Think about all the things that you have learned about wildfire in this module.

In light of climate change, explain how this risk map might look different in 100 years.

Answer:

generalization

Explanation:

Please mark me brainliest I need to level up

Answer:

[tex]3.32\ \text{kW}[/tex]

Explanation:

[tex]T_c[/tex] = Outside temperature = [tex]-5^{\circ}\text{C}[/tex]

[tex]T_h[/tex] = Temperature of room = [tex]21^{\circ}\text{C}[/tex]

[tex]Q_h[/tex] = Heat loss = 135000 kJ/h = [tex]\dfrac{135000}{3600}=37.5\ \text{kW}[/tex]

Coefficient of performance of heat pump

[tex]\text{COP}=\dfrac{1}{1-\dfrac{T_c}{T_h}}\\\Rightarrow \text{COP}=\dfrac{1}{1-\dfrac{273.15-5}{273.15+21}}\\\Rightarrow \text{COP}=11.3[/tex]

Input power

[tex]W_i=\dfrac{Q_h}{\text{COP}}\\\Rightarrow W_i=\dfrac{37.5}{11.3}\\\Rightarrow W_i=3.32\ \text{kW}[/tex]

The minimum power required to drive this heat pump is [tex]3.32\ \text{kW}[/tex].

You've asked an Incomplete question, lacking options. I answered based on the existing O*NET report.

Answer:

high school diploma

Explanation:

According to the Occupational Information Network (O*NET), most people who are Licensing Examiners and Inspectors typically have a high school diploma.

In other words, they do not seek to acquire a post-secondary school education.

Answer:

B

Explanation:

According to edge its answer B

associate's degree or on-the-job experience

got it right as a lucky guess as the O*net site is updated but edge doesn't bother to update their questions or links.

Click this link to view O*NET’s Education section for Licensing Examiners and Inspectors. According to O*NET, what is the most common level of education among Licensing Examiners and Inspectors?  

bachelor’s degree

associate's degree or on-the-job experience .......This is the correct answer.

some college, no degree

associate degree

Answer:

Explanation:

The cylinder head sits on the engine and closes off the combustion chamber. The gap that remains between the cylinder head and the engine is completed by the head gasket. Another task of the cylinder head is to ensure the constant lubrication of the cylinder        

Answer:

a) for back pressures of (a) 500 kPa

- mass flow rate is 2.127 kg/s

- exit Mach number is 1.046

b) for back pressures of (a) 784 kPa

- mass flow rate is 1.793 kg/s

- exit Mach number is 0.6

Explanation:

Given that;

A₂ = 0.001 m²

P₁ = 1 MPa

T₁ = 360 K

k = 1.4

P₂ = 500 Kpa

(1000/500)^(1.4-1 / 1.4) = 360 /T₂

2^(0.4/1.4) = 360/T₂

1.219 = 360 / T₂

T₂ = 360 / 1.219

T₂ = 295.32 K

CpT₁ + V₁²/2000 = CpT₂ + V₂²/2000

we substitute

CpT₁ + V₁²/2000 = CpT₂ + V₂²/2000

1.005 × 360 =  1.005 × 295.32 + v₂²/2000

v₂ = 360.56 m/s²

p₂v₂ = mRT₂

500 × (0.001 × 360.56) = m × 0.287 × 295.32

m = 2.127 kg/s

so Mach Number = V₂ / Vc

Vc = √( kRT) = √( 1.4 × 287 × 295.32) = 344.47 m/s

So Mach Number =  V₂ / Vc  =  360.56 / 344.47 = 1.046

Therefore for back pressures of (a) 500 kPa

- mass flow rate is 2.127 kg/s

- exit Mach number is 1.046

b)

AT P₂ = 784 kPa

(1000/784)^(1.4-1 / 1.4) = 360/T₂

T₂ = 335.82 K

now

V₂²/2000 = 1.005( 360 - 335.82)

V₂ = 220.45 m/s

P₂V₂ = mRT₂

784 × (0.001 × 220.45) = m( 0.287) ( 335.82)

