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Artyom0805 [142]
3 years ago
7

When welding stick (SMAW) what is the distance between top of bare end of electrode and base metal?

Engineering
1 answer:
agasfer [191]3 years ago
5 0

Answer:

the answer is D) Arc Length

Explanation:

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Independent auto lots usually have ____ finance rates than dealerships
wel

Independent auto lots usually have <u>higher</u> finance rates than dealerships

<u>Explanation:</u>

The finance rates that are charged by the dealers are lower than the finance charges that are charged by the independent auto. In case if you are getting financed through dealerships, you can also negotiate with them to charge finance rates and lower the charges of the finance.

But this negotiation and lowering of the finance rates is not possible with the independent auto lots and thus they charge higher rates compared to the dealerships.

8 0
3 years ago
The enforcement of OSHA standards is provided by federal and state
Gnoma [55]

Answer:

Explanation:

Enforcing OSHA, Occupational Safety and Health Administration, standards is not a job for electricians, lawmakers or tax collectors. The right answer is safety inspectors.

3 0
2 years ago
The three sub regions of South America are the Andes Mountains, the Amazon Rainforest, and the Eastern Highlands. The Atacama De
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Answer:

<:

Explanation:

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2 years ago
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failure to yield the right-of-way to another vehicle or pedestrian is the primary Collision factor in about 20% of fatal and inj
laila [671]

You could just create a turn on green arrow. Have a button for pedestrians. The only way for them to turn is if they have a green arrow and the green arrow will only appear when pedestrians are stopped or finished walking and all cars are clear or fully stopped

8 0
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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