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GaryK [48]
3 years ago
11

Photon Lighting Company determines that the supply and demand functions for its most popular lamp are as follows: S(p) = 400 - 4

p + 0.00002p4 and D(p) = 2800 - 0.0012p3, where p is the price. Determine the price for which the supply equals the demand.
A) $93.24
B) $100.24
C) $96.24
D) $99.24

Physics
2 answers:
torisob [31]3 years ago
6 0

Answer: The correct answer is (C).

Explanation:

Supply function:

S(p)=400-4p+0.00002p^4

Demand function:

D(p)=2800-0.0012p^3

S(p)=D(p), price for which supply is equal to demand:

S(p)=D(p)=400-4p+0.00002p^4=2800-0.0012p^3

After solving this above equation graphically, we will get the values of p.

1) p =96.236

2) p=( -118.258)

We will reject the negative value of p.

So, the value of p that price for which the supply equals the demand is $ 96.236 \approx $96.24. Hence, the correct answer is option (C).

andrey2020 [161]3 years ago
3 0

Answer:

Price equals $96.24

Explanation:

It is given that,

The supply function is given by, S(p)=400-4p+0.00002p^4

The demand function is given by, D(p)=2800-0.0012p^3

We have to find the price for which the supply equals to the demand.

S(p)=D(p)

400-4p+0.00002p^4=2800-0.0012p^3

After solving above equation we get :

So, it is clear from the attached graph that p = 96.24 and q = -118.26

So, the correct option is (c) " $96.24 ".

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An open-end mercury manometer is connected to a low-pressure pipeline that supplies a gas to a laboratory. Because paint was spi
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Answer:

a

P_G  = 14.03 \  psig  

b

h_m =   0.148 \  m

Explanation:

From the question we are told that

The pressure of the manometer when there is no gas flow is P_{m} =  15.5 \  psig  =  15.5 *  6894.76 =  106868.78 \ N/m^2

The level of mercury is h  =  950 \ mm  =  0.950 \  m

The drop in the mercury level at the visible arm is d =  39.0 =  0.039 \  m

Generally when there is no gas flow the pressure of the manometer is equal to the gauge pressure which is mathematically represented as

P_g  =  P_m  =  g *  \delta h  * \rho

Here \rho is the density of mercury with value \rho = 13.6 *10^{3} kg/m^3

and \delta h is the difference in the level of gas in arm one and two

So

\delta h  =  \frac{106868.78}{  13.6 *10^{3} *  9.8 }

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Generally from manometry principle we have that

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Here P_G is the pressure of the gas

P_G +13.6 *10^{3} * 9.8  * 0.039    -  13.6 *10^{3}  *  9.8  * [0.950 - (0.148 + 0.039)] = 0

P_G  =  9.6724 04 *10^{4} \  N/m^2

converting to  psig

P_G  = \frac{ 9.6724 04 *10^{4} }{6894.76}

P_G  = 14.03 \  psig

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