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Blizzard [7]
3 years ago
8

Technician A says independent shops are not affiliated with vehicle manufacturers, but it is easy for technicians who work in th

ese shops to get manufacturer training on new technologies. Technician B says independent shops are not affiliated with vehicle manufacturers, making it harder for independent technicians to access training on new vehicle technology. Who is correct? Technician A Technician A Both A and B Both A and B Neither A nor B Neither A nor B Technician B Technician B
Engineering
1 answer:
KatRina [158]3 years ago
8 0

Answer:

b

Explanation:

i did it yeater dayajsbs

You might be interested in
Which crystal system(s) listed below has (have) the following relationship for the unit cell edge lengths?
lianna [129]

Answer:

Both B and G ( Hexagonal and Tetragonal )

Explanation:

The crystals system listed below has the following relationship for the unit cell edge lengths; a = b ≠ c ( hexagonal and Tetragonal )

hexagonal ; represents  a crystal system  which has three equal axes that have an angle of 60⁰ between them while Tetragonal denotes crystals that have  three axes which have only two of its axes equal in length.

5 0
3 years ago
For a fluid with a Prandtl Number of 1000.0, the hydrodynamic layer is thinner than the thermal boundary layers. a) True b) Fals
kvv77 [185]

Answer:

(b)False

Explanation:

Given:

 Prandtl number(Pr) =1000.

We know that   Pr=\dfrac{\nu }{\alpha }

  Where \nu is the molecular diffusivity of momentum

             \alpha is the molecular diffusivity of heat.

 Prandtl number(Pr) can also be defined as

    Pr=\left (\dfrac{\delta }{\delta _t}\right )^3

Where \delta is the hydrodynamic boundary layer thickness and \delta_t is the thermal boundary layer thickness.

So if Pr>1 then hydrodynamic boundary layer thickness will be greater than thermal boundary layer thickness.

In given question Pr>1 so  hydrodynamic boundary layer thickness will be greater than thermal boundary layer thickness.

So hydrodynamic layer will be thicker than the thermal boundary layer.

8 0
3 years ago
A small metal particle passes downward through a fluid medium while being subjected to the attraction of a magnetic field such t
bekas [8.4K]

Answer:

a)Δs = 834 mm

b)V=1122 mm/s

a=450\ mm/s^2

Explanation:

Given that

s = 15t^3 - 3t\ mm

a)

When t= 2 s

s = 15t^3 - 3t\ mm

s = 15\times 2^3 - 3\times 2\ mm

s= 114 mm

At t= 4 s

s = 15t^3 - 3t\ mm

s = 15\times 4^3- 3\times 4\ mm

s= 948 mm

So the displacement between 2 s to 4 s

Δs = 948 - 114 mm

Δs = 834 mm

b)

We know that velocity V

V=\dfrac{ds}{dt}

\dfrac{ds}{dt}=45t^2-3

At t=  5 s

V=45t^2-3

V=45\times 5^2-3

V=1122 mm/s

We know that acceleration a

a=\dfrac{d^2s}{dt^2}

\dfrac{d^2s}{dt^2}=90t

a= 90 t

a = 90 x 5

a=450\ mm/s^2

4 0
3 years ago
Suppose you have two boxes in front of you. One box contains a Thevenin Equivalent (voltage source in series with a resistor) an
fomenos

Answer:

1. Measure the temperature of the boxes and leave them unconnected.

2. Norton reduces his circuit down to a single resistance in parallel with a constant current source. A real-life Norton equivalent circuit would be continuously wasting power (as heat) as the current source dumps energy into the resistor, even when externally unconnected, while a Thevenin equivalent circuit would sit there doing nothing.

3. The Norton equivalent box would get warm and eventually run out of power. The Thevenin equivalent box would stay at ambient temperature.

8 0
3 years ago
Tubing cutters for copper tube are only available in one stlye true or false
Ratling [72]

Answer:

please how are you death for me account

3 0
3 years ago
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