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Elenna [48]
3 years ago
8

What is the most likely cause a brake light staying on?

Engineering
1 answer:
Monica [59]3 years ago
8 0
The answer is either a or b
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Can some one plz give me brainlys
Brums [2.3K]

Answer:

how do u do that?

Explanation:

confusion

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2 years ago
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Which of the following types of protective equipment protects workers who are passing by from stray sparks or metal while anothe
lawyer [7]

A protective equipment which protects workers who are passing by from stray sparks or metal while another worker is welding is: E. Welding Screens.

A wielder refers to an individual who is saddled with responsibility of joining two or more metals together by wielding.

During the process of wielding, sparks and minute metallic objects are produced, which are usually hazardous to both the wielder and other workers within the vicinity.

Hence, the following protective equipment are meant to be worn or used directly by a wielder (worker) who is wielding:

  • Visors.
  • Goggles.
  • Protective clothing.
  • Dark walls.

However, a protective equipment which protects other workers who are passing by from stray sparks or metallic objects while wielder (worker) is welding is referred to as welding screens.

Find more information: brainly.com/question/15442363

4 0
3 years ago
Salvage ethnography is the effort to ensure that ethnography remains an important part of anthropology. recording of linguistic
fenix001 [56]

Answer:

                D

Explanation:

                            D

6 0
3 years ago
A car travells at 67.5 km\h in 120 km.how long will it take to reach the destination
Kruka [31]

mark me the brainiest here

average speed (in km/h) of a car stuck in traffic that drives 12 kilometers in 2 hours.

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3 years ago
A spring-mass-damper instrument is employed for acceleration measurements. The spring constant is 12000 N/m. The mass is 5 g. Th
shepuryov [24]

Answer:

a) 246.56 Hz

b) 203.313 Hz

c) Add more springs

Explanation:

Spring constant = 12000 N/m

mass = 5g = 5 * 10^-3 kg

damping ratio = 0.4

<u>a) Calculate Natural frequency </u>

Wn = √k/m = \sqrt{12000 /  5*10^{-3}  }

                   = 1549.19 rad/s  ≈ 246.56 Hz

<u>b) Bandwidth of instrument </u>

W / Wn = \sqrt{1-2(0.4)^2}

W / Wn = 0.8246

therefore Bandwidth ( W ) = Wn * 0.8246 = 246.56 * 0.8246 = 203.313 Hz

C ) To increase the bandwidth we have to add more springs

5 0
3 years ago
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