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Zielflug [23.3K]

4 years ago

5

Use the orange points (square symbol) to plot the initial short-run industry supply curve when there are 20 firms in the market.

(Hint: You can disregard the portion of the supply curve that corresponds to prices where there is no output since this is the industry supply curve.) Next, use the purple points (diamond symbol) to plot the short-run industry supply curve when there are 30 firms. Finally, use the green points (triangle symbol) to plot the short-run industry supply curve when there are 40 firms.

1 answer:

Anton [14]4 years ago

3 0

Answer:

Check the explanation

Explanation:

Kindly check the attached image below to see the step by step explanation to the question above.

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Example 1: the two dimensional points P1(0,0) and P2(1,0) and the two tangents P', (1,1) and P2 (0,-1).find the equation of the

Stells [14]

Answer: (0,0)+ (1,0)= 1 lines upwards( suggesting that this is a line graph not saying it is but as an example) an (1,1) and (0,-1) all make a small square ( as this is a 2 dimensional graph that it has a negative side too,(below the positive side)) i hope this helps and is what you are looking for

Explanation:

6 0

3 years ago

T he area of a circle is pr 2. Define r as 5, then find the area of a circle,using MATLAB®.(b) The surface area of a sphere is 4

aksik [14]

Answer:

Area of Circle = 78.5398

Surface Area of Sphere = 1.2566 x 10^3 = 1256.6 ft

Volume of Sphere = 33.5103 ft

Explanation:

Please find below the written MatLab script used to solve the problem. I had to define r in each case to solve for the Area of the circle, the surface area and the volume of the Sphere.

r=5; % define r as 5

a=pi*r^2;% calculate the area of the circle

AreaOfCircle=a

r=10; % define r and 10 ft

sa=4*pi*r^2; %Calculate the surface area of the sphere

SphereSurfaceArea=sa

r=2;% define r as 2 ft

vs=(4/3)*pi*r^3;% Calculate the volume of the sphere

VolumeShere=vs

3 0

3 years ago

Read 2 more answers

For better thermal control it is common to make catalytic reactors that have many tubes packed with catalysts inside a larger sh

Paul [167]

Answer:

the pressure drop is 0.21159 atm

Explanation:

Given that:

length of the reactor L = 2.5 m

inside diameter of the reactor d= 0.025 m

diameter of alumina sphere $dp$ = 0.003 m

particle density = 1300 kg/m³

the bed void fraction $\in =$ 0.38

superficial mass flux m = 4684 kg/m²hr

The Feed is methane with pressure P = 5 bar and temperature T = 400 K

Density of the methane gas $\rho$ = 0.15 mol/dm ⁻³

viscosity of methane gas $\mu$ = 1.429 x 10⁻⁵ Pas

The objective is to determine the pressure drop.

Let first convert the Density of the methane gas from 0.15 mol/dm ⁻³ to kg/m³

SO; we have :

Density = 0.15 mol/dm ⁻³

Molar mass of methane gas (CH₄) = (12 + (1×4) ) = 16 mol

Density = $0.1 5 *\dfrac{16}{0.1^3}$

Density = 2400

Density $\rho_f$ = 2.4 kg/m³

Density = mass /volume

Thus;

Volume = mass/density

Volume of the methane gas = 4684 kg/m²hr / 2.4 kg/m³

Volume of the methane gas = 1951.666 m/hr

To m/sec; we have :

Volume of the methane gas = 1951.666 * 1/3600 m/sec = 0.542130 m/sec

$Re = \dfrac{dV \rho}{\mu}$

$Re = \dfrac{0.025*0.5421430*2.4}{1.429*10^5}$

$Re=2276.317705$

For Re > 1000

$\dfrac{\Delta P}{L}=\dfrac{1.75 \rho_f(1- \in)v_o}{\phi_sdp \in^3}$

$\dfrac{\Delta P}{2.5}=\dfrac{(1.75 *2.4)(1- 0.38)*0.542130}{1*0.003 (0.38)^3}$

$\Delta P=8575.755212*2.5$

$\Delta = 21439.38803 \ Pa$

To atm ; we have

$\Delta P = \dfrac{21439.38803 }{101325}$

$\Delta P =0.2115903087 \ atm$

ΔP ≅ 0.21159 atm

Thus; the pressure drop is 0.21159 atm

4 0

3 years ago

What are the standard procedures involved in the fixing/securing of cables?

Sladkaya [172]

<u>Cable should be pre-cut and hung suspended for 48 hours to develop its most natural set and lay prior to installation.</u>

<u>Cable should be installed with, not against, its natural set. ... </u>

<u>Strain relief on either end will reduce conductor breakage at the flex points.</u>

4 0

2 years ago

Three identical fatigue specimens (denoted A, B, and C) are fabricated from a nonferrous alloy. Each is subjected to one of the

Law Incorporation [45]

Answer:

B A and C

Explanation:

Given:

Specimen σ $_{max}$ σ $_{min}$

A +450 -150

B +300 -300

C +500 -200

Solution:

Compute the mean stress

σ $_{m}$ = (σ $_{max}$ + σ $_{min}$ )/2

σ $_{mA}$ = (450 + (-150)) / 2

= (450 - 150) / 2

= 300/2

σ $_{mA}$ = 150 MPa

σ $_{mB}$ = (300 + (-300))/2

= (300 - 300) / 2

= 0/2

σ $_{mB}$ = 0 MPa

σ $_{mC}$ = (500 + (-200))/2

= (500 - 200) / 2

= 300/2

σ $_{mC}$ = 150 MPa

Compute stress amplitude:

σ $_{a}$ = (σ $_{max}$ - σ $_{min}$ )/2

σ $_{aA}$ = (450 - (-150)) / 2

= (450 + 150) / 2

= 600/2

σ $_{aA}$ = 300 MPa

σ $_{aB}$ = (300- (-300)) / 2

= (300 + 300) / 2

= 600/2

σ $_{aB}$ = 300 MPa

σ $_{aC}$ = (500 - (-200))/2

= (500 + 200) / 2

= 700 / 2

σ $_{aC}$ = 350 MPa

From the above results it is concluded that the longest fatigue lifetime is of specimen B because it has the minimum mean stress.

Next, the specimen A has the fatigue lifetime which is shorter than B but longer than specimen C.

In the last comes specimen C which has the shortest fatigue lifetime because it has the higher mean stress and highest stress amplitude.

7 0

4 years ago