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professor190 [17]
3 years ago
9

The cashew industry is perfectly competitive and until now each of the identical firms in the industry have been earning zero ec

onomic profits while selling ay units of output each (for a combined industry-wide total of qy units) at a market equilibrium price of P1 per unit. An unexpected increase in the demand for cashews raises the market equilibrium price to P2, which creates a situation in which P2 exceeds MC at 91 units of output.
a. If the firms continued producing 91 units each, would their combined output of cashews be too little, too much, or just right to achieve allocative efficiency?
i. Just right
ii. Too much
iii. Too little
b. In the long run, what will happen to the supply of cashews and the price of cashews?
i. The industry's supply of cashews will exceed Q1 and the price of cashews will equal P1.
ii. The industry's supply of cashews will be less than Q1 and the price of cashews will be less than P1.
iii. The industry's supply of cashews will equal Q1 and the price of cashews will equal P2.
iv. The industry's supply of cashews will exceed Q1 and the price of cashews will equal P2.
Business
1 answer:
gladu [14]3 years ago
8 0

Answer:

a. iii. Too little

b. i. The industry's supply of cashews will exceed Q1 and the price of cashews will equal P1.

Explanation:

Allocative efficiency refers to the point in production where Marginal Revenue equals Marginal cost. As this is a perfectly competitive market, marginal revenue is the same as price which as shown in the question, exceeds Marginal cost. The firms are therefore producing too little to achieve allocative efficiency and need to produce more to make price and marginal cost equal.

In the long run, the firms will produce more such that supply would exceed the original quantity supplied of Q1. This will lead to the price falling back to P1 as there is now less scarcity.

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4 years ago
A metal sphere of radius 15 cm has a net charge of 3.0 $ 10#8 C. (a) What is the electric field at the sphere’s surface? (b) If
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Answer:

E = 1.20 × 10^{4} N/C

V = 1800 V

x = 0.058 m

Explanation:

given data

radius = 15 cm

net charge = 3 × 10^{-8} C

electric potential decreased = 500 V

solution

we get here electric field at the sphere’s surface that is

electric field at the sphere’s surface E  =  \frac{q}{4\pi \epsilon _o R^2}   ............1

put here value

electric field at the sphere’s surface E  = \frac{3\times 10^{-8}\times 8.99\times 10^9}{ 0.15^2}    

E = 1.20 × 10^{4} N/C

and

potential on surface of sphere is

V =  \frac{q}{4\pi \epsilon _o R} ................2

V = \frac{3\times 10^{-8}\times 8.99\times 10^9}{ 0.15}  

V = 1800 V

and

now we get distance that is x

and we know here

ΔV = V(x) - V   ..............3

substitute here value

-500V = \frac{q}{4\pi \epsilon _o } \times (\frac{1}{R+x} - \frac{1}{R})

-500 V = {3\times 10^{-8}\times 8.99\times 10^9} \times (\frac{1}{0.15+x} - \frac{1}{0.15})

solve it we get x

x = 0.058 m

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Explanation:

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