172.83 = 96.38 m

m = 172.83 / 96.38

m = 1.793 kg/s

just like in a)

Vc = √( kRT) = √( 1.4 × 287 × 335.82) = 367.32 m/s

Mach Number = V₂ / Vc = 220.45 / 367.32 = 0.6

Therefore for back pressures of (a) 784 kPa

- mass flow rate is 1.793 kg/s

- exit Mach number is 0.6

Following are the

Given:

[tex]A_2 = 0.001\ m^2\\\\P_1 = 1\ MPa\\\\T_1 = 360\ K\\\\k = 1.4\\\\P_2 = 500\ Kpa\\\\[/tex]

To find:

Flow rate of mass, and Mach number

Solution:

For point a)

Using formula:

[tex]\to P^{\frac{r-1}{r}} \alpha T\\\\\to (\frac{1000}{500})^(\frac{1.4-1}{1.4}) = \frac{360}{T_2}\\\\\to 2^(\frac{0.4}{1.4}) = \frac{360}{T_2}\\\\\to 1.219 = \frac{360}{T_2}\\\\\to T_2 = \frac{360}{1.219}\\\\\to T_2 = 295.32\ K\\\\[/tex]

[tex]\to C_pT_1 + \frac{V_1^2}{2000} = C_pT_2 + \frac{V_2^2}{2000}\\\\\to 1.005 \times 360 = 1.005 \times 295.32 + \frac{v_2^2}{2000}\\\\\to v_2 = 360.56 \ \frac{m}{s^2} \\\\\to p_2v_2 = mRT_2\\\\\to 500 \times (0.001 \times 360.56) = m \times 0.287 \times 295.32\\\\\to m = 2.127\ \frac{kg}{s}\\\\[/tex]

Mach Number [tex]= \frac{V_2}{V_c}\\\\[/tex]

[tex]\to V_c = \sqrt{( kRT)} = \sqrt{( 1.4 \times 287 \times 295.32)} = 344.47 \ \frac{m}{s}\\\\[/tex]

[tex]\to \frac{V_2}{V_c} = \frac{360.56}{344.47} = 1.046[/tex]

For point b)

[tex]\to P_2 = 784\ kPa\\\\\to (\frac{1000}{784})^{(\frac{0.4}{1.4})} = \frac{360}{T_2}\\\\\to T_2 = 335.82\ K\\\\[/tex]

now

[tex]\to \frac{V_2^2}{2000} = 1.005( 360 - 335.82)\\\\\to V_2 = 220.45 \frac{m}{s}\\\\\to V_c = \sqrt{( kRT)} = \sqrt{( 1.4 \times 287 \times 335.82)} = 367.32\ \frac{ m}{s}\\\\\to c= \frac{220.45}{ 367.32} = 0.6\\[/tex]

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Given :

Two children are playing on a seesaw. The child on the left weighs 50 lbs. And the child on the right weighs 100 lbs.

To Find :

If the child on the left is 3 feet from the pivot, how far must the child on the right be from the pivot from the seesaw to be balanced.

Solution :

We know, for seesaw to balance :

[tex]m_1gd_1=m_2gd_2[/tex]

Here, [tex]d_1\ and \ d_2[/tex] is distance from origin from pivot point.

Putting all the values in the equation, we get :

[tex]50\times g\times 3=100\times g\times d_2\\\\d_2=1.5\ feet[/tex]

Therefore, distance of right child from pivot to balance the seesaw is 1.5 feet.

Hence, this is the required solution.

Answer:

1.5510 mm

Explanation:

Plane strain fracture toughness = 45 MPa√m

failing stress ( б ) = 300 MPa

maximum length of surface crack ( a )= 0.95 mm

Determine maximum allowable surface crack length ( in mm )

we will make use of this relationship for  Design stress equation to determine the value of Y

б = [tex]\frac{k}{y\sqrt{\pi *a} }[/tex]    --------- ( 1 )

k = 45 MPa√m

б  = 300 MPa

a = 0.95 mm

y = ?

