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3 years ago

What ways are people and their communities impacted by wildfire?

Engineering

1 answer:

skelet666 [1.2K]3 years ago

3 0

Explanation:

Aren't these all correct? I mean all of these are possibilities, except for the first one and the 5th one. I'll explain them all :)

1) Destruction of wildlife habitat

This doesn't really affect people. Mainly the animals living in the wild, because their homes would get burnt down, or ruined so they need to find a new place to live.

2) Timber Industry Losses

Wildfires can lower job rates in the timber industry, which can leave people struggling, looking for a new job.

3) Transportation Shutdowns

For example, things like trains might not be able to work.

You know those large railroads that are getting held up? Those could get burnt down, or unstable, which could lead to many injuries and preventions to get somewhere.

4) Houses Burned or Damaged

This is probably one of the most well known issues of wildfire.

Because of this, a bunch of citizens might not be able to live anywhere if their home gets burned down. Some people are lucky and have family nearby they can live with, but others don't, forced to live on the streets.

5) Loss of Tourism

This issue really only effects specific people.

Only people who really have any sort of relationship with touring would feel threatened by these changes.

6) Health Problems from Breathing Smoke

Yes, this is a huge issue.

Inhaling harmful smoke can inflame your lungs and airway, causing them to swell and block oxygen. This can lead to acute respiratory distress syndrome and respiratory failure.

7) Loss of places of employment

Just like number 2, this could leave people stranded with nowhere to go. Some might have to become homeless if they can't pay rent, or buy groceries.

8) Town Buildings Burned or Damaged

Think about it-- How would you feel if your town building got burnt down? It would be a huge deal, and will take months; maybe years to rebuild.

I have no clue if I did this the way you wanted, but- here. xd

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A solid titanium alloy [G 114 GPa] shaft that is 720 mm long will be subjected to a pure torque of T 155 N m. Determine the mini

madam [21]

Answer:

Part 1: The diameter of the shaft so that the shear stress is not more than 150 MPa is 17.3 mm.

Part 2: The diameter of the shaft so that the twist angle is not more than 7° is 16.9 mm.

Explanation:

Part 1

The formula is given as

$\dfrac{T}{J}=\dfrac{\tau}{R}$

Here T is the torque which is given as 155 Nm

J is the rotational inertia which is given as $\dfrac{\pi d^4}{32}$

τ is the shear stress which is given as 150 MPa

R is the radius which is given as d/2 so the equation becomes

$\dfrac{T}{J}=\dfrac{\tau}{R}\\\dfrac{155}{\pi d^4/32}=\dfrac{150 \times 10^6}{d/2}\\\dfrac{155 \times 32}{\pi d^4}=\dfrac{300 \times 10^6}{d}\\\dfrac{1578.82}{d^4}=\dfrac{300 \times 10^6}{d}\\d^3=\dfrac{1578.82}{300 \times 10^6}\\d^3=5.26 \times 10^{-6}\\d=0.0173 m \approx 17.3 mm$

So the diameter of the shaft so that the shear stress is not more than 150 MPa is 17.3 mm.

Part 2

The formula is given as

$\dfrac{T}{J}=\dfrac{G\theta}{L}$

Here T is the torque which is given as 155 Nm

J is the rotational inertia which is given as $\dfrac{\pi d^4}{32}$

G is the torsional modulus which is given as 114 GPa

L is the length which is given as 720 mm=0.720m

θ is the twist angle which is given as 7° this is converted to radian as

$\theta=\dfrac{7*\pi}{180}\\\theta=0.122 rad\\$

so the equation becomes

$\dfrac{T}{J}=\dfrac{G\theta}{L}\\\dfrac{155}{\pi d^4/32}=\dfrac{114 \times 10^9\times 0.122}{0.720}\\\dfrac{1578.81}{d^4}=1.93\times 10^{10}\\d^4=\dfrac{1578.81}{1.93\times 10^{10}}\\d=(\dfrac{1578.81}{1.93\times 10^{10}})^{1/4}\\d=0.0169 m \approx 16.9mm$

So the diameter of the shaft so that the twist angle is not more than 7° is 16.9 mm.

