This is the top part of the picture. I need someone to put the correct letter by the word !

This Is The Top Part Of The Picture. I Need Someone To Put The Correct Letter By The Word !

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Answer 1

Answer:

J = power steering pump  

A = hydraulic hoses

D = pitman arm

G = rack and pinion assembly

B = idler arm

C = center link

F = tie rod

E = tie rod end

H = rack and pinion boot

I = outer tie rod end

Explanation:


Related Questions

1) Each of the following would be considered company-confidential except
A) a contract bid B) employee salaries C) your company's strategic plan D) your company's address

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Answer is your company’s address

A 1.00 liter solution contains 0.46 M hydrocyanic acid and 0.35 M potassium cyanide If 25.0 mL of water are added to this system, indicate whether the following statements are true or false. (Note the the volume MUST CHANGE upon the addition of water.)
A. The concentration of HCN will increase.
B. The concentration of CN" will decrease.
C. The equilibrium concentration of Hy0 will remain the same 4
D. The pH will remain the same.
E. The ratio of [HCN]/[CN] will decrease.

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I won leader solution contain 0.46 mL of hydronic I said of 0.3 potassium

The human eye, as well as the light-sensitive chemicals on color photographic film, respond differently to light sources with different spectral distributions. Daylight lighting corresponds to the spectral distribution of the solar disk, which may be approximated as a blackbody at 5800K. Incandescent lighting from the usual household bulb corresponds approximately to the spectral distribution of a black body at 2900K. Calculate the band emission fractions for the visible region, 0.47 mu m to 0.65 mum, for each of the lighting sources. Calculate the wavelength corresponding to the maximum spectral intensity for each of the light sources

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Answer:

a) at T = 5800 k  

  band emission = 0.2261

at T = 2900 k

  band emission = 0.0442

b) daylight (d) = 0.50 μm

    Incandescent ( i ) =  1 μm

Explanation:

To Calculate the band emission fractions we will apply the Wien's displacement Law

The ban emission fraction in spectral range λ1 to λ2 at a blackbody temperature T can be expressed as

F ( λ1 - λ2, T ) = F( 0 ----> λ2,T) - F( 0 ----> λ1,T )

Values are gotten from the table named: blackbody radiation functions

a) Calculate the band emission fractions for the visible region

at T = 5800 k  

  band emission = 0.2261

at T = 2900 k

  band emission = 0.0442

attached below is a detailed solution to the problem

b)calculate wavelength corresponding to the maximum spectral intensity

For daylight ( d ) = 2898 μm *k / 5800 k  = 0.50 μm

For Incandescent ( i ) = 2898 μm *k / 2900 k = 1 μm

You are designing a package for 200 g of snack food that is sensitive to oxygen, and fails when it absorbs 120 ppm of oxygen (by weight). Marketing tells you it wants the snack to be in a plastic pouch measuring 6 inches by 6 inches (ignore seems), so it will have a total surface area available for permeation of 72 in(6" x 6" x 2 sides). You need to recommend an appropriate plastic material for this product, to provide a minimum of 70 days shelf life. Follow these steps:
a. Calculate the allowable oxygen gain, in cm at STP. (5 pts)
b. At this point, you do not know the material you will use, so you do not know the permeability coefficient or the thickness. The better the barrier the plastic you choose, the thinner the material can be to provide the appropriate barrier. Rather than simple trial and error, a sensible approach is to solve for the ratio of P/L that is required. We can solve the basic permeability equation for this ratio: P = 9 At Ap L Use the information you have to determine the required value for P/L, expressing your answer in cm/(100 in? d atm). (5 pts)
c. Use the information in the textbook (chapters 4 and 14 or in another reliable source; provide reference if you have used chapter 4, 14 or any other source) on oxygen permeability coefficients for various polymers to select a polymer that would be suitable, and calculate the required thickness. (Be sure this is reasonable; for example, if the required thickness is more than 20 mils, you need to choose a different polymer!) Note that chapter 4 presents these values in the units you used in (b) while chapter 14 presents values with different units, so unit conversion would be required. In your answer, state the material you have chosen, its oxygen permeability coefficient, and the minimum thickness you recommend. (Be sure to express the thickness with no more than one decimal place.) Obviously, there is more than one solution to this problem, but you only need one. (10 pts)

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Snack food is the sensitivity to auction and avails win the George 120

In a device to produce drinking water, humid air at 320C, 90% relative humidity and 1 atm is cooled to 50C at constant pressure. If the duty on the unit is 2,200 kW of heat is removed from the humid air, how much water is produced and what is the volumetric flow rate of air entering the unit

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Answer: hello your question lacks some data below is the missing data

Air at 32C has H = 0.204 kJ/mol and at 50C has H = -0.576 kJ/mol

H of steam can be found on the steam tables – vapor at 32C and 1 atm; vapor at 5C and liquid at 5C. Assume the volume of the humid air follows the ideal gas law.