From equation 1 make Y subject of the equation ( also substitute values into equation 1 above )

hence ; y = 2.7457

Now determine maximum allowable surface crack when component is exposed to a stress of 300 MPa and made from another alloy with plane-strain fracture toughness of 57.5 MPa√m

we will apply the equation

б = [tex]\frac{k}{y\sqrt{\pi *a} }[/tex]    --------- ( 2 )

K = 57.5 MPa√m

б = 300 MPa

y = 2.7457

a ( maximum allowable surface crack ) = ?

from equation make a subject of the equation

a = [tex]\frac{1}{\pi } (\frac{k}{\alpha y} )^{2}[/tex]  

a = [tex]\frac{1}{\pi } (\frac{57.5}{300*2.7457} )^2[/tex]  =   1.5510 mm

Answer:

a) 4.1 kw

b) 4.68 tons

c) 4.02

Explanation:

Saturated vapor enters compressor at ( p1 ) = 2.6 bar

Saturated liquid exits the condenser at ( p2 ) = 12 bar

Isentropic compressor efficiency = 80%

Mass flow rate = 7 kg/min

A) Determine compressor power in KW

compressor power = m ( h2 - h1 )

                                = 7 / 60 ( 283.71 - 248.545 )

                                = 4.1 kw

B) Determine refrigeration capacity in tons = m ( h1 - h4 )

                                                                       = 7/60 ( 248.545 - 107.34 )

                                                                       = 16.47 kw = 4.68 tons

C) coefficient of performance ( COP )

= Refrigeration capacity / compressor power

= 16.47 / 4.1 = 4.02

Attached below is the beginning part of the solution

Answer:

B: Team has the ball near center line

Explanation:

i guessed so...

Connecting rods undergo a process to alleviate manufacturing stresses from forging, a process known as cold forging.​

Forging is a metalworking process that involves shaping metal with localized compressive forces. A hammer or a die is used to deliver the blows.

Forging is frequently classified according to temperature: cold forging, warm forging, or hot forging.

The goal of forging is to make metal parts. Metal forging produces some of the most durable manufactured parts available when compared to other manufacturing methods.

Minor cracks and empty spaces in the metal are filled as the metal is heated and pressed.

Cold forging is the process of deforming a metal material at room temperature using extremely high pressure. The slug is placed in a die and compressed by a press until it fits into the desired shape.

Thus, the answer is cold forging.

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Answer:

D, meter.

Explanation:

Rhythm is associated with meter, which identifies units of stressed and unstressed syllables.

If a poem has a regular rhythm throughout the poem, it has a meter. Option D is correct.

Meter, which distinguishes between stressed and unstressed syllables, is related to rhythm. The fundamental rhythmic framework of a stanza or a line of poetry is known as meter.

The number of feet in the poem serves as a measure of the poem's meter, which is the rhythm of the language.

Many traditional poem forms call for a certain verse meter or a group of meters that alternate in a specified pattern. Prosody refers to both the study of meters and other types of versification, as well as their practical application.

Therefore, option D is correct.

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Answer:

FOLLOWED BY...  2 6 5 3 5 8 9

UwU

Answer: A. The output DC voltage will be lower

Explanation:

Using a resistor in a power-supply filter instead of an inductor will lead to a lower DC voltage output as resistors reduce voltage.

It would therefore be ideal to use an inductor as it does not lower DC voltage but inductors are expensive and can be quite large which is why it is more common to see resistors used in power-supply filter circuits.

Answer:

waste stream

Explanation:

i got it right on sp2

70.40mol cuz

1g sodium sulfate = 0.00704mol

take 10kg × 1000 = 10,000g

10,000g × 0.00704

final answer 70.40mol

(as per my thinking)

Answer:

70.40mol cuz

1g sodium sulfate = 0.00704mol

take 10kg × 1000 = 10,000g

10,000g × 0.00704

final answer 70.40mol

Answer:

7.47 lb. s/ft

22.42  lb. s/ft

Explanation:

Without mincing words let us dive straight into the solution to the above problem.

STEP ONE: calculate or determine the frequency.

