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3 years ago

Technician A says that 5W-30 would be better to use than 20W-50 in most vehicles in

Engineering

2 answers:

shtirl [24]3 years ago

5 0

Technician is correct sorry if im wronghg

Ulleksa [173]3 years ago

5 0

And I don’t want to make it happen anymore I’m sorry for making

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As a human resources professional for a large company, Tommy is involved with the ___________ process. His goal is to find the r

zaharov [31]

Answer:

Recruitment

Explanation:

The process of recruitment involves several activities whose main objective is to get the right people with relevant qualifications at the appropriate time to meet the needs of an organisation whenever there’s need. The process involves screening a pool of candidates through several steps to obtain the best fit for the job.

4 0

3 years ago

A motor vehicle engine operating with a diesel engine takes in atmospherics air at a temperature of 30°C and pressure of 1 bar.

nydimaria [60]

Answer:

η=0.568

Explanation:

At inlet condition

temperature = 30 C and pressure P=1 bar

The maximum temperature = 13456 C

Compression ratio r= 12

We know that for process 1-2

$\dfrac{T_2}{T_1}=r^{\gamma -1}$

$\dfrac{T_2}{303}=12^{1.4 -1}$

$T_2=818.68 K$

Now for process 2-3

$\dfrac{T_3}{T_2}=\dfrac{V_3}{V_2}$

$\dfrac{273+1345}{818.86}=\dfrac{V_3}{V_2}$

$\dfrac{V_3}{V_2}=1.97$

So the cut off ratio ρ=1.97

Efficiency of diesel engine

$\eta =1-\dfrac{\rho ^{\gamma}-1}{r^{\gamma -1}\gamma \left (\rho -1\right )}$

Now put the values

$\eta =1-\dfrac{1.97 ^{1.4}-1}{12^{1.4-1}\times 1.4\times \left (1.97 -1\right )}$

⇒η=0.568

So the efficiency is 56.8%.

4 0

3 years ago

E total kinetic energy of a 2500 lbm car when it is moving at 80 mph (in BTU)?

Galina-37 [17]

Answer:

KE= 687.21 BTU

Explanation:

Given that

Mass of car = 2500 lbm

We know that 1 lb=0.45 kg

So the mass of car m =1133.98 kg

Velocity of car= 80 mph

We know that 1 mph =0.44 m/s

So velocity of car = 35.76 m/s

As we know that kinetic energy (KE) is given as follows

$KE=\dfrac{1}{2}mv^2$

Now by putting the values

$KE=\dfrac{1}{2}\times 1133.98\times 35.76^2$

KE=725.05 KJ

We know that 1 KJ = 0.94 BTU

So KE= 687.21 BTU

3 0

4 years ago

g Two Standard 1/2" B18.8.2 dowel pins are to be installed in part B with an LN1 fit. The thickness of plate A is .750 +/- .005"

allochka39001 [22]

Answer:

nmuda mudaf A certain vehicle loses 3.5% of its value each year. If the vehicle has an initial value of $11,168, construct a model that represents the value of the vehicle after a certain number of years. Use your model to compute the value of the vehicle at the end of 6 years.

Explanation:

5 0

3 years ago

Two cars are traveling on level terrain at 55 mi/h on a road with a coefficient of adhesion of 0.75. The driver of car 1 has a 2

Vlad1618 [11]

Answer:

0.981

Explanation:

velocity of cars ( v1 , v2 ) = 55 mi/h

coefficient of adhesion ( u ) = 0.75

Reaction time of driver of car 1 = 2.3 -s

Reaction time of driver of car 2 = 1.9 -s

breaking efficiency of car 2 ( n2 ) = 0.80

Determine the braking efficiency of car 1

First determine the distance travelled during reaction time ( dr )

dr = v * tr ------- ( 1 )

tr ( reaction time )

v = velocity

note : 1 mile = 1609 m , I hour = 60 * 60 secs

back to equation 1

for car 1

dr1 =( 55 * 2.3 * 1609 ) / ( 60 * 60 )

= 56.53 m

for car 2

dr2 = ( 55 * 1.9 * 1609 ) / ( 60 * 60 )

= 46.70 m

next we calculate the stopping distances ( d ) using the relation below

ds = d + dr

d = distance travelled during break

dr = distance travelled during reaction time

where : d = $\frac{v^2intial}{2ugn}$

for car 1

d1 = $\frac{(55)^2}{2*0.75*9.81* n1} * (\frac{1609}{3600} )^2$

∴ d1 = $\frac{41.10}{n1}$

for car 2

d2 = $\frac{(55)^2}{2*0.75*9.8*0.8} * (\frac{1609}{3600} )^2$

∴ d2 = 51.38

since the stopping distance for both cars are the same

d1 + dr1 = d2 + dr2

( 41.10 /n1 ) + 56.53 = 51.38 + 46.70

solve for n1

hence n1 = 0.981 ( braking efficiency of car 1 )

4 0

3 years ago