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dolphi86 [110]
3 years ago
10

If most wildfires are suppressed (all fires are put out) for many years, how does the risk of wildfire in the area change in the

future?
There will be a higher risk of wildfires
There will be a lower risk of wildfires
There is no difference in risk

Explain your answer.
Engineering
1 answer:
Serjik [45]3 years ago
5 0

Answer:

there is no difference in risk

Explanation:

even though they put out the fires, if they don't get rid of the source from which the wildfires comes from then wildfires wont stop coming

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Resolver em c as equações:<br><br> Z elevado à 2 igual a 3+4i
Marrrta [24]

Answer:

Quando um (ou todos) os outros coeficientes de uma equação do segundo grau são iguais a zero, essa equação é chamada incompleta. Neste artigo, analisaremos os métodos que podem ser usados para resolver equações incompletas, no caso em que o coeficiente C = 0, ou seja, o coeficiente é nulo.

Explanation:

Um lado da equação será a potência e outro, o número inteiro. De outro modo, transforme a equação deixando-a isolada em um dos lados. Reescreva a equação. Prepare-a a fim de extrair o logaritmo de ambos os lados, que é o inverso da potência. Você pode calcular o logaritmo de base.

3 0
3 years ago
A rectangular car-top carrier of 1.7-ft height, 5.0-ft length (front to back), and 4.2-ft width is attached to the top of a car.
Nataliya [291]

Answer:

\Delta P =1.2 \frac{1.3}{2}(26.822m/s)^2 (4.2*1.7*(0.3048)^2)=13.88 hp

Explanation:

We can assume that the general formula for the drag force is given by:

D= C_D \frac{\rho}{2}V^2 A

And we can see that is proportional to the area. On this case we can calculate the area with the product of the width and the height. And we can express the grad force like this:

D_1 = C_{D1} \frac{\rho}{2}V^2 (wh)

Where w is the width and h the height.

The last formula is without consider the area of the carrier, but if we use the area for the carrier we got:

D_2 = C_{D2} \frac{\rho}{2}V^2 (wh+ A_{carrier})

If we want to find the additional power added with the carrier we just need to take the difference between the multiplication of drag force by the velocity (assuming equal velocities for both cases) of the two cases, and we got:

\Delta P = C_{D2} \frac{\rho}{2}V^2 (wh+ A_{carrier}) V-  C_{D1} \frac{\rho}{2}V^2 (wh) V

We can assume the same drag coeeficient C_{D1}=C_{D2}=C_{D} and we got:

\Delta P = C_{D} \frac{\rho}{2}V^2 (wh+ A_{carrier}) V-  C_{D} \frac{\rho}{2}V^2 (wh) V

\Delta P = C_{D} \frac{\rho}{2}V^3 (A_{carrier})

1.7 ft =0.518 m

60 mph = 26.822 m/s

In order to find the drag coeffcient we ned to estimate the Reynolds number first like this:

R_E= \frac{Vl}{v}= \frac{26.822m/s*0.518 m}{1.58x10^{-4} Pa s}= 8.79 x10^{4}

And the value for the kinematic vicosity was obtained from the table of physical properties of the air under standard conditions.

Now we can find the aspect ratio like this:

\frac{l}{h}=\frac{5}{1.7}2.941

And we can estimate the calue of C_D = 1.2 from a figure.

And we can calculate the power difference like this:

\Delta P =1.2 \frac{1.3}{2}(26.822m/s)^2 (4.2*1.7*(0.3048)^2)=13.88 hp

8 0
3 years ago
What is your Favorite Diesel Truck?
jok3333 [9.3K]

Answer:

Race truck go vroom

Explanation:

Truck go fast

3 0
3 years ago
Read 2 more answers
What are some common work contexts for Licensing Examiners and Inspectors? Select four options.
Akimi4 [234]

According to O*NET, the common work contexts for Licensing Examiners and Inspectors include:

  1. Telephone
  2. Face-to-face discussions
  3. Contact with others
  4. Importance of being exact or accurate.

O*NET is an acronym for occupational information network and it refers to a free resource center or online database that is updated from time to time with several occupational definitions, so as to help the following categories of people understand the current work situation in the United States of America:

  • Workforce development professionals
  • Students
  • Human resource (HR) managers
  • Job seekers
  • Business firms

On O*NET, work contexts are typically used to describe the physical and social elements that are common to a particular profession or occupational work. Also, the less common work contexts are listed toward the bottom while common work contexts are listed toward the top.

According to O*NET, the common work contexts for Licensing Examiners and Inspectors include:

1. Telephone

2. Face-to-face discussions

3. Contact with others

4. Importance of being exact or accurate.

Read more on work contexts here: brainly.com/question/22826220

6 0
3 years ago
Read 2 more answers
An air-standard Carnot cycle is executed in a closed system between the temperature limits of 350 and 1200 K. The pressures befo
SVEN [57.7K]

This question is incomplete, the complete question is;

An air-standard Carnot cycle is executed in a closed system between the temperature limits of 350 and 1200 K. The pressures before and after the isothermal compression are 150 and 300 kPa, respectively. If the net work output per cycle is 0.5 kJ, determine (a) the maximum pressure in the cycle, (b) the heat transfer to air, and (c) the mass of air. Assume variable specific heats for air.

Answer:

a) the maximum pressure in the cycle is 30.01 Mpa

b) the heat transfer to air is 0.7058 KJ

c) mass of Air is 0.002957 kg

Explanation:

Given the data in the question;

We find the relative pressure of air at 1200 K (T1) and 350 K ( T4)

so from the "ideal gas properties of air table"

Pr1 = 238

Pr4 = 2.379

we know that Pressure P1 is only maximum at the beginning of the expansion process,

so

now we express the relative pressure and pressure relation for the process 4-1

P1 = (Pr2/Pr4)P4

so we substitute

P1 = (238/2.379)300 kPa

P1 = 30012.6 kPa = 30.01 Mpa

Therefore the maximum pressure in the cycle is 30.01 Mpa

b)

the Thermal heat efficiency of the Carnot cycle is expressed as;

ηth = 1 - (TL/TH)

we substitute

ηth = 1 - (350K/1200K)

ηth = 1 - 0.2916

ηth = 0.7084

now we find the heat transferred

Qin = W_net.out / ηth

given that the net work output per cycle is 0.5 kJ

we substitute

Qin = 0.5 / 0.7084

Qin = 0.7058 KJ

Therefore, the heat transfer to air is 0.7058 KJ

c)

first lets express the change in entropy for process 3 - 4

S4 - S3 = (S°4 - S°3) - R.In(P4/P3)

S4 - S3 = - (0.287 kJ/Kg.K) In(300/150)kPa

= -0.1989 Kj/Kg.K = S1 - S2

so that; S2 - S1 = 0.1989 Kj/Kg.K

Next we find the net work output per unit mass for the Carnot cycle

W"_netout = (S2 - S1)(TH - TL)

we substitute

W"_netout = ( 0.1989 Kj/Kg.K )( 1200 - 350)K

= 169.065 kJ/kg

Finally we find the mass

mass m = W_ net.out /  W"_netout

we substitute

m = 0.5 / 169.065

m = 0.002957 kg

Therefore, mass of Air is 0.002957 kg

5 0
3 years ago
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