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3 years ago

How is engine power expressed?

2 answers:

6 0

Engine power is the power that an engine can put out. It can be expressed in power units, most commonly kilowatt, pferdestärke (metric horsepower), or horsepower.

mars1129 [50]3 years ago

5 0

Answer: Engine power is the power that an engine can put out. It can be expressed in power units, most commonly kilowatt, pferdestärke (metric horsepower), or horsepower.

Explanation: (I hope this helped!! ^^)

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Do the coil resistances have any effect on the plots?

PolarNik [594]

Because of the skin depth effect, the current at high frequency tends to flow at very low depth from radius. Then at high frequency the effective cross section of the wire is narrower than at DC.

Fro example skin depth at 100 kHz is 0.206 mm (0.008”), a wire more thicker than AWG26 could be a waste of copper, better use a bunch of thin wire (Litz wire) to rise the Q factor.

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3 years ago

A gas metal arc welder is also known as a _____ welder.<br> A) TIGB) GTAWC) GMAWD) Resistance spot

nignag [31]

Answer:

GMAW

Explanation:

It's literally the initials of that type of welding

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3 years ago

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kakasveta [241]

Answer:

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Explanation:

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3 years ago

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U 4. Find 2 bridges in the US and answer the following:

Zarrin [17]

Answer:

Im guessing this is for CEA for PLTW, if so look up the exact assignment number and look at online examples of the exact same assignment.

Explanation:

6 0

3 years ago

(a) The lower yield point for an iron that has an average grain diameter of 1 x 10-2 mm is 230 MPa. At a grain diameter of 6 x 1

olya-2409 [2.1K]

Answer:

The answer is " $4.35 \times 10^{-3}\ mm$ and 157.5 MPa".

Explanation:

In point A:

The strength of its products with both the grain dimension is linked to this problem. This formula also for grain diameter of 310 MPA is represented as its low yield point

$y = yo + \frac{k}{\sqrt{x}}$

Here y is MPa is low yield point, x is mm grain size, and k becomes proportionality constant.

Replacing the equation for each condition:

$y = y_o + \frac{k}{\sqrt{(1 \times 10^{-2})}}\\\\\ \ \ \ \ \ \ 230 = yo + 10k\\\\ y = yo + \frac{k}{\sqrt{(6\times 10^{-3})}}\\\\275 = yo + 12.90k$

People can get yo = 275 MPa with both equations and k= 15.5 Mpa $mm^{\frac{1}{2}}$ .

To substitute the answer,

$310 = 275 + \frac{(15.5)}{\sqrt{x}}\\\\x = 0.00435 \ mm = 4.35 \times 10^{-3}\ mm$

In point b:

The equation is $\sigma y = \sigma 0 + k y d^{\frac{1}{2}}$

equation is:

$75 = \sigma o+4 ky \\\\175 = \sigma o+12 ky\\\\ky = 12.5 MPa (mm)^{\frac{1}{2}} \\\\ \sigma 0 = 25 MPa\\\\d= 8.9 \times 10^{-3}\\\\d^{- \frac{1}{2}} =10.6 mm^{-\frac{1}{2}}\\$

by putting the above value in the formula we get the $\sigma y$ value that is= 157.5 MPa

5 0

3 years ago