Technician A says that when the malfunction indicator light or service engine light is on you should retrieve the diagnostic trouble code and follow the manufacturers recommended procedure. Technician B says that all obd-ll monitors must have the enabled criteria achieved before a test is performed. Who is correct?

Answer:

b. What is the operator's efficiency for the day?

                                                      AND

e. What would be the proper piece rate (rate expressed in money) for this job, assuming that the above time standard is correct?

Explanation:

Answer:

the torque required to RAISE the load is Tr = 18.09 Nm

the torque required to LOWER the load is Tl = 10.069 ≈ 10.07 Nm

the Overall Efficiency e = 0.2199 ≈ 0.22

Explanation:

Given that; F = 5 kN, p = 5mm, d = 35mm

Dm = d - p/2

Dm = 35 - ( 5/2) = 35 - 2.5

DM = 32.5mm

So the torque required to RAISE the load is

Tr = ( 5 × 32.5)/2 [(5 + (π × 0.09 × 32.5)) / ( (π × 32.5) - ( 0.09 × 5))] + [( 5 × 0.06 × 45)/2]

Tr = 81.25 × (14.1892 / 101.6518) + 6.75

Tr = 11.3414 + 6.75

Tr = 18.09 Nm

the torque required to LOWER the load is

Tl =  ( 5 × 32.5)/2 [(π × 0.09 × 32.5) - 5) / ( (π × 32.5) + ( 0.09 × 5))] + [( 5 × 0.06 × 45)/2]

Tl = 81.25 × 4.1892 / 102.5518 + 6.75

Tl = 3.3190 + 6.75

Tl = 10.069 ≈ 10.07 Nm

So since torque required to LOWER the load is positive

that is, the thread is self locking

Therefore the efficiency is

e = ( 5 × 5 ) / ( 2π × 18.09 )

e = 25 / 113.6628

e = 0.2199 ≈ 0.22

Answer:

a) the minimum time (minutes) for the aluminium casting to solidify is 2.86 min

b) the minimum time (minutes) for the grey iron casting to solidify is 2.13 min. Therefore solidification of grey iron cast will take less time (2.13 min) compared to the solidification of the aluminium cast (2.86 min)

Explanation:

Given that; diameter of Disk = 500 mm, thickness t = 20, mold constant Cm = 2.0 sec/mm²

first we find the volume and Area;

Volume V = πD²t / 4

Volume V = π × (500)² × 20 / 4 = 3,926,991 mm³

Area A = 2πD²/ 4 + πDt

Area A = {[π × (500)²] / 2} +{ π × (500) × (20)}

Area A = 392,699.08 + 31,415.93

Area A = 424,115 mm²

a)

Chvorinov’s rule

T(aluminium) = Cm (V/A) ²

T(aluminium) =  2.0 × (3,926,991 / 424,115) ²

T(aluminium) = 171.5 s = 2.86 min

∴ the minimum time (minutes) for the casting to solidify is 2.86 min

b)

For cast iron

Cm (mold constant = 1.488 sec/mm²)

Chvorinov’s rule

T(iron) = Cm (V/A) ²

T(iron) = 1.488 × (3,926,991 / 424,115) ²

T(iron) = 127.5720s = 2.13 min

Therefore solidification of grey iron cast will take less time (2.13 min) compared to the solidification of the aluminium cast (2.86 min)

Answer:

C) Decreases with an increase in temperature

Explanation:

As the temperature of a transistor increases, the thermal runaway property of the transistor becomes more significant and the transistors, conducting more freely as a result of the rise in temperature, causes an increase in the collector current or leakage current. The transistor base-emitter voltage decreases as a result.

With increased heating due to heavy current flow, the transistor is damaged.

Answer:

  leakage

Explanation:

That current is "leakage current."

Answer:

True

Explanation:

I looked it up

Answer:

true

Explanation:

Answer: hello attached below is the diagram which is part of your question

Total entropy change  = entropy change in cold reservoir + entropy change in hot reservoir = -0.166 + 0.083 = -0.0837 kj/k  it violates Clausius increase of entropy which is Sgen > 0

Explanation:

Clausius statement states that it is impossible to transfer heat energy from a cooler body to a hotter body in a cycle or region without any other external factors affecting it .  

applying the increase in entropy principle to prove this

temp of cold reservoir (t hot)= 600 k

temp of hot reservoir(t cold) = 1220 k

energy (q) = 100 kj

total entropy change  = entropy change in cold reservoir + entropy change in hot reservoir = -0.166 + 0.083 = -0.0837 kj/k

entropy change in cold reservoir = Q/t cold = 100 / 600 = -0.166 kj/k

entropy change in hot reservoir = Q / t hot = 100 / 1220 = 0.083 kj/k

hence it violates  Clausius inequality of increase of entropy principle which is states that generated entropy has to be > 0

Answer:

Correct Answer:

A. water pump

Explanation:

Timing belt in a vehicle helps to ensure that crankshaft, pistons and valves operate together in proper sequence. Timing belts are lighter, quieter and more efficient than chains that was previously used in vehicles.