The frequency can be calculated by making use of the formula given below;

frequency, w = √k/m. Thus, frequency = √ 12 × 15/ [40/32.2] = 12 rad/s.

STEP TWO: Determine or calculate for the viscous damping coefficient, c  for damping ratio of 0.5 and 1.5 respectively.

The viscous damping coefficient, c for 0.5 = [ 12 × 0.5 × 2 ×[ 40/32.2] / 2= 7.47 lb. s/ft.

The viscous damping coefficient, c for 1.5= [ 12 × 1.5 × 2 ×[ 40/32.2] / 2 = 22.42  lb. s/ft.

Answer:

A turbo blade has the best features from both other types of blade. The continuous, serrated edge makes for fast cutting while remaining smooth and clean. They are mostly used to cut a variety of materials, such as tile, stone, marble, granite, masonry, and many other building materials.

Explanation:

Answer:

Masonry Saw

Explanation:

Masonry Saws can be used to cut brick, ceramic, tile and or stone.

Answer:

hello below is missing piece of the complete question

minimum size = 0.3 cm

answer : 0.247 N/mm2

Explanation:

Given data :

section span : 10.9 and 13.4 cm

minimum load applied evenly to the top of span  : 13 N

maximum load for each member ; 4.5 N

lets take each member to be 4.2 cm

Determine the max value of P before truss fails

Taking average value of section span ≈ 12 cm

Given minimum load distributed evenly on top of section span = 13 N  

we will calculate the value of   by applying this formula

= [tex]\frac{Wl^2}{12} = (0.013 * 0.0144 )/ 12[/tex]  =  1.56 * 10^-5

next we will consider section ; 4.2 cm * 0.3 cm

hence Z (section modulus ) = BD^2 / 6  

                                             = ( 0.042 * 0.003^2 ) / 6  = 6.3*10^-8

Finally the max value of P( stress ) before the truss fails

= M/Z = ( 1.56 * 10^-5 ) / ( 6.3*10^-8 )  

          = 0.247 N/mm2

The car that is used to warn drivers about oversized loads is the Pilot Car.

The Pilot Car is also called Escort Car. It is a vehicle used to warn other vehicles of the presence of an over-sized vehicle.

The role of Pilot vehicle operators is to warn road users (motorists) to be cautious of over-sized loads or vehicles.

The cars are used to guide motorists that are making use of roads in construction sites.

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Answer:

false

Explanation:

Answer:

c

Explanation:

Thermoplastic parts are formed by forcing a molten solution into a mold.

Thus option D is correct.

Here,

Thermoplastic are plastic parts made from thermoplastic materials. These materials have the ability to be melted and remolded several times without undergoing any chemical change.

The thermoplastic parts are commonly used for different purposes such as in automotive industries, construction, medical, consumer goods, and much more. These parts are easily moldable and can be made into different shapes and sizes. They are also lightweight, strong, and durable, making them ideal for a wide range of applications.

They are commonly used for applications that require high strength and durability, such as in the automotive and aerospace industries.

Therefore option D is correct.

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Answer:

Air speed in the wind-tunnel [tex]v_{2}[/tex] = 27.5 m/s

Explanation:

Given data :

Manometer reading ; p1 - p2 = 45 mm of water

Pressure at section ( I ) p1 = 100 kPa ( abs )

temperature ( T1 ) = 25°C

Pw ( density of water ) = 999 kg/m3

g = 9.81 m/s^2

next we apply Bernoulli equation at section 1 and section 2

p1 - p2 = [tex]\frac{PairV^{2} _{2} }{2}[/tex]     ----------  ( 1 )

considering  ideal gas equation

Pair ( density of air ) = [tex]\frac{P}{RT}[/tex] ------------------- ( 2 )

R ( constant ) = 287 NM/kg.k

T = 25 + 273.15 = 298.15 k

P1 = 100 kN/m^2 = 100 * 10^3 or N/m^2

substitute values into equation ( 2 )

= 100 * 10^3 / (287 * 298.15)

= 1.17 kg/m^3

Also note ; p1 - p2 = PwgΔh  ------- ( 3 )

finally calculate the Air speed in the wind-tunnel test section by equating equation ( 1 ) and ( 3 )

