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3 years ago

Plz help me

1 answer:

3 0

As much as the human and physical capital in economy increases, there is a decrease in the marginal gain in economic growth that will diminish.

<u>Explanation:</u>

Low-income countries might have an advantage achieving greater worker productivity and economic growth in the future as their economic growth is faster than the high - income countries.

As much as the human and physical capital in economy increases, there is a decrease in the marginal gain in economic growth that will diminish. And this is called, the laws of diminishing returns.

Secondly, low - income countries find it easier in developing technologies than the high - income technologies especially countries like India and China.

High - income countries put effort in inventing new technologies, whereas low - income countries just improve and improvise their technology.

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In tropical climates, the water near the surface of the ocean remains warm throughout the year as a result of solar energy absor

mote1985 [20]

Answer:

7.07%

Explanation:

Thermal efficiency can be by definition seen as the ratio of the heat utilized by a heat engine to the total heat units in the fuel consumed.

We will determine the thermal efficiency of the given problem at the attached file.

7 0

3 years ago

1. The equilibrium number of vacancies in Ni at 1123 K is 4.7x1022m-3. The atomic weight and density of Ni at 1123 K are 58.69 g

Ahat [919]

Answer:

vacancy formation energy of Ni is 1.400 eV

Explanation:

given data

number of vacancies in Ni = 4.7 x $10^{22}$ $m^{-3}$

atomic weight = 58.69 g/mol

density = 8.8 g/cm³

solution

we get here N that is

N = $\frac{N_A \times \rho}{A}$ ...........1

N = $\frac{6.023\times 10^{23} \times 8.8 \times 10^6}{58.69}$

N = $9.030 \times 10^{28}$

and here no of vacancy will be

Nv = $N \times e^{\frac{-Qv}{kT}}$ .................2

put here value

$4.7 \times 10^{22} = 9.030 \times 10^28 \times e^{\frac{-Qv}{8.62\times 10^{-5}\times 1123}}$

$10^{-7} \times 5.20487 = e^{\frac{-Qv}{0.0968}}$

take ln both side

$ln (10^{-7} \times 5.20487 ) = ln (e^{\frac{-Qv}{0.0968}})$

$-14.468 = \frac{-Qv}{0.0968}$

Qv = 1.400 eV

so vacancy formation energy of Ni is 1.400 eV

3 0

3 years ago

A reservoir manometer has vertical tubes of diameter D518 mm and d56 mm. The manometer liquid is Meriam red oil. Develop an alge

Nat2105 [25]

Answer:

Explanation:

Given that :

the diameter of the reservoir D = 18 mm

the diameter of the manometer d = 6 mm

For an equilibrium condition ; the pressure on both sides are said to be equal

∴

$\Delta \ P = \rho _{water} g \Delta h_{water} = \rho _{oil} g \Delta h_{oil}$

$\Delta \ P = \rho _{oil} g (x+L) ----- (1)$

According to conservation of volume:

$A*x = a*L$

$\dfrac{\pi}{4}D^2x = \dfrac{\pi}{4}d^2 L$

$x = ( \dfrac{d}{D})^2L$

Replacing x into (1) ; we have;

$\Delta \ P = \rho _{oil} g ( ( \dfrac{d}{D})^2L+L)$

$\Delta \ P = \rho _{oil} g \ L ( ( \dfrac{d}{D})^2+1)$

$L = \dfrac{\Delta \ P}{\rho _{oil} g \ ( ( \dfrac{d}{D})^2+1)}$

Thus; the liquid deflection is : $L = \dfrac{\Delta \ P}{\rho _{oil} g \ ( ( \dfrac{d}{D})^2+1)}$

when the applied pressure is equivalent to 25 mm of water (gage); the liquid deflection is:

$L = \dfrac{\Delta \ P}{\rho _{oil} g \ ( ( \dfrac{d}{D})^2+1)}$

$L = \dfrac{\rho_{water} \g \Delta \ h}{\rho _{water} SG_{oil}g \ ( ( \dfrac{d}{D})^2+1)}$

$L = \dfrac{\g \Delta \ h}{SG_{oil}g \ ( ( \dfrac{d}{D})^2+1)}$

$L = \dfrac{25}{0.827 ( ( \dfrac{6}{18})^2+1)}$

L = 27.21 mm

4 0

3 years ago

If carpet costs $3.25 a square foot, what would it<br> cost to replace a 200 square foot room?

KatRina [158]

Answer:

650$=200 X 3.25

Explanation:

5 0

3 years ago

Read 2 more answers

: Suppose you’re trying to measure an unknown sensor resistance, ????????????????????o???? , by converting it to a voltage. For

statuscvo [17]

Answer:

(a) R₁ / (R₁ + $R_{sensor}$ )

(b) - $R_{sensor}$ / (R₁ + $R_{sensor}$ )

Explanation:

(a)

i - Using voltage division rule =

v₀ = v₁ [ $R_{sensor}$ / (R₁+ $R_{sensor}$ )]

The sensitivity is

ii - $\int\limits^{vo}_{Rsensor}$ = dv₀/d $R_{sensor}$ * $R_{sensor}$ /v₀

= R₁ / (R₁ + $R_{sensor}$ )

(b)

i - Using voltage divisor rule = v₀ = v₁ [ $R_{sensor}$ / (R₁+ $R_{sensor}$ )]

ii - $\int\limits^{vo}_{Rsensor}$ = dv₀/d $R_{sensor}$ * $R_{sensor}$ /v₀

= - $R_{sensor}$ / (R₁ + $R_{sensor}$ )

7 0

3 years ago