Most car manufacturers recommended that, when replacing timing belt, tension assembly, water pump, camshaft oil seal should also be replaced with it at same time.

Answer:

B. Cutting planes are used to identify their locations.

Explanation:

Revolved view is a cross section view of revolved 90 degrees around a cutting plane projections. The revolved view of print will differ from a cross sectional view. It includes a line nothing the axis of revolution for the view. The correct answer is B. The revolved section in the prints has cutting planes that will be used to identify their location.

Answer:

30 feet

Explanation:

Given data :

design speed = 30 miles/h

super elevation = 0.08

determine the width of the turning roadway

calculate the value of R = V^2 / 15( e + p)

e = 0.08 , p = 0.2 , v = 30

R = (30)^2 / 15 ( 0.08 + 0.2 )

  = 900 / 15 ( 0.28 )

  ≈ 215 ft

pavement width from the calculation above = 28 ft

width of the turning roadway = pavement width + 2 = 30 feet ( because there are two vertical widths joining up the main road at the T junction )

Answer:

Move the buret clamp to a ring stand with a larger base.

Explanation:

A right stand is used for titration experiments in the laboratory. It holds the burette firmly during experiments so that accurate readings can be taken.

The right stand is made up of support base, vertical stainless steel, clamp with adjustable screw that holds on to the vertical rod.

The clamp is used to hold the burette in place.

If after clamping a buret to a ring stand, you notice that the set-up is tippy and unstable, the best action will be to move the buret clamp to a ring stand with a larger base.

The larger base provides a better center of gravity and stabilises the setup

Answer:

b. inversely proportional to radius of curvature

Explanation:

In curvilinear motions, the normal acceleration which is also called the centripetal acceleration is always directed towards the center of the circular path of motion. This acceleration has a magnitude that is directly proportional to the square of the speed of the body undergoing the motion and inversely proportional to the radius of the curvature of the motion path. The centripetal or normal acceleration a,  can be given by;

a = [tex]\frac{v^2}{r}[/tex]

Where;

v = speed of the body

r = radius of curvature.

Answer:

The answer is below

Explanation:

An AC voltage is represented by the relation v= 12 sin(500πt). Determine the:

The equation of an AC voltage is given as:

[tex]V=V_msin(2\pi ft)[/tex]

Where Vm is the maximum value of voltage and f is the frequency

From V= 12 sin(500πt), Vm = 12, 2πft = 500πt

(a) The peak to peak voltage is total amplitude (both the negative and positive amplitude) of the voltage, it is the difference between the positive amplitude and the negative amplitude. The peak to peak voltage ([tex]V{p-p}[/tex]) is given as:

[tex]V_{p-p}=2V_m=2*12=24\ V[/tex]

b) The frequency is the number of oscillation per second. The frequency (f) is gotten from:

2πft = 500πt

2f = 500

f = 500/2

f = 250 Hz

c) The root mean square voltage is the dc value of the voltage. It is given by:

[tex]V_{rms}=\frac{V_m}{\sqrt{2} }=\frac{12}{\sqrt{2} }=8.5\ V[/tex]

d) The period (T) is the time taken to complete one oscillation, it is given by:

[tex]T=\frac{1}{f}\\ \\T=\frac{1}{250} =0.004\ s[/tex]

Answer: Tl = - 13.3°C

the lowest outdoor temperature is - 13.3°C

Explanation:

Given that;

Temperature of Th = 21°C = 21 + 273 = 294 K

the rate at which heat lost is Qh = 5400 kJ/h°C

the power input to heat pump Wnet = 6 kw

The COP of a reversible heat pump depends on the temperature limits in the cycle only, and is determined by;

COPhp = Th/(Th - Tl)

COPhp = Qh/Wnet

Qh/Wnet = Th/(Th -Tl)

the amount of heat loss is expressed as

Qh = 5400/3600(294 - Tl)

the temperature of sink

( 5400/3600(294 - Tl)) / 6 = 294 / ( 294 - Tl)

now solving the equation

Tl = 259.7 - 273

Tl = - 13.3°C

so the lowest outdoor temperature is - 13.3°C

Answer:

Explanation:

In a particular application involving airflow over a heated surface, the boundary layer temperature distribution, T(y), may be approximated as:

[ T(y) - Ts / T∞ - Ts ] = 1 - e^( -Pr (U∞y / v) )

where y is the distance normal to the surface and the Prandtl number, Pr = Cpu/k = 0.7, is a dimensionless fluid property. a.) If T∞ = 380 K, Ts = 320 K, and U∞/v = 3600 m-1, what is the surface heat flux? Is this into or out of the wall? (~-5000 W/m2 , ?). b.) Plot the temperature distribution for y = 0 to y = 0.002 m. Set the axes ranges from 380 to 320 for temperature and from 0 to 0.002 m for y. Be sure to evaluate properties at the film temperature.

Answer:

"Process capability" is the correct answer.

Explanation:

Answer: y' = - x'

Explanation:

Let f(x) = 2x + y

then f'(x) = 2 + y'

Let f(y) = x - 2y

then f'(y) = x' - 2

Given:  f'(x) + f'(y) = 0

         2 + y' + x' - 2 = 0

                y' + x'= 0

                 y' = -x'

This can also be written as:       [tex]\dfrac{dy}{dx}=-\dfrac{d}{dx}[/tex]

Answer:

  resistance

Explanation:

A strain gauge changes resistance with applied strain.

Answer:

moment of inertia = 4.662 * 10^6 [tex]mm^4[/tex]

Explanation:

Given data :

Mass of machine = 400 kg = 400 * 9.81 = 3924 N

length of span = 3.2 m

E = 200 * 10^9 N/m^2

frequency = 9.3 Hz

Wm ( angular frequency ) = 2 [tex]\pi f[/tex] = 58.434 rad/secs

also Wm = [tex]\sqrt{\frac{g}{t} }[/tex]  ------- EQUATION 1

g = 9.81

deflection of simply supported beam

t = [tex]\frac{wl^3}{48EI}[/tex]

insert the value of t into equation 1

W[tex]m^2[/tex] = [tex]\frac{g*48*E*I}{WL^3}[/tex]   make I the subject of the equation

I ( Moment of inertia about the neutral axis ) = [tex]\frac{WL^3* Wn^2}{48*g*E}[/tex]

I = [tex]\frac{3924*3.2^3*58.434^2}{48*9.81*200*10^9}[/tex]  = 4.662 * 10^6 [tex]mm^4[/tex]

Answer:

pay off the parking tickets

Explanation:

In the scenario being described, the best thing to do would be to pay off the parking tickets. The parking tickets stay under your name, and if they are not paid in time can cause problems down the road. For starters, if they are not paid in time the amount will increase largely which will be harder to pay. If that increased amount is also not paid, then the government will suspend your licence indefinitely which can later lead to higher insurance rates.

Answer:

a) Mt = 0.0023229

b) = U1 = 214.07

c) = V₁  = 0.861 m³/kg

d) = Vr1 = 621.2

Explanation:

Given that

R = 0.287 KJ/kg.K, T1 = 300 K , P1 = 100 kPa , V1 = 500 cm³, r = 10 , Q = 60 kW , Speed N = 5600 RPM, Number of cylinders K = 4

specific heat at constant volume Cv = 0.7174 kJ/kg.K

Specific heat at constant pressure is 1.0045 Kj/kg.K

a)  To determine the total mass (kg) of air in the engine.

we say

P1V1 = mRT1

we the figures substitute

(100 x 10³) ( 500 x  10⁻⁶) = m ( 0.287 x  10³) ( 300 )

50 = m x 86100

m = 0.00005 / 86100 = 0.0005807 ( mass of one cylinder)

Total mass of 4 cylinder

Mt = m x k

Mt = 0.0005807 x 4

Mt = 0.0023229

b) To determine the specific internal energy (kJ/kg) at state 1

i.e at T1 = 300

we obtain the value of specific internal energy U1 at 300 K ( state 1) from the table ideal gas properties of air.

U1 = 214.07

c) To determine the specific volume (m³/kg) at state 1.

we say

V₁ = V1/m

V₁ = (500 x  10⁻⁶) / 0.0005807

V₁  = 0.861 m³/kg

d) To determine the relative specific volume at state 1.

To obtain the value of relative specific volume at 300 K ( i.e state 1) from the table ideal gas properties of air.