[tex]\frac{PairV^{2} _{2} }{2}[/tex]   =  PwgΔh  

[tex]V^{2} _{2}[/tex] = [tex]\frac{2*999* 9.81* 0.045}{1.17}[/tex]  =  753.86

[tex]v_{2}[/tex] = 27.5 m/s

Answer:

the actual specific work required by the compressor to operate is 2425.88 kJ/kg

Explanation:

Given that;

h₁ = 1350 kJ/kg

h₂₅ = 3412 kJ/kg

compressor efficiency П_ise = 0.85

we know that;

compressor efficiency П_ise = isentropic work / actual work

П_ise = (h₂₅ - h₁) / (h₂ - h₁ )

so

0.85 =  (h₂₅ - h₁) / (actual work )

Actual work = (h₂₅ - h₁) / 0.85

Actual work = (3412 - 1350) / 0.85

Actual work = 2062 / 0.85

Actual work = 2425.88 kJ/kg

Therefore the actual specific work required by the compressor to operate is 2425.88 kJ/kg

A power washer is being used to clean the siding of a house. Water at the rate of 0.1 kg/s enters at 20°c and 1 atm, with the velocity 0.2m/s. The jet of water exits at 23°c, 1 atm with a velocity 20m/s at an elevation of 5m. At steady state, the magnitude of the heat transfer rate from power unit to the surroundings is 10% of the power input. Determine the power input to the motor in kW.

Answer:

Net power of 1.2 KW is being extracted

Explanation:

We are given;

Mass flow rate; m' = 0.1kg/s

Inlet temperature; T1 = 20°C = 293K

Inlet pressure; P1 = 1 atm = 10^(5) pa

Inlet velocity; v1 = 0.2 m/s

Exit Pressure; P2 = 1 atm = 10^(5) pa

Exit Temperature; T2 = 1 atm = 296K

Exit velocity; V2 = 20m/s

Change in elevation; h = Z2 - Z1 = 5m

We are told that the magnitude of the heat transfer rate from the power unit to the surroundings is 10% of the power input.

Thus;

Q = -0.1W

From Bernoulli equation;

Q - W = ∆Potential energy + ∆Kinetic energy + ∆Pressure energy

Where;

∆Potential energy = mg(z2 - z1)

∆Kinetic energy = ½m(v2² - v1²)

∆Pressure energy = mc_p(T2 - T1)

Thus;

-0.1W - W = [m'g(z2 - z1)] + [½m'(v2² - v1²)] + [m'c_p(T2 - T1)]

Where C_p is specific heat capacity of water = 4200 J/Kg.k

Plugging in the relevant values, we have;

-1.1W = (0.1 × 9.81 × 5) + (½ × 0.1(20² - 0.2²)) + (0.1 × 4200 × (296 - 293))

-1.1W = 4.905 + 19.998 + 1260

-1.1W = 1284.903

W = -1284.903/1.1

W ≈ -1168 J/s ≈ -1.2 KW

The negative sign means that work is extracted from the system.

Answer:

joule

Explanation:

heat is express in joules in x Ray equipment

Answer:

The answer is "Both A and B" are right

Explanation:

During the previous twenty years car producers have made significant advances in planning vehicle structures that give more noteworthy tenant insurance in planar accidents (Lund and Nolan 2003). Be that as it may, there has been little agreement with respect to the significance of rooftop strength in rollover crashes, just as the best strategy for surveying that strength. In 2006 one-fourth of lethally harmed traveler vehicle tenants were associated with crashes where vehicle rollover was considered the most hurtful occasion (Protection Establishment for Expressway Wellbeing, 2007). Numerous lethally harmed tenants in rollovers are unbelted, and some are totally or mostly launched out from the vehicle (Deutermann 2002).

There is difference concerning how underlying changes could influence launch hazard or the danger of injury for inhabitants who stay in the vehicle, paying little mind to belt use.

Answer:A Only

Explanation:

I Took the test

Answer:

lol

Explanation:r

Answer:a

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