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Alona [7]
3 years ago
13

For what two reasons do countries specialize? Countries specialize so that opportunity costs can be increased. Countries special

ize to excel in the production of specific goods and services. Countries specialize to make the most efficient use of their unique set of resources. Countries specialize to increase the number of their imported products.
Engineering
1 answer:
serg [7]3 years ago
8 0

Answer:

Countries specialize to excel in the production of specific goods and services;   Countries specialize to make the most efficient use of their unique set of resources.

Explanation:

Countries specialise to increase total gains from trade & specialisation

  • They specialise if they have <u>absolute advantage</u> in production of a good, in terms of their capability to produce a good at lower cost.
  • They specialise if they have <u>comparative advantage</u> in production of a good, in terms of their capacity to produce a good at a lower opportunity cost (less other goods sacrifised)
  • They specialise if the advantage in producing a good is due to the product using their <u>abundant factor (resource) intensively</u>.
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A compound sliding miter saw can be used to make
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Water is pumped steadily through a 0.10-m diameter pipe from one closed pressurized tank to another tank. The pump adds 4.0 kW o
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Complete Question

Complete Question is attached below.

Answer:

V'=5m/s

Explanation:

From the question we are told that:

Diameter d=0.10m

Power P=4.0kW

Head loss \mu=10m

 \frac{P_1}{\rho g}+\frac{V_1^2}{2g}+Z_1+H_m=\frac{P_2}{\rho g}+\frac{V_2^2}{2g}+Z_2+\mu

 \frac{300*10^3}{\rho g}+35+Hm=\frac{500*10^3}{\rho g}+15+10

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Generally the equation for Power is mathematically given by

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Therefore

 Q=\frac{P}{\rho g H_m}

 Q=\frac{4*10^4}{1000*9.81*10.9}

 Q=0.03935m^3/sec

Since

 Q=AV'

Where

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Therefore

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3 years ago
Horizontal shear forces and, consequently, horizontal shear stresses are caused in a flexural member at those locations where th
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A thin rim with a mean diameter of 1.2 m cross-section of 15 mm x 200 mm is subjected to an internal pressure of 10 MPa and rota
Soloha48 [4]

Answer:

The net centrifugal force over the rim is 30000N, the radial stress is 397887 Pa and the total change in diameter is 4.98 mm.

Explanation:

Lets first calculate the force in the rim due to the centrifugal force. For doing this we may assume that the centrifugal force is constant along with thick because of the thin thick.

Fc = m.ω^2/R

Where m is the mass, w the angular speed and R the mean radius.  The mass is computing by the rim density and its volume:

m=p.V

m=p*(A*R)

Where A is the cross-sectional area in meters:

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The angular speed in rad/s is:

ω = 800r/m . 1m/60s = 133.33 r/s

Thus the centrifugal force is:

Fc = (28.08 kg)*(133.33 rad/s)^2*(0.6m) = 299505N = 30000N

Note that the calculating value is the net contribution to the whole rim but the centrifugal force is distributed along the rim's external area:

fc = Fc / (2π .R .b)

Where b is rim's with equal to 200mm :

fc = 300000 N / (2π*0.6m*0.2m) = 397887 N/m^2

The centrifugal force can be taken as internal pressure:

Pfc = 397887 N/m^2 = 3978787 Pa

As both pressures act expanding the rim it can be summed:

Pt=Pi+Pfc

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Then for a thinner thick the stress is calculated by:

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σr = Pt.d/(2*t)

σr = 10397887 Pa / (2*0.015m*0.2m) = 415915480Pa = 415MPa

We know that the radial specific deformation ε is:

σr = E / εr

εr = σr / E

For a young modulus of 200GPa:

εr = 415MPa / 200GPa

εr = 415MPa / 200000MPa=0.002075

By definition the specific deformation is written in terms of the total change in the radius:

εr = Δr / R

Δr = R / εr =0.002075 * 1.2 m = 0.00249m

As we need the change in diameter:

Δd = 2Δr =0.00498m= 4.98mm [/tex]

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