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trasher [3.6K]
3 years ago
5

1. What's the maximum overall length of the part? 2. What material is used to fabricate the part

Engineering
1 answer:
kiruha [24]3 years ago
8 0

Answer:

1. The maximum length overall length of the part is 4 inches

2. The material with which the part is fabricated is 0.25 inch thick Hot Rolled Steel

Explanation:

The information given in the drawing are;

The angle of projection of the drawing = Third angle projection

The thickness of the material = 0.25" plate

The material with which the part is fabricated = 0.25 inch thick HRS

HRS is the acronym for Hot Rolled Steel

Therefore;

<u>The material with which the part is fabricated is 0.25 inch thick Hot Rolled Steel</u>

The scale of the drawing = 1:1 = which is 1 is to 1 drawing

The maximum length overall length of the part, l = 4 units

By visually testing the 1:1 scale drawing measurements of 4 units is comparable 4 inches measurement with an online ruler

Therefore;

<u>The maximum length overall length of the part, l = 4 inches</u>

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A lake has a carrying capacity of 10,000 fish. At the current level of fishing, 2,000 fish per year are taken with the catch uni
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Answer:

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Explanation:

So in this problem we are going to be examining the application of a  population dynamics a fishing pond and stock fishing and objective would be to obtain the maximum sustainable yield and and the population of the fish at the obtained maximum sustainable yield,  so basically we would be applying an engineering solution to fishing

 

    So the current  yield which is mathematically represented as

                               \frac{dN}{dt} =   \frac{2000}{1 \ year }

 Where dN is the change in the number of fish

            and dt is the change in time

So in order to obtain the solution we need to obtain the  rate of growth

    For this we would be making use of the growth rate equation which is

                                      r = \frac{[\frac{dN}{dt}] }{N[1-\frac{N}{K} ]}

  Where N is the population of the fish which is given as 4,000 fishes

          and  K is the carrying capacity which is given as 10,000 fishes

             r is the growth rate

        Substituting these values into the equation

                              r = \frac{[\frac{2000}{year}] }{4000[1-\frac{4000}{10,000} ]}  =0.833

The maximum sustainable yield would be dependent on the growth rate an the carrying capacity and this mathematically represented as

                      Max Yield  = \frac{rK}{4} = \frac{(10,000)(0.833)}{4} = 2082 \ fishes \ per \ year

So since the maximum sustainable yield is 2082 then the the population need to be higher than 4,000 so in order to ensure a that this maximum yield the population size denoted by p' would be

                          p' = \frac{K}{2}  = \frac{10,000}{2}  = 5000\ fishes          

7 0
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Answer: Hello the question is incomplete below is the missing part

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Explanation:

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h2 = [ h1 + ( V2^2 - V1^2 ) / 2 ] * 1 / 32.2 * 1 / 778

∴ h2 = 136.17 Btu/Ibm

From Table A-17

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Answer:

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3 years ago
700.0 liters of a gas are prepared at 760.0 mmHg and 100.0 °C. The gas is placed into a tank under high pressure. When the tank
ololo11 [35]

Answer:

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Explanation:

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V₁ = 700.0 L

P₁ = 760.0 mmHg = 1 atm

T₁ = 100.0 °C

When the gas is in the thank we have:

V₂ =?

P₂ = 20.0 atm

T₂ = 32.0 °C      

Now, we can find the volume of the gas in the thank by using the Ideal Gas Law:

PV = nRT

V_{2} = \frac{nRT_{2}}{P_{2}}    (1)

Where R is the gas constant

With the initials conditions we can find the number of moles:

n = \frac{P_{1}V_{1}}{RT_{1}}    (2)

By entering equation (2) into (1) we have:

V_{2} = \frac{P_{1}V_{1}}{RT_{1}}*\frac{RT_{2}}{P_{2}} = \frac{1 atm*700.0 L*32.0 ^{\circ}}{100.0 ^{\circ}*20.0 atm} = 11.2 L

Therefore, When the gas is placed into a tank the volume of the gas is 11.2 L.

I hope it helps you!                                                                                                                                                                                

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3 years ago
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