6 0

3 years ago

For methyl chloride at 100°C the second and third virial coefficients are: B = −242.5 cm 3 ·mol −1 C = 25,200 cm 6 ·mol −2 Calcu

bogdanovich [222]

Answer:

a)W=12.62 kJ/mol

b)W=12.59 kJ/mol

Explanation:

At T = 100 °C the second and third virial coefficients are

B = -242.5 cm^3 mol^-1

C = 25200 cm^6 mo1^-2

Now according isothermal work of one mole methyl gas is

W=- $\int\limits^a_b {P} \, dV$

a= $v_2\\$

b= $v_1$

from virial equation

$\frac{PV}{RT}=z=1+\frac{B}{V}+\frac{C}{V^2}\\ \\P=RT(1+\frac{B}{V} +\frac{C}{V^2})\frac{1}{V}\\$

And

$W=-\int\limits^a_b {RT(1+\frac{B}{V} +\frac{C}{V^2}\frac{1}{V} } \, dV$

a= $v_2\\$

b= $v_1$

Now calculate V1 and V2 at given condition

$\frac{P1V1}{RT} = 1+\frac{B}{v_1} +\frac{C}{v_1^2}$

Substitute given values $P_1\\$ = 1 x 10^5 , T = 373.15 and given values of coefficients we get

$10^5(v_1)/8.314*373.15=1-242.5/v_1+25200/v_1^2$

Solve for V1 by iterative or alternative cubic equation solver we get

$v_1=30780 cm^3/mol$

Similarly solve for state 2 at P2 = 50 bar we get

$v_1=241.33 cm^3/mol$

Now

$W=-\int\limits^a_b {RT(1+\frac{B}{V} +\frac{C}{V^2}\frac{1}{V} } \, dV$

a=241.33

b=30780

After performing integration we get work done on the system is

W=12.62 kJ/mol

(b) for Z = 1 + B' P +C' P^2 = PV/RT by performing differential we get

dV=RT(-1/p^2+0+C')dP

Hence work done on the system is

$W=-\int\limits^a_b {P(RT(-1/p^2+0+C')} \, dP$

a= $v_2\\$

b= $v_1$

by substituting given limit and P = 1 bar , P2 = 50 bar and T = 373 K we get work

W=12.59 kJ/mol

The work by differ between a and b because the conversion of constant of virial coefficients are valid only for infinite series

8 0

3 years ago

What is the approximate theoretical maximum efficiency of a heat engine receiving heat at 627°C and rejecting heat to 27°c? a)-2

AysviL [449]

Answer:

The correct option is (e) η = 0.66

Explanation:

given data:

temperature at entry is 627 degree C

temperature at exit is 27 degree C

the efficiency of a engine is given as η

$\eta = (1-\frac{T_{L}}{T_{H}})*100$

where $T_{L}$ is temperature in kelvin at exit point

$T_{H}$ is temperature in kelvin at entry point

$\eta = (1-\frac{27+273}{627+273})*100$

η = 66.66 %

5 0

3 years ago

A method used to control density and overcrowding by restricting setbacks, building height, or ratio of open area is known as __

Mamont248 [21]

Answer:

Bulk zoning.

Explanation:

Zoning is a method used by municipal authorities or government authorities to divide a land area into zones. It is a urban planning method. There are several types of zoning; bulk zoning being one of them.

Bulk zoning can be defined as a set of regulations used to control lot size, ratio of floor area, open space, yards, setbacks, bulding height, etc. These setbacks and restrictions, then, help authorities to determine maximun size of a building on a zoning area, and also the positioning of the building.

Therefore, bulk zoning is the correct answer.

7 0

3 years ago

me that both a triaxial shear test and a direct shear test were performed on a sample of dry sand. When the triaxial test is per

stepan [7]

Answer:

shear strength = 2682.31 Ib/ft^2

Explanation:

major principal stress = 100 Ib / in2

minor principal stress = 20 Ib/in2

Normal stress = 3000 Ib/ft2

<u>Determine the shear strength when direct shear test is performed </u>

To resolve this we will apply the coulomb failure criteria relationship between major and minor principal stress a

for direct shear test

use Mohr Coulomb criteria relation between normal stress and shear stress

Shear strength when normal strength is 3000 Ib/ft = 2682.31 Ib/ft^2

attached below is the detailed solution

8 0

3 years ago