At T1 = 300 k

Vr1 = 621.2

Answer:

Explanation:

f = 50.0 Hz, L = 0.650 H, π = 3.14

C = 4.80 μF, R = 301 Ω resistor. V = 120volts

XL = wL = 2πfL

= 2×3.14×50* 0.650

= 204.1 Ohm

Xc= 1/wC

Xc = 1/2πfC

Xc = 1/2×3.14×50×4.80μF

= 1/0.0015072

= 663.48Ohms

1. Total impedance, Z = sqrt (R^2 + (Xc-XL)^2)= √ 301^2+ (663.48Ohms - 204.1 Ohm)^2

√ 90601 + (459.38)^2

√ 90601+211029.98

√ 301630.9844

= 549.209

Z = 549.21Ohms

2. I=V/Z = 120/ 549.21Ohms =0.218Ampere

3. P=V×I = 120* 0.218 = 26.16Watt

Note that

I rms = Vrms/Xc

= 120/663.48Ohms

= 0.18086A

4. I(max) = I(rms) × √2

= 0.18086A × 1.4142

= 0.2557

= 0.256A

5. V=I(max) * XL

= 0.256A ×204.1

=52.2496

= 52.250volts

6. V=I(max) × Xc

= 0.256A × 663.48Ohms

= 169.85volts

7. Xc=XL

1/2πfC = 2πfL

1/2πfC = 2πf× 0.650

1/2×3.14×f×4.80μF = 2×3.14×f×0.650

1/6.28×f×4.8×10^-6 = 4.082f

1/0.000030144× f = 4.082×f

1 = 0.000030144×f×4.082×f

1 = 0.000123f^2

f^2 = 1/0.000123048

f^2 = 8126.922

f =√8126.922

f = 90.14 Hz

I inferred you want literal editing of the text above.

Explanation:

Here's a correction of the sentences:

1. The author, an expert in cybersecurity will speak via Zoom on Wednesday.

In this sentence, punctuation mark ( , ) was added and the word  'this' was replaced with 'on'.

2. Williams' book contains many illustrations, which makes it easy to read.

Added punctuation and made a revision of the sentence.

3. Based on the available evidence, the university administrators have opted for a hybrid format for the fall, which begins September 20.

Mainly added punctuations to make the senstence clarer.

4. (Thesis statement) I believe Free laptops should be offered to all students who need them.

Made a few additions.

Answer:

a.  the temperature (K) at state 2 is  [tex]\mathbf{T_2 =270.76 \ K}}[/tex]

b.  the pressure (kPa) at state 2 is   [tex]\mathtt{ \mathbf{ p_2 = 81.79 \ kPa }}[/tex]

c.  the specific enthalpy (kJ/kg) at state 2 is [tex]\mathbf{h_2 = 271.84 \ kJ/kg}}[/tex]

d.  the temperature (K) at state 3 is   [tex]\mathbf{ T_3 = 532.959 \ K}[/tex]

Explanation:

From the given information:

T1 (K) = 249

P1 (kPa) = 61

V1 (m/s) = 209

rp = 10.7

rc = 1.8

The objective is  to determine the following:

a. Determine the temperature (K) at state 2.

b. Determine the pressure (kPa) at state 2.

c. Determine the specific enthalpy (kJ/kg) at state 2.

d. Determine the temperature (K) at state 3.

To start with the specific enthalpy (kJ/kg) at state 2.

By the relation of steady -flow energy balance equation for diffuser (isentropic)

[tex]h_1 + \dfrac{V_1^2}{2}=h_2+\dfrac{V^2_2}{2}[/tex]

[tex]h_1 + \dfrac{V_1^2}{2}=h_2+0[/tex]

[tex]h_2=h_1 + \dfrac{V_1^2}{2}[/tex]

For ideal gas;enthalpy is only a function of temperature, hence [tex]c_p[/tex]T = h

where;

[tex]h_1[/tex] is the specific enthalpy at inlet  = [tex]c_pT_1[/tex]

[tex]h_2[/tex] is the specific enthalpy at  outlet = [tex]c_pT_2[/tex]

[tex]c_p[/tex]  = 1.004  kJ/kg.K or 1004 J/kg.K

Given that:

[tex]T_1[/tex] (K) = 249

[tex]V_1[/tex] (m/s) = 209

∴

[tex]h_2=C_pT_1+ \dfrac{V_1^2}{2}[/tex]

[tex]h_2=1004 \times 249+ \dfrac{209^2}{2}[/tex]

[tex]h_2 = 249996+21840.5[/tex]

[tex]\mathbf{\mathtt{h_2 = 271836.5 \ J/kg}}[/tex]

[tex]\mathbf{h_2 = 271.84 \ kJ/kg}}[/tex]

Determine the temperature (K) at state 2.

