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Dahasolnce [82]
3 years ago
13

A front wheel drive vehicle with four wheel disc brakes is pulling to the left. Tech A says an external kink or internal restric

tion in the LF brake line will result in this condition. Tech B says to use a compression fitting to repair a section of brake line. Who is correct? Tech A Tech A Tech B Tech B Both Both Neither
Engineering
1 answer:
STALIN [3.7K]3 years ago
5 0

Answer:

Tech A is correct.          

Explanation:

A front-wheel-drive pulling to the left can result from several factors. One of them is definitely a faulty break.

A correct diagnosis linking the problem to the brakes is when there is an internal restriction and the pull is constant to one side and gets worse when the brakes are applied.

To confirm this, one would need to lift the vehicle and rotate each wheel by hand to check for excessive friction.

So the restriction may be caused by:

  • brake calipers that are sticky to the drum
  • too much brake fluid in the brake master cylinder - this prevents the caliper pistons from retracting when the brakes are released
  • misadjusted drum brakes and or parking brakes.

Cheers

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The cold drawn AISI 1040 steel bar with 25-mm width and 10-mm thick has a 6- mm diameter thru hole in the center of the plate. T
4vir4ik [10]

Answer:

A)  ( N ) = 1.54

B)  N ( Goodman ) = 1.133,  N ( Morrow) = 1.35

Explanation:

width of steel bar = 25-mm

thickness of steel bar = 10-mm

diameter = 6-mm

load on plate = between 12 kN AND 28 kN

notch sensitivity = 0.83

A ) Fatigue factor of safety based on yielding criteria

= δa + δm = \frac{Syt}{n}   =  91.03 + 227.58 = 490 / N

therefore Fatigue number of safety ( N ) = 1.54

δa (amplitude stress ) = kf ( Fa/A) = 2.162 * ( 8*10^3 / 190 ) = 91.03 MPa

A = area of steel bar = 190 mm^2 , Fa = amplitude load = 8 KN , kf = 2.162

δm (mean stress ) = kf ( Fm/A ) = (2.162 * 20*10^3 )/ 190 = 227.58 MPa

Fm = mean load  = 20 *10^3

B) Fatigue factor of safety based on Goodman and Morrow criteria

δa / Se + δm / Sut = 1 / N

= 91.03 / 183.15 + 227.58 / 590 = 1 /N

Hence N = 1.133 ( based on Goodman criteria )

note : Se = endurance limit (calculated) = 183.15 , Sut = 590

applying Morrow criteria

N =   1 / ( δa/Se) + (δm/ δf )

   = 1 / ( 91.03 / 183.15 ) + (227.58 / 935 )  

   = 1.35

6 0
3 years ago
Consider a plane composite wall that is composed of two materials of thermal conductivities kA = 0.1 W/m*K and kB = 0.04 W/m*K a
nadya68 [22]

Answer:

q=39.15 W/m²

Explanation:

We know that

Thermal resistance due to conductivity given as

R=L/KA

Thermal resistance due to heat transfer coefficient given as

R=1/hA

Total thermal resistance

R_{th}=\dfrac{L_A}{AK_A}+\dfrac{L_B}{AK_B}+\dfrac{1}{Ah_1}+\dfrac{1}{Ah_2}+\dfrac{1}{Ah_3}

Now by putting the values

R_{th}=\dfrac{0.01}{0.1A}+\dfrac{0.02}{0.04A}+\dfrac{1}{10A}+\dfrac{1}{20A}+\dfrac{1}{0.3A}

R_{th}=4.083/A\ K/W

We know that

Q=ΔT/R

Q=\dfrac{\Delta T}{R_{th}}

Q=A\times \dfrac{200-40}{4.086}

So heat transfer per unit volume is 39.15 W/m²

q=39.15 W/m²

4 0
3 years ago
Intravenous infusions are usually driven by gravity by hanging the bottle at a sufficient height to counteract the blood pressur
Bingel [31]

Answer:

(a) BP = 11.99 KPa

(b) h = 2 m

Explanation:

(a)

Since, the fluid pressure and blood pressure balance each other. Therefore:

BP = ρgh

where,

BP = Blood Pressure

ρ = density of fluid = 1020 kg/m³

g = acceleration due to gravity = 9.8 m/s²

h = height of fluid = 1.2 m

Therefore,

BP = (1020 kg/m³)(9.8 m/s²)(1.2 m)

<u>BP = 11995.2 Pa = 11.99 KPa</u>

(b)

Again using the equation:

P = ρgh

with data:

P = Gauge Pressure = 20 KPa = 20000 Pa

ρ = density of fluid = 1020 kg/m³

g = acceleration due to gravity = 9.8 m/s²

h = height of fluid = ?

Therefore,

20000 Pa = (1020 kg/m³)(9.8 m/s²)h

<u>h = 2 m</u>

7 0
3 years ago
Give me top 5 British snacks
tia_tia [17]

Answer:

1. Mini Cheddars

2. Sausage Roll

3. Monster Munch

4. Cheese Twists

5. Flapjacks

6 0
2 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
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