5. Which of these materials in a shop contain metals and toxins and can pollute the environment? A) Antifreeze B) Solvents C) Batteries D) All of the above

I say it's D) All of the above​

Answer:

Engine

Explanation:

Semi-independent suspension is the most common type of suspension system used on body over frame vehicles.

Semi-independent suspension give the front wheels some individual movement.

This suspension only used in rear wheels.

Thus, the correct option is Semi-independent suspension

Learn more about Semi-independent suspension

https://brainly.com/question/23838001

#SPJ2

Answer:

D

Explanation:

Got it wrong so i could answer

Complete question:

If it took a 30m³ capacity tank mediated by 2cm waterproof water faucet for 10 hours, calculate the flow speed (exit) of water from the tank.

Answer:

the flow rate of the water from the tank is 0.05 m³/min

Explanation:

Given;

volume of water in the tank, v = 30 m³

length of the waterproof faucet, L = 2cm = 0.02 m

duration of water flow through the tank, t = 10 hours

The flow rate of the water from the tank is calculated as;

[tex]Q = \frac{V}{t} = \frac{30 \ m^3}{10\ h \ \times \ 60 \min} = 0.05 \ m^3/ \min[/tex]

Therefore, the flow rate of the water from the tank is 0.05 m³/min

Answer:

<❤)structural analysi(❤>

Explanation:

(♨️)BRAINLEIST PLEASE(♨️)

Answer:

Structural analysis

Explanation:

Answer:

Not seeing any other information, the best answer I can give is 2m.

Explanation:

M = magnitude

You see, if they have an equal charge, and you add them, it'd be 2 * m, or 2m.

Answer: hello your question is incomplete below is the missing part

question :Determine the temperature of the cooled water exiting the cooling tower

answer : T  = 43.477° C

Explanation:

Temp of water at exit = 45°C

mass flow rate of cooling tower = 15,000 kg/s

Temp of makeup water = 20°C

Assuming an atmospheric pressure of = 101.3 kPa

Determine temperature of the cooled water exiting the cooling tower

Water entering cooling tower at 45°C

Given that Latent heat of water at 45°C = 43.13 KJ/mol

Cp(wet air) = 1.005+ 1.884(y1)

where: y1 - Inlet mole ratio = (0.01257) / (1 - 0.01257) = 0.01273

Hence : Cp(wet air) = 29.145 +  (0.01273) (33.94) = 29.577 KJ/kmol°C

First step : calculate the value of Q

Q = m*Cp*(ΔT) + W(latent heat)

Q = 321.6968 (29.577) (40-30) +  43.13 (18.26089)

Q =  95935.8547 KJ/s

Given that mass rate of water = 15000 kg/s

Hence the temperature of the cooled water can be calculated using the equation below

Q = m*Cp*∆T

Cp(water) = 4.2 KJ/Kg°C

95935.8547 = (15000)*(4.2)*(45 - T)

( 45 - T ) = 95935.8547/ 63000.    ∴ T  = 43.477° C

Answer:

The free body diagram of the system is, 558 368 368 508 O ?? O, Consider the equilibrium of horizontal forces. F

Explanation:

I hope this helps you but I think and hope this is the right answer sorry if it’s wrong.

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

Answer:

a

Explanation:

Answer:

LOS = A

Explanation:

Given all the parameters the level of service as seen from the attached graph

is LOS =  A

To determine the LOS from the attached graph

calculate the trial value of Vp

Vp = V / PHF

     = (100 + 150) / 0.95  =  263 pc/h

since the trial value of Vp = ( 0 to 600 ) pc/h . hence E.T = 1.7 , ER = 1

next we will calculate the flow rate

flow rate = 1 / [ ( 1 + PT(ET - 1 ) + PR ( ER - 1 ) ]

             Fhr  = 1 / 1.035 = 0.966 ≈ 1

next calculate the real value of Vp

Vp = V / ( PHF * N * Fhr * Fp )

     = ( 100 + 150 ) / ( 0.95 * 2 * 1 * 1 )

Vp ≈ 126 pc/h/In

Next calculate the density

D = Vp /  S  =  126 / ( 45 * 1.61 )  = 1.74 pc/km/In

Answer:

the critical flaw length is 10.06 mm

Explanation:

Given the data in the question;

plane strain fracture toughness [tex]K_{tc[/tex] = 92 Mpa√m

yield strength σ[tex]_y[/tex] = 900 Mpa

design stress is one-half of the yield strength ( 900 Mpa / 2 ) 450 Mpa

Y = 1.15

we know that;

