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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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Answer: I hope this helps I read this in a book once

Explanation:

Shower meteors come from the dust released by comets as they travel through our solar system. The dust spreads out along the comet's orbit and forms an elliptical trail of debris that passes around the sun and crosses the orbits of the planets.

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3 years ago
The density of gasoline is 730 kg/m3 at 0°C. Its average coefficient of volume expansion is 9.60 10-4(°C)−1. Assume 1.00 gal of
kipiarov [429]

Answer: 0.4911 kg

Explanation:

We have the following data:

\rho_{0\°C}= 730 kg/m^{3} is the density of gasoline at 0\°C

\beta=9.60(10)^{-4} \°C^{-1} is the average coefficient of volume expansion

We need to find the extra kilograms of gasoline.

So, firstly we need to transform the volume of gasoline from gallons to m^{3}:

V=8.50 gal \frac{0.00380 m^{3}}{1 gal}=0.0323 m^{3} (1)

Knowing density is given by: \rho=\frac{m}{V}, we can find the mass m_{1} of 8.50 gallons:

m_{1}=\rho_{0\°C}V

m_{1}=(730 kg/m^{3})(0.0323 m^{3})=23.579 kg (2)

Now, we have to calculate the factor f by which the volume of gasoline is increased with the temperature, which is given by:

f=(1+\beta(T_{f}-T_{o})) (3)

Where T_{o}=0\°C is the initial temperature and T_{f}=21.7\°C is the final temperature.

f=(1+9.60(10)^{-4} \°C^{-1}(21.7\°C-0\°C)) (4)

f=1.020832 (5)

With this, we can calculate the density of gasoline at 21.7\°C:

\rho_{21.7\°C}=730 kg/m^{3} f=(730 kg/m^{3})(1.020832)

\rho_{21.7\°C}=745.207 kg/m^{3} (6)

Now we can calculate the mass of gasoline at this temperature:

m_{2}=\rho_{21.7\°C}V (7)

m_{2}=(745.207 kg/m^{3})(0.0323 m^{3}) (8)

m_{2}=24.070 kg (9)

And finally calculate the mass difference \Delta m:

\Delta m=m_{2}-m_{1}=24.070 kg-23.579 kg (10)

\Delta m=0.4911 kg (11) This is the extra mass of gasoline

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