H of water liquid at 5C = 21 kJ/kg; vapor at 5C = 2510.8 kJ/kg; H of water vapor at 32C = 2560.0 kJ/kg

Answer :

a) 34.98 lit/min

b) 1432.53 m^3/min

Explanation:

a) Calculate how much water is produced

density of water = 1 kg/liter

First we will determine the mass of condensed water using the relation below

inlet mass - outlet vapor mass =  0.0339508 * n * 18/1000 ----- ( 1 )

where : n = 57241.57

hence equation 1 = 34.98 Kg/min

∴ volume of water produced =  mass of condensed water / density of water

                                                =  34.98 Kg/min / 1 kg/liter

                                                = 34.98 lit/min

b) calculate the Volumetric flow rate of air entering the unit

applying the relation below

Pv = nRT

101325 *V = 57241.57 * 8.314 * 305  

∴ V = 1432.53 m^3/min

Determine the slopes and deflections at points B and C for the beam shown below by the moment-area method. E=constant=70Gpa I=500 (10^6)mm^4

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Answer:

hello your question is incomplete attached below is the complete question

answer :

Slopes : B = 180 mm , C = 373 mm

Deflection: B = 0.0514 rad ,  C = 0.077 rad

Explanation:

Given data :

I = 500(10^6) mm^4

E = 70 GPa

The M / EI  diagram is attached below

Deflection angle at B

∅B = ∅BA = [ 150 (6) + 1/2 (300)*6 ] / EI

                 = 1800 / ( 500 * 70 ) = 0.0514 rad

slope at B

ΔB = ΔBA = [ 150(6)*3 + 1/2 (300)*6*4 ] / EI

                 = 6300 / ( 500 * 70 ) = 0.18 m = 180 mm

Deflection angle at C

∅C = ∅CA = [ 1800 + 300*3 ] / EI

                 = 2700 / ( 500 * 70 )

                 = 2700 / 35000 = 0.077 rad

Slope at C

ΔC = [ 150 * 6 * 6 + 1/2 (800)*6*7 + 300(3) *1.5 ]

     = 13050 / 35000 = 373 mm

Air initially at 120 psia and 500o F is expanded by an adiabatic turbine to 15 psia and 200o F. Assuming air can be treated as an ideal gas and has variable specific heat. a) Determine the specific work output of the actual turbine (Btu/lbm). b) Determine the amount of specific entropy generation during the irreversible process (Btu/lbm R). c) Determine the isentropic efficiency of this turbine (%).

Answers

Answer:

a) specific work output of the actual turbine is 73.14 Btu/lbm

b) the amount of specific entropy generation during the irreversible process is 0.050416 Btu/lbm°R

c) Isentropic efficiency of the turbine is  70.76%

Explanation:

Given the data in the question;

For an adiabatic turbine; heat loss Q = 0

For Initial State;

p₁ = 120 psia

T₁ = 500°F = 959.67°R

from table; { Gas Properties of Air }

At T₁ = 959.67°R

[tex]s_1^0[/tex] = 0.74102 Btu/lbm°R

[tex]h_1[/tex] = 230.98 Btu/lbm

For Finial state;

p₂ = 15 psia

T₂ = 200°F = 659.67°R

[tex]s^0_{2a[/tex] = 0.64889 Btu/lbm°R

[tex]h_{2a[/tex] = 157.84 Btu/lbm

we know that R for air is 0.06855 Btu/lbm.R

a)

The specific work output of the actual turbine Wₐ is;

W[tex]_a[/tex] = [tex]h_1[/tex]  - [tex]h_{2a[/tex]

we substitute

W[tex]_a[/tex] = 230.98 - 157.84

W[tex]_a[/tex] = 73.14 Btu/lbm

Therefore, specific work output of the actual turbine is 73.14 Btu/lbm

b)

amount of specific entropy generation during the irreversible process.

