Ask question

Ask question

All categories

3 years ago

9

Two technicians are discussing a vehicle that will not start. Tech A states that a problem with the immobilizer system may be th

e cause. Tech B states that an open in the starter relay may be the cause. Which technician is correct?

1 answer:

Anestetic [448]3 years ago

3 0

Technician B.

Explanation:

The starter motor is what turns the engine, until enough electricity is generated with momentum to keep the engine running and cycling. If there is a problem with the immobilizer, the engine may shake or come off the mount at higher RPMs, however, will not have any problems initially starting the engine. The starter is most important for starting the engine, therefore, Technician B is correct In this case.

You might be interested in

What is the tolerance of number 4?

Kamila [148]

Answer:

Answer: ±0.02 units or 20±0.02 units or 19.98-20.02 units depending on how they prefer its written (typically the first or second one)

Explanation:

says on the sheet. Unless otherwise stated 0.XX = ±0.02 tolerance

(based on image sent in other post)

5 0

3 years ago

Two balanced Y-connected loads in parallel, one drawing 15kW at 0.6 power factor lagging and the other drawing 10kVA at 0.8 powe

NemiM [27]

Answer:

(a) attached below

(b) $pf_{C}=0.85$ $lagging$

(c) $I_{C} =32.37 A$

(d) $X_{C} =49.37$ Ω

(e) $I_{cap} =9.72 A$ and $I_{line} =27.66 A$

Explanation:

Given data:

$P_{1}=15 kW$

$S_{2} =10 kVA$

$pf_{1} =0.6$ $lagging$

$pf_{2}=0.8$ $leading$

$V=480$ $Volts$

(a) Draw the power triangle for each load and for the combined load.

$\alpha_{1}=cos^{-1} (0.6)=53.13$ °

$\alpha_{2}=cos^{-1} (0.8)=36.86$ °

$S_{1}=P_{1} /pf_{1} =15/0.6=25 kVA$

$Q_{1}=P_{1} tan(\alpha_{1} )=15*tan(53.13)=19.99$ ≅ $20kVAR$

$P_{2} =S_{2}*pf_{2} =10*0.8=8 kW$

$Q_{2} =P_{2} tan(\alpha_{2} )=8*tan(-36.86)=-5.99$ ≅ $-6 kVAR$

The negative sign means that the load 2 is providing reactive power rather than consuming

Then the combined load will be

$P_{c} =P_{1} +P_{2} =15+8=23 kW$

$Q_{c} =Q_{1} +Q_{2} =20-6=14 kVAR$

(b) Determine the power factor of the combined load and state whether lagging or leading.

$S_{c} =P_{c} +jQ_{c} =23+14j$

or in the polar form

$S_{c} =26.92$ °

$pf_{C}=cos(31.32) =0.85$ $lagging$

The relationship between Apparent power S and Current I is

$S=VI^{*}$

Since there is conjugate of current I therefore, the angle will become negative and hence power factor will be lagging.

(c) Determine the magnitude of the line current from the source.

Current of the combined load can be found by

$I_{C} =S_{C}/\sqrt{3}*V$

$I_{C} =26.92*10^3/\sqrt{3}*480=32.37 A$

(d) Δ-connected capacitors are now installed in parallel with the combined load. What value of capacitive reactance is needed in each leg of the A to make the source power factor unity?Give your answer in Ω

$Q_{C} =3*V^2/X_{C}$

$X_{C} =3*V^2/Q_{C}$

$X_{C} =3*(480)^2/14*10^3$ Ω

(e) Compute the magnitude of the current in each capacitor and the line current from the source.

Current flowing in the capacitor is

$I_{cap} =V/X_{C} =480/49.37=9.72 A$

Line current flowing from the source is

$I_{line} =P_{C} /3*V=23*10^3/3*480=27.66 A$

8 0

3 years ago

The elementary liquid-phase series reaction

liraira [26]

Answer:

Concentration of A: $\frac{C_{A} }{C_{Ao} } =e^{-k_{1}t }$

Concentration of B: $\frac{C_{B} }{C_{Ao} } =\frac{k_{1} }{k_{2}-k_{1} } (e^{-k_{1}t } -e^{-k_{2}t } )$

Concentration of C: $\frac{C_{C} }{C_{Ao} } =1+\frac{k_{1} }{k_{2}-k_{1} } e^{-k_{2}t } -\frac{k_{2} }{k_{2}-k_{1} } e^{-k_{1} t}$

the image shows the graphs of the three concentrations

Explanation:

We have the reaction:

A ------->k1--------->B------------->k2--------->C

Each reaction:

$r_{A} =-k_{1} C_{A} \\r_{B} =k_{1} C_{A} -k_{2} C_{B} \\r_{C} =k_{2} C_{C}$

Where Cn is the concentration of each specie (A,B,C)

The mass balance for A:

$-\frac{dC_{A} }{dt} =-r_{A} \\-\frac{dC_{A} }{dt}=k_{1} C_{A} \\-\int\limits^y_x {\frac{dC_{A} }{dt} } \,=k_{1} t\\\frac{C_{A} }{C_{Ao} } =e^{-k_{1}t }$

Where x=CAo and y=CA

The mass balance for B:

$-\frac{dC_{B} }{dt} =-r_{B} \\-\frac{dC_{B} }{dt}=k_{2} C_{B} -k_{1} C_{A} \\\frac{dC_{B} }{dt}+k_{2} C_{B}=k_{1} C_{A}\\\frac{C_{B} }{C_{Ao} } =\frac{k_{1} }{k_{2}-k_{1} } (e^{-k_{1}t }-ex^{-k_{2}t } )$

The mass balance for C:

$\frac{C_{C} }{C_{Ao} } =1-\frac{C_{A} }{C_{Ao} } -\frac{C_{B} }{C_{Ao} } \\\frac{C_{C} }{C_{Ao} }=1+\frac{k_{1} }{k_{2}-k_{1} } e^{-k_{2} t}-\frac{k_{2} }{k_{2}-k_{1} } e^{-k_{1}t }$

The maximum concentration of C is:

$C_{Cmax} =C_{Ao} (\frac{k_{2} }{k_{1} } )^{\frac{k_{2} }{k_{2}-k_{1} }} =1.6(\frac{0.01}{0.4} )^{\frac{0.01}{0.01-0.4} } =1.76mol/dm^{3}$

and the maximum time is:

$t_{max} =\frac{ln\frac{k_{2} }{k_{1} } }{k_{2}-k_{1} } =\frac{ln\frac{0.01}{0.4} }{0.01-0.4} =9.4 h$

6 0

3 years ago

Sarah is developing a Risk Assessment for her organization. She is asking each department head how long can they be without thei

Natali [406]

Answer:

Sarah is asking each department head how long they can be without their primary system. Sarah is trying to determine the Recovery Time Objective (RTO) as this is the duration of time within which the primary system must be restored after the disruption.

Recovery Point Objective is basically to determine the age of restoration or recovery point.

Business recovery and technical recovery requirements are to assess the requirements to recover by Business or technically.

Hence, Recovery Time Objective (RTO) is the correct answer.

8 0

3 years ago

2. Recycled tires are frequently turned into?

Leya [2.2K]

Answer:vmmjhnv

Explanation:

b m bn

6 0

3 years ago