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igor_vitrenko [27]
3 years ago
12

What is the main incentive for using a traditional economy? A. profit B. stability C. progress D. economic equality Please selec

t the best answer from the choices provided A B C D
Physics
2 answers:
VladimirAG [237]3 years ago
8 0

Answer:

the answer is B, stability

Explanation:

A traditional economy is a system that is based on honorable customs, history, and beliefs. Tradition guides economic decisions, such as production and distribution. Traditional economies depend on agriculture, fishing, hunting, gathering or some combination above. They use exchange instead of money. Most traditional economies operate in emerging markets and developing countries. They are often in Africa, Asia, Latin America and the Middle East. But you can find scholarships from traditional economies scattered all over the world. Economists and anthropologists believe that all other economies started out as traditional economies. Thus, they expect the remaining traditional economies to evolve into market, command or mixed economies over time.

timama [110]3 years ago
5 0

Answer:

b stability

Explanation:

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A standing wave of the third overtone is induced in a stopped pipe, 2.5 m long. The speed of sound is The frequency of the sound
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Answer:

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Explanation:

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f_{3}=\frac{(3)(340m/s)}{4(2.5m)}=102\ Hz

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3 0
2 years ago
the jet plane travels along the vertical parabolic path. when it is at point a it has speed of 200 m/s, which is increasing at t
givi [52]

Explanation:

Here is the complete question i guess. The jet plane travels along the vertical parabolic path defined by y = 0.4x². when it is at point A it has speed of 200 m/s, which is increasing at the rate .8 m/s^2. Determine the magnitude of acceleration of the plane when it is at point A.

→ The tangential component of acceleration is rate of increase in the speed of plane so,

a_{t} = v = 0.8 m/s^{2}

→ Now we have to find out the radius of curvature at point A which is 5 Km (from the figure).

dy/dx = d(0.4x²)/dx

         = 0.8x

Take the derivative again,

d²y/dx² = d(0.8x)/dx

          = 0.8

at x= 5 Km

dy/dx = 0.8(5)

         = 4

p = \frac{[1+ (\frac{dy}{dx})^{2}]^{\frac{3}{2} }   }{\frac{d^{2y} }{dx^{2} } }

now insert the values,

p = \frac{[1+(4)^{2}]^{\frac{3}{2} }  }{0.8}  = 87.62 km

→ Now the normal component of acceleration is given by

a_{n} = \frac{v^{2} }{p}

    = (200)²/(87.6×10³)

aₙ = 0.457 m/s²

→ Now the total acceleration is,

a = [(a_{t})^{2} +(a_{n} )^{2} ]^{0.5}

a = [(0.8)^{2} + (0.457)^{2}]^{0.5}

a = 0.921 m/s²

4 0
2 years ago
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