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marin [14]
3 years ago
8

Technician A says that a defective crankshaft position sensor can cause a no spark condition technician B says that a faulty ign

ition control module can cause a no spark condition which technician is correct
Engineering
1 answer:
Aliun [14]3 years ago
5 0
Both tech distributor ignition
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(a) Determine the dose (in mg/kg-day) for a bioaccumulative chemical with BCF = 103 that is found in water at a concentration of
solmaris [256]

Answer:

0.064 mg/kg/day

6.25% from water, 93.75% from fish

Explanation:

Density of water is 1 kg/L, so the concentration of the chemical in the water is 0.1 mg/kg.

The BCF = 10³, so the concentration of the chemical in the fish is:

10³ = x / (0.1 mg/kg)

x = 100 mg/kg

For 2 L of water and 30 g of fish:

2 kg × 0.1 mg/kg = 0.2 mg

0.030 kg × 100 mg/kg = 3 mg

The total daily intake is 3.2 mg.  Divided by the woman's mass of 50 kg, the dosage is:

(3.2 mg/day) / (50 kg) = 0.064 mg/kg/day

b) The percent from the water is:

0.2 mg / 3.2 mg = 6.25%

And the percent from the fish is:

3 mg / 3.2 mg = 93.75%

3 0
2 years ago
The bulk density of a compacted soil specimen (Gs = 2.70) and its water content are 2060 kg/m^3 and 15.3%, respectively. If the
yaroslaw [1]

Answer:

the saturated density should be

Explanation:

4 0
3 years ago
simple Brayton cycle using air as the working fluid has a pressure ratio of 10. The minimum and maximum temperatures in the cycl
Irina18 [472]

Answer:

a) 764.45K

b) 210.48 kJ/kg

c) 30.14%

Explanation:

pressure ratio = 10

minimum temperature = 295 k

maximum temperature = 1240 k

isentropic efficiency for compressor = 83%

Isentropic efficiency for turbine = 87%

<u>a) Air temperature at turbine exit </u>

we can achieve this by interpolating for enthalpy

h4 = 783.05 kJ/kg ( calculated in the background ) at state 4 using Table A-17  for  Ideal gas properties of air

T4 ( temperature at Turbine exit ) = 760 + ( 780 - 760 ) (\frac{783.05-778.18}{800.13-778.18} ) = 764.45K

<u>b) The net work output </u>

first we determine the actual work input to compressor

Wc = h2 - h1  ( calculated values )

     = 626.57 - 295.17 =  331.4 kJ/kg

next determine the actual work done by Turbine

Wt = h3 - h4  ( calculated values )

     = 1324.93 - 783.05 = 541.88 kJ/kg

finally determine the network output of the cycle

Wnet = Wt - Wc

         = 541.88 - 331.4  = 210.48 kJ/kg

<u>c) determine thermal efficiency </u>

лth = Wnet / qin  ------ ( 1 )

where ; qin = h3 - h2

<em>equation 1 becomes </em>

лth = Wnet / ( h3 - h2 )

      = 210.48 / ( 1324.93 - 626.57 )

      = 0.3014  =  30.14%

6 0
2 years ago
Select the correct answer.
olya-2409 [2.1K]

Answer:

screw is the answer of the question

6 0
3 years ago
Read 2 more answers
Two Technicians are discussing ShopKey Pro. Technician
Leno4ka [110]

Answer:

Technician B

Explanation:

i took he test already

6 0
3 years ago
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