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zheka24 [161]
3 years ago
10

Establishes general guidelines concerning licensing and vehicle

Engineering
2 answers:
gavmur [86]3 years ago
5 0

Answer:

A. National Highway Safety Act

Explanation:

The National Highway Safety Act establishes general guidelines concerning licensing, vehicle registration and inspection, and traffic laws for state regulations. The act was made in 1966 to reduce the amount of death on the highway as a result of increase in deaths by 30% between 1960 and 1965

National Traffic and Motor Vehicle Safety Act regulates vehicle manufacturers  by ensuring national safety standards and issuance recalls for defective vehicles

Uniform Traffic Control Devices Act  defines shapes, colors and locations for road signs, traffic signals, and road markings

Nikitich [7]3 years ago
5 0
The answer would be A
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Answer:

Wat part 2

Explanation:

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3 years ago
Explain with schematics the operating principle of solid state lasers.
alina1380 [7]

Explanation:

A solid state laser contains a cavity like structure fitted with spherical mirrors or plane mirrors at the end filled with a rigidly bonded crystal. It uses solid as the medium. It uses glass or crystalline materials.

    It is known that active medium used for this type of laser is a solid material. This lasers are pumped optically by means of a light source which is used as a source of energy for the laser. The solid materials gets excited by absorbing energy in the form of light from the light source. Here the pumping source is light energy.  

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3 years ago
Technician a says that if a tapered roller bearing is adjusted to loose
Effectus [21]
The technician is true
5 0
3 years ago
You are working as an electrical technician. One day, out in the field, you need an inductor but cannot find one. Looking in you
telo118 [61]

Answer:

a) the inductance of the coil is 6 mH

b) the emf generated in the coil is 18 mV  

Explanation:

Given the data in the question;

N = 570 turns

diameter of tube d = 8.10 cm = 0.081 m

length of the wire-wrapped portion l =  35.0 cm = 0.35 m

a) the inductance of the coil (in mH)

inductance of solenoid

L = N²μA / l

A = πd²/4  

so

L = N²μ(πd²/4) / l

L = N²μ(πd²) / 4l

we know that μ = 4π × 10⁻⁷ TmA⁻¹

we substitute

L = [(570)² × 4π × 10⁻⁷× ( π × (0.081)² )] / 4(0.35)

L =  0.00841549 / 1.4

L = 6 × 10⁻³ H    

L = 6 × 10⁻³ × 1000 mH

L = 6 mH

Therefore, the inductance of the coil is 6 mH

b)

Emf ( ∈ ) = L di/dt

given that; di/dt = 3.00 A/sec

{∴ di = 3 - 0 = 3 and dt = 1 sec}

Emf ( ∈ ) = L di/dt

we substitute

⇒ 6 × 10⁻³ ( 3/1 )

= 18 × 10⁻³ V

= 18 × 10⁻³ × 1000

= 18 mV  

Therefore, the emf generated in the coil is 18 mV  

7 0
2 years ago
Determine the hydraulic radius for the following rectangular open channel width =23m water depth =3m
Romashka-Z-Leto [24]

Answer:

2.379m

Explanation:

The width = 23m

The depth = 3m

The radius is denoted as R

The wetted area is = A

The perimeter perimeter = P

Hydraulic radius

R = A/P

The area of a rectangular channel

= Width multiplied by Depth

A = 23x3

A = 69m²

Perimeter = (2x3)+23

P = 6+23

P= 29

Hydraulic radius R = 69/29

= 2.379m

This answers the question

Thank you!

8 0
3 years ago
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