_(s) = (\rho_(p))/(density of water, \rho_(w))$

where

$g_(s) = (8790.76)/(1000) = 8.79076$

Now, for water at $20^(\circ)c$ :

Specific density of water = $998.2 kg/m^(3)$

Specific gravity of water = $0.998 kg/m^(3)$

Specific weight of water at $20^(\circ)c$ :

$w_(w) = \rho_(20^(\circ))* g = 998.2* 9.8 = 9782.36 N/m^(3)$

A test bottle containing just seeded dilution water has its DO level drop by 0.8 mg/L in a 5-day test. A 300 mL BOD bottle filled with 30 mL of wastewater and the rest with seeded dilution water experiences a drop of 7.3 mg/L in the same period (5-day). Calculate the BOD5 of the wastewater.

Answers

Answer:

BOD_5 = =65.8 mg/l

Explanation:

dilution water DO level = 0.8 m/l

BOD level drop to 7.3 mg/l

we know that BOD at 5th day can be clculated by using following relation $BOD_5 = ((D_1 -D_2) -(B_1-B_2)(1-P))/(P)$

$D_1 -D_2$ - DO drop in BOD bottle

$B_1-B_2$ - dilution water drop

P= 30/300 = 0.1

$BOD_5 = ((7.3) -(0.8)(1-0.1))/(0.1)$

$BOD_5 = =65.8 mg/l$

Describe the difference between design guidelines or criteria and design performance. Explain the relationship between the use of guideline/criteria tools and performance tools during the design process

Answers

Answer:

PART A

Design guidelines are sets of procedures to be followed in order to enhance the designing of an object or other things.

Design Performance is the actual process of carrying out the design process of an object using the design guidelines or criteria.

PART B

(1) Design guidelines tools helps to enhance design Performance.

(2) Design guidelines tools helps the designing performance tools to be effective.

Explanation:Design guidelines are the various steps which has special tools used to guide the designer in order to enhance the designing performance tools and ensure that the design process is done devoid of errors.

Design Performance tools are tools which helps to enhance the actual design Activities.

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