Critical crack length [tex]a_c[/tex] = 1/π( [tex]K_{tc[/tex] / Yσ )²

we substitute

[tex]a_c[/tex] = 1/π( 92 Mpa√m / (1.15 × 450 Mpa  )²

[tex]a_c[/tex] = 1/π( 92 Mpa√m / (517.5 Mpa  )²

[tex]a_c[/tex] = 1/π( 0.177777  )²

[tex]a_c[/tex] = 1/π( 0.03160466 )

[tex]a_c[/tex] = 0.01006 m = 10.06 mm

Therefore, the critical flaw length is 10.06 mm

{ [tex]a_c[/tex] = ( 10.06 mm ) > 3 mm

The critical flow is subject to detection

Answer:

Constante de resorte = 1.1 N/m

Explanation:

Dados los siguientes datos;

Fuerza = 50N

Extensión = 60cm - 15cm = 45cm

Para encontrar la constante del resorte;

Matemáticamente, la fuerza ejercida para estirar un resorte viene dada por la fórmula;

Fuerza = constante de resorte * extensión

Sustituyendo en la fórmula, tenemos;

50 = constante de resorte * 45

Constante de resorte = 50/45

Constante de resorte = 1.1 N/m

Answer:

the ratio of the etched to the original crack tip radius is 30.24

Explanation:

Given the data in the question;

we determine the initial fracture stress using the following expression;

(σf)₁ = 2(σ₀)₁ [tex][[/tex] α₁/([tex]p_t[/tex])₁ [tex]]^{1/2[/tex] ----- let this be equation 1

where; (σ₀)₁ is the initial fracture strength

([tex]p_t[/tex])₁ is the original crack tip radius

α₁ is the original crack length.

first, we determine the final crack length;

α₂ = α₁ - 16% of α₁

α₂ = α₁ - ( 0.16 × α₁)

α₂ = α₁ - 0.16α₁

α₂ = 0.84α₁

next, we calculate the final fracture stress;

the fracture strength is increased by a factor of 6;

(σ₀)₂ = 6( σ₀ )₁

Now, expression for the final fracture stress

(σf)₂ = 2(σ₀)₂ [tex][[/tex] α₂/([tex]p_t[/tex])₂ [tex]]^{1/2[/tex] ------- let this be equation 2

where ([tex]p_t[/tex])₂ is the etched crack tip radius

value of fracture stress of glass is constant

Now, we substitute 2(σ₀)₁ [tex][[/tex] α₁/([tex]p_t[/tex])₁ [tex]]^{1/2[/tex] from equation for (σf)₂  in equation 2.

0.84α₁ for α₂.

6( σ₀ )₁ for (σ₀)₂.

∴

2(σ₀)₁ [tex][[/tex] α₁/([tex]p_t[/tex])₁ [tex]]^{1/2[/tex]  = 2(6( σ₀ )₁) [tex][[/tex] 0.84α₁/([tex]p_t[/tex])₂ [tex]]^{1/2[/tex]  

divide both sides by 2(σ₀)₁

[tex][[/tex] α₁/([tex]p_t[/tex])₁ [tex]]^{1/2[/tex]  =  6 [tex][[/tex] 0.84α₁/([tex]p_t[/tex])₂ [tex]]^{1/2[/tex]

[tex][[/tex] 1/([tex]p_t[/tex])₁ [tex]]^{1/2[/tex]  =  6 [tex][[/tex] 0.84/([tex]p_t[/tex])₂ [tex]]^{1/2[/tex]

[tex][[/tex] 1/([tex]p_t[/tex])₁ [tex]][/tex]  =  36 [tex][[/tex] 0.84/([tex]p_t[/tex])₂ [tex]][/tex]

1 / ([tex]p_t[/tex])₁ = 30.24 / ([tex]p_t[/tex])₂

([tex]p_t[/tex])₂ = 30.24([tex]p_t[/tex])₁

([tex]p_t[/tex])₂/([tex]p_t[/tex])₁ = 30.24

Therefore, the ratio of the etched to the original crack tip radius is 30.24

Answer:

Va / Vb = 0.5934

Explanation:

First step is to determine total head losses at each pipe

at Pipe A

For 1/4 open gate valve head loss = 17 *Va^2 / 2g

elbow loss = 0.75 Va^2 / 2g

at Pipe B

For 1/3 closed ball valve head loss = 5.5 *Vb^2 / 2g

elbow loss = 0.75 * Vb^2 / 2g

Given that both pipes are parallel

17 *Va^2/2g +  0.75*Va^2 / 2g = 5.5 *Vb^2 / 2g  + 0.75 * Vb^2 / 2g

∴ Va / Vb = 0.5934

Answer:

ummm ok?