SInce; [tex]\mathtt{h_2 = c_pT_2 = 271.84 \ kJ/kg}[/tex]

[tex]\mathtt{ c_pT_2 = 271.84 \ kJ/kg}[/tex]

[tex]\mathtt{T_2 = \dfrac{271.84 \ kJ/kg}{ c_p}}[/tex]

[tex]\mathtt{T_2 = \dfrac{271.84 \ kJ/kg}{1.004 \ kJ/kg.K}}[/tex]

[tex]\mathbf{T_2 =270.76 \ K}}[/tex]

Determine the pressure (kPa) at state 2.

For isentropic condition,

[tex]\mathtt{ \dfrac{T_2}{T_1}= \begin {pmatrix} \dfrac{p_2}{p_1} \end {pmatrix} ^\dfrac{k-1}{k}}[/tex]

where ;

k = specific heat ratio = 1.4

[tex]\mathtt{ \dfrac{270.76}{249}= \begin {pmatrix} \dfrac{p_2}{61} \end {pmatrix} ^\dfrac{1.4-1}{1.4}}[/tex]

[tex]\mathtt{ 1.087389558= \begin {pmatrix} \dfrac{p_2}{61} \end {pmatrix} ^\dfrac{0.4}{1.4}}[/tex]

[tex]\mathtt{ 1.087389558 \times 61 ^ {^ \dfrac{0.4}{1.4} }}=p_2} ^\dfrac{0.4}{1.4}}[/tex]

[tex]\mathtt{ 3.519487255=p_2} ^\dfrac{0.4}{1.4}}[/tex]

[tex]\mathtt{ \mathbf{ p_2 = \sqrt[0.4]{3.519487255^{1.4}} }}[/tex]

[tex]\mathtt{ \mathbf{ p_2 = 81.79 \ kPa }}[/tex]

d. Determine the temperature (K) at state 3.

For the isentropic process

[tex]\mathtt{\dfrac{T_3}{T_2} = \begin {pmatrix} \dfrac{p_3}{p_2} \end {pmatrix}^{\dfrac{k-1}{k}}}[/tex]

where;

[tex]\mathtt{\dfrac{p_3}{p_2} }[/tex] is the compressor ratio [tex]\mathtt{r_p}[/tex]

Given that ; the compressor ratio [tex]\mathtt{r_p}[/tex] = 10.7

[tex]\mathtt{\dfrac{T_3}{T_2} = \begin {pmatrix} r_p \end {pmatrix}^{\dfrac{k-1}{k}}}[/tex]

[tex]\mathtt{\dfrac{T_3}{270.76} = \begin {pmatrix} 10.7 \end {pmatrix}^{\dfrac{1.4-1}{1.4}}}[/tex]

[tex]\mathtt{\dfrac{T_3}{270.76} = \begin {pmatrix} 10.7 \end {pmatrix}^{^ \dfrac{0.4}{1.4}}}[/tex]

[tex]\mathtt{{T_3}{} =270.76 \times\begin {pmatrix} 10.7 \end {pmatrix}^{^ \dfrac{0.4}{1.4}}}[/tex]

[tex]\mathbf{ T_3 = 532.959 \ K}[/tex]

Answer:

New electric field = 18 N/C

Explanation:

Given:

Length (E1) = 2 cm

New length (E2) = 4 cm

Electric field =  36 N/C

Find:

New electric field

Computation:

New electric field = 36 [2 / 4]

New electric field = 36 [1/2]

New electric field = 18 N/C

Answer:

There are five (5) unknown support reactions in this problem.

Explanation:

A roller joint rotates and translates along the surface on which the roller rests. The resulting reaction force is always a single force that is perpendicular to, and away from, the surface. This allows the roller to move in a single plane along the surface where it rests.

A cable support provides support in one direction, parallel, and in opposite direction to the load on it. There exists a single reaction from the cable pointed upwards.

A ball-and-socket joint have  reaction forces in all 3 cardinal  directions. This allows it to move in the x-y-z plane.

The total unknown reactions on the member are five in number.

Answer:

d. all of the above

Explanation:

There are two components of acceleration for a particle moving in a circular path, radial and tangential acceleration.

The radial acceleration is given by;

[tex]a_r = \frac{V^2}{R}[/tex]

Where;

V is the velocity of the particle

R is the radius of the circular path

This radial acceleration is always directed towards the center of the path, perpendicular to the tangential acceleration and negative.

Therefore, from the given options in the question, all the options are correct.

d. all of the above

There are different type of acceleration.  The radial component of acceleration of a particle moving in a circular path is always negative, directed towards the center of the path and perpendicular to the transverse component of acceleration.

This is due to because of the centripetal force. So centripetal force is the reason for a radial acceleration.

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