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3 years ago

Around the 1760s what allowed for more supplies to be moved around?

Engineering

1 answer:

Licemer1 [7]3 years ago

3 0

Around the 1760s the factor that allowed for more supplies to be moved around-A free trade agreement

Explanation:

A free trade agreement is a contract between two or more countries to reduce the export and import barriers . As per the free trade agreement the goods and services of a nation can be bought and sold across international borders of other countries and the restriction imposed by the government (Like tariffs, quotas, subsidies ) is restricted /limited in order to encourage free trade.

The canton system of china is an example

So the statement that Around the 1760s the factor that allowed for more supplies to be moved around-A free trade agreement

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Air enters a compressor steadily at the ambient conditions of 100 kPa and 22°C and leaves at 800 kPa. Heat is lost from the comp

telo118 [61]

Answer:

a) 358.8K

b) 181.1 kJ/kg.K

c) 0.0068 kJ/kg.K

Explanation:

Given:

P1 = 100kPa

P2= 800kPa

T1 = 22°C = 22+273 = 295K

q_out = 120 kJ/kg

∆S_air = 0.40 kJ/kg.k

T2 =??

a) Using the formula for change in entropy of air, we have:

∆S_air = $c_p In \frac{T_2}{T_1} - Rln \frac{P_2}{P_1}$

Let's take gas constant, Cp= 1.005 kJ/kg.K and R = 0.287 kJ/kg.K

Solving, we have:

[/tex] -0.40= (1.005)ln\frac{T_2}{295} ln\frac{800}{100}[/tex]

$-0.40= 1.005(ln T_2 - 5.68697)- 0.5968$

Solving for T2 we have:

$T_2 = 5.8828$

Taking the exponential on the equation (both sides), we have:

[/tex] T_2 = e^5^.^8^8^2^8 = 358.8K[/tex]

b) Work input to compressor:

$w_in = c_p(T_2 - T_1)+q_out$

$w_in = 1.005(358.8 - 295)+120$

= 184.1 kJ/kg

c) Entropy genered during this process, we use the expression;

Egen = ∆Eair + ∆Es

Where; Egen = generated entropy

∆Eair = Entropy change of air in compressor

∆Es = Entropy change in surrounding.

We need to first find ∆Es, since it is unknown.

Therefore ∆Es = $\frac{q_out}{T_1}$

$\frac{120kJ/kg.k}{295K}$

∆Es = 0.4068kJ/kg.k

Hence, entropy generated, Egen will be calculated as:

= -0.40 kJ/kg.K + 0.40608kJ/kg.K

= 0.0068kJ/kg.k

3 0

3 years ago

python Write a program that asks a user to type in two strings and that prints •the characters that occur in both strings. •the

Yuliya22 [10]

Answer:

see explanation

Explanation:

#we first get the elements as inputs

x = input("enter string A :")

y = input("enter string B :")

#then we make independent sets with each

x = set(x)

y = set(y)

#then the intersection of the two sets

intersection = set.intersection(x,y)

#another set for the alphabet

#we use set.difference to get the elements present in x and not in y, and

#viceversa, finally we get the difference between the alphabet and the #intersection of the elements in our strings

z = set('abcdefghijklmnopqrstuvwxyz')

print('\nrepeated :\n',intersection)

print('differences :\n',' Items in A and not B\n',

set.difference(x,y),'\nItems in B and not A\n',

set.difference(y,x))

print('\nItems in neither :\n',set.difference(z,intersection))

8 0

4 years ago

A plate and frame heat exchanger has 15 plates made of stainless steel that are 1 m tall. The plates are 1 mm thick and 0.6 m wi

hodyreva [135]

Answer:

14.506°C

Explanation:

Given data :

flow rate of water been cooled = 0.011 m^3/s

inlet temp = 30°C + 273 = 303 k

cooling medium temperature = 6°C + 273 = 279 k

flow rate of cooling medium = 0.02 m^3/s

Determine the outlet temperature

we can determine the outlet temperature by applying the relation below

Heat gained by cooling medium = Heat lost by water

= ( Mcp ( To - 6 ) = Mcp ( 30 - To )

since the properties of water and the cooling medium ( water ) is the same

= 0.02 ( To - 6 ) = 0.011 ( 30 - To )

= 1.82 ( To - 6 ) = 30 - To

hence To ( outlet temperature ) = 14.506°C

6 0

3 years ago

An aluminum oxide component must not fail when a tensile stress of 12.5 MPa is applied. Determine the maximum allowable surface

aivan3 [116]

Answer:

1.44 mm

Explanation:

Compute the maximum allowable surface crack length using

$C=\frac {2E\gamma}{\pi \sigma_c^{2}}$ where E is the modulus of elasticity, $\gamma$ is surface energy and $\sigma_c$ is tensile stress

Substituting the given values

$C=\frac {2\times 393\times 10^{9}\times 0.9}{\pi\times (16\times 10^{6})^{2}= 0.001441103 m\approx 1.44mm$

The maximum allowable surface crack is 1.44 mm

4 0

3 years ago

Technician A says that the neutral safety switch must be bad and should be replaced. Technician B says that the neutral safety s

lukranit [14]

Answer:

Technician A is correct

Explanation:

The neutral safety switch when bad must be replaced and not adjusted as suggested by technician B because if the neutral safety switch is bad the Engine might not crank when put in neutral but it will crank when put in park and this is very bad for the life of the Engine it is better to replace it. A test for a bad/faulty neutral safety switch will be required to ascertain the level of damage it might cause to the Engine and prompt replacement is essential as well. because the neutral helps to prevent the car from starting when it is already engaged in a gear position therefore protecting the car from sudden collisions

3 0

3 years ago

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