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Jet001 [13]
3 years ago
13

Technician A says that a voltage drop of 0.8 volts on the starter ground circuit is within specifications. Technician B says tha

t high starter draw current could be caused by a spun main bearing in the engine. Who is correct?
Engineering
1 answer:
Romashka-Z-Leto [24]3 years ago
8 0

Answer:

Technician A is wrong

Technician B is right

Explanation:

voltage drop of 0.8 volts on the starter ground circuit is not within specifications. Voltage drop should be within the range of 0.2 V to 0.6 V but not more than that.

A spun bearing can seize itself around the crankshaft journal causing it not to move. As the car ignition system is turned on, the stater may draw high current in order to counter this seizure.

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In an aircraft jet engine at takeoff, the combustion products expand adiabatically in the exhaust nozzle. At entrance to the noz
tensa zangetsu [6.8K]

At entrance to the nozzle, the pressure is 0.180 MPa and the temperature is 1200 K. The kinetic energy of the gas entering the nozzle is very much smaller than ... The specific heat of the exhaust gas varies with temperature approximately as follows: ... Problem 4P: In an aircraft jet engine at takeoff, the combustion product.

5 0
3 years ago
Match each titration term with its definition.
Illusion [34]

Answer:

1) titration

2) titrand

3) equivalence point

4) titrant

5) Burette

6) Indicator

Explanation:

The process in which a known volume of a standard solution is added to another solution so that the standard solution can react with the solution of unknown concentration such that its concentration is determined  can be referred to as titration.

The solution which is added to another solution  is called the titrant.  The titrand is the solution of unknown concentration

A burette is a glassware used to slowly add a known volume of the titrant to the titrand.

The indicator used signals the point when the reaction is complete by a color change. At this point, a stoichiometric amount of titrant has been added to the titrand. This is also referred to as the equivalence point.

3 0
3 years ago
A three-point bending test was performed on an aluminum oxide specimen having a circular cross section of radius 5.0 mm (0.20 in
Sonja [21]

The load is 17156 N.

<u>Explanation:</u>

First compute the flexural strength from:  

σ = FL / πR^{3}

   = 3000 \times (40 \times 10^-3) / π (5 \times 10^-3)^3

σ = 305 \times 10^6 N / m^2.

We can now determine the load using:

F = 2σd^3 / 3L

  = 2(305 \times 10^6) (15 \times 10^-3)^3 / 3(40 \times 10^-3)

F = 17156 N.  

7 0
3 years ago
21. How long can food that requires time-temperature control be left in the danger zone?
dimulka [17.4K]

Answer: A maximum of 1 hour

Explanation:

Read your lesson buddy!!

8 0
3 years ago
Read 2 more answers
You are working in a lab where RC circuits are used to delay the initiation of a process. One particular experiment involves an
Ymorist [56]

Answer:

t'_{1\2} = 6.6 sec

Explanation:

the half life of the given circuit is given by

t_{1\2} =\tau ln2

where [/tex]\tau = RC[/tex]

t_{1\2} = RCln2

Given t_{1\2} = 3 sec

resistance in the circuit is 40 ohm and to extend the half cycle we added new resister of 48 ohm. the net resitance is 40+48 = 88 ohms

now the new half life is

t'_{1\2} =R'Cln2

Divide equation 2 by 1

\frac{t'_{1\2}}{t_{1\2}} = \frac{R'Cln2}{RCln2} = \frac{R'}{R}

t'_{1\2} = t'_{1\2}\frac{R'}{R}

putting all value we get new half life

t'_{1\2} = 3 * \frac{88}{40}  = 6.6 sec

t'_{1\2} = 6.6 sec

7 0
3 years ago
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