Explanation:

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Answer:

a) converting Martensite to spheroidite

The heat treatment procedure for converting Martensite to spheroidite  involves heating Martensite of  0.76 wt% C steel for approximately 24 hours at a temperature of 700°C

b) Converting Spheroidite to martensite

The heat treatment entails heating Spheroidite of 0.76 wt% C steel to a temperature of 760°C  to austenization then it will  be quenched at temperature > 140°C

c) Converting Bainite to Pearlite

The heat treatment involves  heating Bainite of 0.76 wt% C steel to a temperature of 720°C  until austenization then it will  be quenched at temperature < 35°C

d) Converting Pearlite to Bainite

The heat treatment involves heating Pearlite of  0.76 wt% C steel to a temperature of 720°C until austenization then it will  be cooled at a temperature range 220°C to 540°C. the temperature is maintained at this range until the complete formation of Bainite

e) Converting Spheroidite to perlite

The heat treatment involves  heating Spheroidite of 0.76 wt% C steel to a temperature of 720°C  until austenization then it will  be quenched at temperature < 35°C

f) Perlite to Spheroidite

The heat treatment procedure for converting Perlite to spheroidite  involves heating Perlite of  0.76 wt% C steel for approximately 24 hours at a temperature of 700°C

g) Tempered martensite to martensite

The heat treatment entails heating Tempered martensite   of 0.76 wt% C steel to a temperature of 760°C  until austenization then it will  be quenched at temperature > 140°C

h) Bainite to spheroidite

The heat treatment procedure for converting Perlite to spheroidite  involves heating Bainite of  0.76 wt% C steel for approximately 24 hours at a temperature of 700°C

Explanation:

The heat treatment procedure is simply the heating of a metal to a high temperature  and cooling the metal back. during this process the metal will undergo certain mechanical changes

a) converting Martensite to spheroidite

The heat treatment procedure for converting Martensite to spheroidite  involves heating Martensite of  0.76 wt% C steel for approximately 24 hours at a temperature of 700°C

b) Converting Spheroidite to martensite

The heat treatment entails heating Spheroidite of 0.76 wt% C steel to a temperature of 760°C  to austenization then it will  be quenched at temperature > 140°C

c) Converting Bainite to Pearlite

The heat treatment involves  heating Bainite of 0.76 wt% C steel to a temperature of 720°C  until austenization then it will  be quenched at temperature < 35°C

d) Converting Pearlite to Bainite

The heat treatment involves heating Pearlite of  0.76 wt% C steel to a temperature of 720°C until austenization then it will  be cooled at a temperature range 220°C to 540°C. the temperature is maintained at this range until the complete formation of Bainite

e) Converting Spheroidite to perlite

The heat treatment involves  heating Spheroidite of 0.76 wt% C steel to a temperature of 720°C  until austenization then it will  be quenched at temperature < 35°C

f) Perlite to Spheroidite

The heat treatment procedure for converting Perlite to spheroidite  involves heating Perlite of  0.76 wt% C steel for approximately 24 hours at a temperature of 700°C

g) Tempered martensite to martensite

The heat treatment entails heating Tempered martensite   of 0.76 wt% C steel to a temperature of 760°C  until austenization then it will  be quenched at temperature > 140°C

h) Bainite to spheroidite

The heat treatment procedure for converting Perlite to spheroidite  involves heating Bainite of  0.76 wt% C steel for approximately 24 hours at a temperature of 700°C

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

Answer:

The algorithm is as follows:

0. Start

1. MyArray = []

2. Input n

3. if n <1 or n > 20:

3.1print("Invalid Size")

4. else:

4.1 for [tex]i = 0[/tex] to n - 1:

 4.1.1 Input MyArray[i]

 4.2 even = []; odd = []

4.3 enum = 0; onum = 0

4.4 for i = 0 to n - 1:

 4.4.1 if i%2 == 0:

  4.4.1.1 even[enum]=list[i]

  4.4.1.2 enum = enum + 1

 4.4.2 else:

  4.4.2.1 odd[onum]=list[i]

  4.4.2.2 onum= onum + 1  

4.5 MyArray.clear()

4.6 enum=0

4.7 while even:

 4.7.1 minm = even[0]

 4.7.2 for x in even:  

  4.7.2.1 if x < minm:

   4.7.2.1.1 minm = x

 4.7.3 MMyArray[enum] = minm

 4.7.4 even.remove(minm)

 4.7.5 enum = enum + 1  

4.8 while odd:

 4.8.1 maxm = odd[0]

 4.8.2 for x in odd:  

  4.8.2.1 if x > maxm:

   4.8.2.1.1 maxm = x

 4.8.3 MMyArray[enum] = maxm

 4.8.4 odd.remove(maxm)

 4.8.5 enum = enum + 1

4.9 for i = 0 to n - 1:

 4.9.1 print MyArray[i]

5. Stop

Explanation:

This starts the algorithm

0. Start

This creates an empty array

1. MyArray = []

This gets input for n (the length of the array)

2. Input n

If n is is less than 1 or greater than 20, then the input is invalid

3. if n <1 or n > 20:

3.1print("Invalid Size")

For valid values of n, we have:

4. else:

The italicized gets input into the array

4.1 for [tex]i = 0[/tex] to n - 1:

 4.1.1 Input MyArray[i]

This creates empty arrays for even index and for odd index

 4.2 even = []; odd = []

This initializes even index and odd index to 0

4.3 enum = 0; onum = 0

This iterates through the array indices

4.4 for i = 0 to n - 1:

If index is even, add array element to even

 4.4.1 if i%2 == 0:

  4.4.1.1 even[enum]=list[i]

  4.4.1.2 enum = enum + 1

If otherwise, add array element to odd

 4.4.2 else:

  4.4.2.1 odd[onum]=list[i]

  4.4.2.2 onum= onum + 1  

Clear elements of MyArray

4.5 MyArray.clear()

Set index to 0

4.6 enum=0

Iterate through the even array

4.7 while even:

Set minimum to the first index

 4.7.1 minm = even[0]

Sort the array in ascending order

 4.7.2 for x in even:  

  4.7.2.1 if x < minm:

   4.7.2.1.1 minm = x

Add the sorted array into MyArray

 4.7.3 MMyArray[enum] = minm

Delete the elements of even array

 4.7.4 even.remove(minm)

 4.7.5 enum = enum + 1  

Iterate through the odd array

4.8 while odd:

Set maximum to the first index

 4.8.1 maxm = odd[0]

Sort the array in descending order

 4.8.2 for x in odd:  

  4.8.2.1 if x > maxm:

   4.8.2.1.1 maxm = x

Add the sorted array into MyArray

 4.8.3 MMyArray[enum] = maxm

Delete the elements of odd array

 4.8.4 odd.remove(maxm)

 4.8.5 enum = enum + 1

Iterate through the indices of the double sorted MyArray

4.9 for i = 0 to n - 1:

Print each array element

 4.9.1 print MyArray[i]

End algorithm

5. Stop

See attachment for the program implemented in Python

Answer:

refer to aja

Explanation:

Answer:

L = spindle

M = lower ball joint

part without the letter showing = steering knuckle

Explanation:

Answer:

- 14.943 W/m^2K  ( negative sign indicates cooling )

Explanation:

Given data:

Area of FPC = 4 m^2

temp of water = 60°C

flow rate = 0.06 l/s

ambient temperature = 8°C

exit temperature = 49°C

Calculate the overall heat loss coefficient

Note : heat lost by water = heat loss through convection

m*Cp*dT  = h*A * ( T - To )

∴ dT / T - To = h*A / m*Cp  ( integrate the relation )

In ( [tex]\frac{49-8}{60-8}[/tex] ) =  h* 4 / ( 0.06 * 10^-3 * 1000 * 4180 )

In ( 41 / 52 ) = 0.0159*h

hence h = - 0.2376 / 0.0159

              = - 14.943  W/m^2K  ( heat loss coefficient )

Answer:

Explanation:

From the given information:

Strain fracture toughness [tex]K_k[/tex]= 75 MPa[tex]\sqrt{m}[/tex]

Tensile stress [tex]\sigma[/tex] = 361 MPa

Value of Y = 1.03

Thus, the minimum length of the critical interior surface crack which will result to fracture can be determined by using the formula:

[tex]a_c = \dfrac{1}{\pi} ( \dfrac{k_k}{\sigma Y})^2 \\ \\ a_c = \dfrac{1}{\pi} \Big [ \dfrac{75 \times \sqrt{10^3}}{361 \times 1.03 } \Big]^2 \\ \\ a_c = \dfrac{1}{\pi} \Big [ 6.378474693\Big]^2 \\ \\ \mathbf{ a_c = 12.95 \ mm}[/tex]