To determine the entropy generation [tex]S_{gen[/tex];

[tex]S_{gen[/tex] = ΔS = [tex]s_{2a[/tex] - [tex]s_1[/tex] =  [tex]s^0_{2a[/tex]  - [tex]s_1^0[/tex] - R ln([tex]\frac{p_2}{p_1}[/tex])

we substitute in our values

[tex]S_{gen[/tex] = 0.64889 - 0.74102 - 0.06855 ln([tex]\frac{15}{120}[/tex])

[tex]S_{gen[/tex] = 0.64889 - 0.74102 + 0.1425457

[tex]S_{gen[/tex] = 0.050416 Btu/lbm°R

Therefore, the amount of specific entropy generation during the irreversible process is 0.050416 Btu/lbm°R

c)

Isentropic efficiency of turbine η[tex]_{is[/tex]

η[tex]_{is[/tex] = {actual work output] / [ ideal work output ] = ([tex]h_1[/tex]  - [tex]h_{2a[/tex] ) / ( [tex]h_1[/tex]  - [tex]h_{2s[/tex] )

Now, for an ideal turbine;

ΔS = 0 = [tex]s_{2s[/tex] - [tex]s_1[/tex]

so, [tex]s_{2s[/tex] - s₁  = [tex]s^0_{2s[/tex] - [tex]s_1^0[/tex] - R ln([tex]\frac{p_2}{p_1}[/tex])

0 =  [tex]s^0_{2s[/tex] - [tex]s_1^0[/tex] - R ln([tex]\frac{p_2}{p_1}[/tex])

[tex]s^0_{2s[/tex]  = [tex]s_1^0[/tex] + R ln([tex]\frac{p_2}{p_1}[/tex])

we substitute

[tex]s^0_{2s[/tex]  = 0.74102 + 0.06855 ln([tex]\frac{15}{120}[/tex])

[tex]s^0_{2s[/tex]  = 0.74102 - 0.1425457

[tex]s^0_{2s[/tex]  = 0.59847 Btu/lbm°R

Now, from table; { Gas Properties of Air }

At [tex]s^0_{2s[/tex]  = 0.59847 Btu/lbm°R; [tex]h_{2s[/tex]  = 127.614 Btu/lbm

η[tex]_{is[/tex]  = [( [tex]h_1[/tex]  - [tex]h_{2a[/tex] ) / ( [tex]h_1[/tex]  - [tex]h_{2s[/tex]  )] × 100%

we substitute

η[tex]_{is[/tex]  = [( 230.98 - 157.84 ) / ( 230.98 - 127.614 )] × 100%

η[tex]_{is[/tex]  = [ 73.14 / 103.366] × 100%

η[tex]_{is[/tex]  = 0.70758 × 100%

η[tex]_{is[/tex]  = 70.76%

Therefore, Isentropic efficiency of the turbine is  70.76%

You are using a Jupyter Notebook to explore data in a DataFrame named productDF. You want to write some inline SQL by using the following code, and visualize the results as a scatter plot: %%sql SELECT cost, price FROM product What should you do before running a cell with the %%sql magic? a. Create a new DataFrame named product from productDF.select("cost", "price") b. Persist the productDF DataFrame using productDF.createOrReplaceTempView("product") c. Filter the productDF dataframe using productDF.filter("cost == price") d. Rename the columns in the productDF DataFrame using productDF.withColumnRenamed("cost", "price")

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You need to explain it more simple as everyone is clueless

The substance xenon has the following properties:
normal melting point: 161.3 K
normal boiling point: 165.0 K
triple point: 0.37 atm, 152.0 K
critical point: 57.6 atm, 289.7 K
A sample of xenon at a pressure of 1.00 atm and a temperature of 204.0 K is cooled at constant pressure to a temperature of 163.7 K.Which of the following are true?
a. One or more phase changes will occur.
b. The final state of the substance is a liquid.
c. The final state of the substance is a solid.
d. The sample is initially a gas.
e. The liquid initially present will vaporize.

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Answer:

the liquid woulriekwvhrnsshsnekwb ndrhwmoadi

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