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irakobra [83]
3 years ago
11

Economic growth can be illustrated by: a. ​ an inward shift of the production possibilities curve. b. ​ a movement along the pro

duction possibilities curve. c. ​ a movement from a point on the production possibilities curve to a point inside the production possibilities curve. d. ​ an outward shift of the production possibilities curve.
Physics
1 answer:
erastovalidia [21]3 years ago
6 0

Answer:

Economic growth can be illustrated by:

d.  an outward shift of the production possibilities curve.

Explanation:

Economic growth is the process of increasing the economy's ability to produce goods and services. It is achieved by increasing the quantity or quality of resources.

Production Possibilities refers to the ability of a country to produce goods or services given the limited resources and technology.  It is therefore possible to increase production of both goods at the same time as long as resources allow it.

The Production Possibilities Curve, also known as the production possibilities frontier, is a graph that shows the maximum number of possible units a company can produce if it only produces two products using all of its resources efficiently. Firstly, and most commonly, growth is defined as an increase in the output that an economy produces over a period of time, the minimum being two consecutive quarters. An increase in an economy's productive potential can be shown by an outward shift in the economy's production possibility frontier (PPF).

Each point on the curve shows how much of each good will be produced when resources shift from making more of one good and less of the other. The curve measures the trade-off between producing one good versus another.PPC or production possibility curve is a curve whose basic purpose is to show the different possible combinations of two goods that can be produced within the given available resource.

The two main characteristics of PPC are: slopes downwards to the right: PPC slopes downwards from left to right. It is because in a situation of fuller utilization of the given resources, production of both the goods cannot be increased simultaneously.

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Which of the following would illustrate a quadratic relation between the dependent and independent variables when graphed?
Kitty [74]

Answer: option A. a graph of the area of a circle vs. its radius r (A = πr²).



Explanation:



A quadratic relation between the dependent and independent variables shows the independent variable raised to the power of 2.



This is it is a polynomial with general form ax² + bx + c, whewre a, b, and c, named coeficients,  are constants.



The function is y =  ax² + bx + c, where x is the independent variable and y is the dependent variable.



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The equation a = 1/b  is an inverse relation, not a quadratic relation.



The relation of distance vs. time for a car moving at constant speed is a linear relation of the kind v = u + st.



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Therefore, the only quadratic relation is shown by  a graph of the area of a circle vs. its radius r.

3 0
3 years ago
Which of the following is true about elements? A. Atoms of all elements contain the same number of protons, but the number of ne
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3 years ago
A drag racing car with a weight of 1600 lbf attains a speed of 270 mph in a quarter-mile race. Immediately after passing the tim
Kaylis [27]

Answer:

15.065ft

Explanation:

To solve this problem it is necessary to consider the aerodynamic concepts related to the Drag Force.

By definition the drag force is expressed as:

F_D = -\frac{1}{2}\rho V^2 C_d A

Where

\rho is the density of the flow

V = Velocity

C_d= Drag coefficient

A = Area

For a Car is defined the drag coefficient as 0.3, while the density of air in normal conditions is 1.21kg/m^3

For second Newton's Law the Force is also defined as,

F=ma=m\frac{dV}{dt}

Equating both equations we have:

m\frac{dV}{dt}=-\frac{1}{2}\rho V^2 C_d A

m(dV)=-\frac{1}{2}\rho C_d A (dt)

\frac{1}{V^2 }(dV)=-\frac{1}{2m}\rho C_d A (dt)

Integrating

\int \frac{1}{V^2 }(dV)= - \int\frac{1}{2m}\rho C_d A (dt)

-\frac{1}{V}\big|^{V_f}_{V_i}=\frac{1}{2m}(\rho)C_d (\pi r^2) \Delta t

Here,

V_f = 60mph = 26.82m/s

V_i = 120.7m/s

m= 1600lbf = 725.747Kg

\rho = 1.21 kg/m^3

C_d = 0.3

\Delta t=7s

Replacing:

\frac{-1}{26.82}+\frac{1}{120.7} = \frac{1}{2(725.747)}(1.21)(0.3)(\pi r^2) (7)

-0.029 = -5.4997r^2

r = 2.2963m

d= r*2 = 4.592m \approx 15.065ft

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Only 5 questions plz answer.
Masja [62]
Question 18: a
question 19: b
question 20: c
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2 years ago
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