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4 years ago

4. Which of the following drive roll designs is used

1 answer:

5 0

Knurled drive rolls are used with gas and self shielded flux cored and metal cored wires, which are softer due to the flux inside and the tubular design.

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An 1800-W toaster, a 1400-W electric frying pan, and a 75-W lamp are plugged into the same outlet in a 15-A, 120-V circuit. The

Mila [183]

Answer:

a) Current drawn by the toaster = 15A

Current drawn by the electric frying pan = 11.67A

Current drawn by the lamp = 0.625A

b) This combination will blow the 15A fuse as the total current requirement for this setup exceeds the 15A rating of the fuse.

Explanation:

a) For parallel connection, there exists, the same voltage and different currents across all the devices.

Voltage cross each of the 3 devices = outlet voltage of 120V

From their respective power rating, current drawn by each device will be calculated.

P = IV

For the toaster, P = 1800 W, V = 120V

I = 1800/120 = 15A

For the electric frying pan, P = 1400 W, V = 120 V

I = 1400/120 = 11.67 A

For the lamp, P = 75 W, V = 120V

I = 75/120 = 0.625 A

b) Total current in a parallel connection setup = Sum total of all the currents.

Total current drawn by all 3 devices = 15 + 11.67 + 0.625 = 27.295A = 27.3 A

This total current requirement surpasses the 15A current rating of the fuse, therefore, this combination will blow the fuse.

Hope this Helps!!!

6 0

4 years ago

Ninety-five percent of the acetone vapor in an 85 vol.% air stream is to be absorbed by countercurrent contact with pure water i

Sati [7]

Answer:

Explanation:

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5 0

3 years ago

Using the Rayleigh criterion, calculate the minimum feature size that can be resolved in a system with a 0.18 NA lens when g-lin

Vladimir79 [104]

Answer:

# for a g line, R = 1.211 μm

# for an I-line, R = 1.013 μm

# for a g line, R = 0.726 μm

# for an I-line, R = 0.243 μm

# for a g line, R = 0.605 μm

# for an I-line, R = 0.608 μm

Explanation:

We know that;

Rayleigh Resolution R = 0.5 × λ/NA

for a g line, λ = 436 nm

for an I-line λ = 365 nm

Now when NA = 0.18

# for a g line, λ = 436 nm

R = 0.5 × 436/0.18 = 1.211 μm

# for an I-line λ = 365 nm

R = 0.5 × 365/0.18 = 1.013 μm

when NA = 0.30

# for a g line, λ = 436 nm

R = 0.5 × 436/0.30 = 0.726 μm

# for an I-line λ = 365 nm

R = 0.5 × 365/0.30 = 0.243 μm

when NA = 0.36

# for a g line, λ = 436 nm

R = 0.5 × 436/0.36 = 0.605 μm

# for an I-line λ = 365 nm

R = 0.5 × 365/0.30 = 0.608 μm

6 0

3 years ago

Soils with low percolation rates do not need special attention during site engineering. select one: true false

saveliy_v [14]

It is accurate to say that site engineering does not require particular consideration for soils with low percolation rates.

<h3>What are percolation rates?</h3>

The rate at which water percolates through the soil is a measure of its ability to absorb and treat effluent, or wastewater that has undergone preliminary treatment in a septic tank.
Minutes per inch are used to measure percolation rate (mpi).
The process of a liquid gently moving through a filter is called percolation. This is how coffee is typically brewed.

The Latin verb percolare, which meaning "to strain through," is the source of the word "percolation." When liquid is strained through a filter, such as when making coffee, percolation occurs.

To learn more about percolation rates, refer to:

brainly.com/question/28170860

#SPJ4

7 0

2 years ago

What types of problems might an electrical engineer need to solve?

aleksandr82 [10.1K]

Answer:

sectores industriales, comerciales o públicos, o para el uso doméstico.

Explanation:

8 0

4 years ago