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2 years ago

Which of the following is NOT an ASE certification? Select one:

Engineering

1 answer:

stiks02 [169]2 years ago

8 0

The option that is not an ASE certification is . A/C and Refrigerants handling certification (609).

<h3>What is ASE certification?</h3>

The term ASE is known to be a body that tends to promotes excellence in regards to vehicle repair, service as well as parts distribution.

Note that in the world today more than a quarter of million of people are known to possess ASE certifications.

Since ASE Certified professionals work in in all areas of the transportation industry. one can say that The option that is not an ASE certification is. A/C and Refrigerants handling certification (609).

Learn more about ASE certification from

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What is the force in kN of work done is 1.2 ms moves through 120m

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Answer:

$\frac{1.2}{120}$

0.01

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Answer:

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Explanation:

given data

length L = 80 cm = 800 mm

width W = 30 cm

height H = 15 cm

make grove length = 80 cm

width = 8 cm

depth = 10 cm

mill toll diameter = 4 mm

axial cutting depth = 20 mm

to find out

How much material removed in 3 minutes

solution

first we find time taken for length of advance that is

time = $\frac{length}{advance}$

here advance is given as 0.001166 mts / sec

so time = $\frac{800}}{0.001166*1000}$

time = 686.106 seconds

now we find material remove rate that is

remove rate = mill toll rate × axial cutting depth × advance

remove rate = 4 × 20×0.001166 ×1000

remove rate = 93.28 mm³/s

material remove in 3 minute = 3 × 60 = 180 sec

so material remove in 3 min = 180 × 93.28

material remove in 3 min is 16790.4 mm³/s

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3 years ago

The homogeneous, reversible, exothermic, liquid phase reaction: A근R Is being carried out in a reactor system consisting of an id

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3 years ago

(a)Compute the electrical conductivity of a cylindrical silicon specimen 7.0 mm (0.28 in.) diameter and 57 mm (2.25 in.) in leng

igor_vitrenko [27]

Answer:

a) $\sigma = 12.2$ (Ω-m)^{-1}

b) Resistance = 121.4 Ω

Explanation:

given data:

diameter is 7.0 mm

length 57 mm

current I = 0.25 A

voltage v = 24 v

distance between the probes is 45 mm

electrical conductivity is given as

$\sigma = \frac{I l}{V \pi r^2}$

$\sigma = \frac{0.25 \times 45\times 10^{-3}}{24 \pi [\frac{7 \times 10^{-3}}{2}]^2}$

$\sigma = 12.2$ (Ω-m)^{-1}[/tex]

$Resistance = \frac{l}{\sigma A}$

$= \frac{l}{ \sigma \pi r^2}$

$= \frac{57 \times 10^{-3}}{12.2 \times \pi [\frac{7 \times 10^{-3}}{2}]^2}$

Resistance = 121.4 Ω

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