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bija089 [108]
4 years ago
5

(01.06 LC)

Physics
1 answer:
Marta_Voda [28]4 years ago
4 0

Answer:

Equilibrium is reached when demand equals supply.

Explanation:

Demand is the amount that consumers want and can buy of a certain product or service in a specific period of time and at a certain price. On the other hand, the supply is the amount that producers want and can sell of a certain product or service in a specific period of time and at a certain price.

In market equilibrium, the quantity demanded of the product or service equals the quantity supplied, so the price also equals. In other words, when market equilibrium is reached, demand and supply are the same, with their corresponding equilibrium price and quantity.

Two situations can occur:

  • When the quantity demanded is greater than the quantity supplied, the market is in a situation of excess demand.
  • On the other hand, it may happen that the price at which the products are being offered is greater than the equilibrium price and the quantity supplied is greater than the quantity demanded. Then there is an excess supply.

<u><em> Equilibrium is reached when demand equals supply.</em></u>

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Oil having a specific gravity of 0.9 is pumped as illustrated with a water jet pump. The water volume flowrate is 1 m3 /s. The w
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Answer:

The rate at which the pump moves oil is 1 m³/s

Explanation:

Assumptions:

  • there is steady-state flow
  • oil and water are incompressible
  • first fluid is water, second fluid is oil and third fluid is the mixture of oil and water.

\rho_1Q_1 + \rho_2Q_2 = \rho_3Q_3 -------equation (i)

where;

ρ is the fluid density

Q is the volumetric flow rate

Q_1 + Q_2 = Q_3--------equation (ii)

Substitute in Q₃ in equation i

\rho_1Q_1 + \rho_2Q_2 = \rho_3(Q_1 +Q_2)

divide through by ρ₁

\frac{\rho_1Q_1}{\rho_1}+ \frac{\rho_2Q_2}{\rho_1} =\frac{ \rho_3(Q_1 +Q_2)}{\rho_1}\\\\Note; \frac{\rho_2}{\rho_1} = \gamma_2 \ and \ \frac{\rho_3}{\rho_1} = \gamma_3\\\\Q_1 + \gamma_2Q_2 = \gamma_3(Q_1+Q_2)

Make Q₂ the subject of the formula

Q_2 = \frac{Q_1(1- \gamma_3)}{\gamma_3-\gamma_2} = \frac{1 (\frac{m^3}{s}) (1-0.95)}{0.95-0.9} = 1 \ \frac{m^3}{s}

Therefore, the rate at which the pump moves oil is 1 m³/s

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The area of a circular trampoline is 108.94 square feet. What is the radius of the trampoline? Round to the nearest hundredth.
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The correct answer to this question is this one: B. 5.89 feet
<span>The area of a circular trampoline is 108.94 square feet. The radius of the trampoline is 5.89 feet.
</span>
For a circle,
A = (pi)r^2
108.94 ft^2 = (pi)r^2
r^2 = (108.94 ft^2)/(pi)
r = sqrt(108.94/pi) ft
r = 5. 89 ft
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