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blagie [28]
3 years ago
13

The complex power of a load is 10-10j VA. What component should be added in parallel with the load so that the new load has a un

ity power factor
Engineering
1 answer:
Korolek [52]3 years ago
7 0
Need points bdjdjdhdhd
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For some metal alloy, a true stress of 345 MPa (50040 psi) produces a plastic true strain of 0.02. How much will a specimen of t
Strike441 [17]

Answer:

the elongation of the metal alloy is 21.998 mm

Explanation:

Given the data in the question;

K = σT/ (εT)ⁿ

given that metal alloy true stress σT = 345 Mpa, plastic true strain εT = 0.02,

strain-hardening exponent n = 0.22

we substitute

K = 345 / 0.02^{0.22

K = 815.8165 Mpa

next, we determine the true strain

(εT) = (σT/ K)^1/n

given that σT = 412 MPa

we substitute

(εT) = (412 / 815.8165 )^(1/0.22)

(εT) = 0.04481 mm

Now, we calculate the instantaneous length

l_i = l_0e^{ET

given that l_0 = 480 mm

we substitute

l_i =480mm × e^{0.04481

l_i =  501.998 mm

Now we find the elongation;

Elongation = l_i - l_0

we substitute

Elongation = 501.998 mm - 480 mm

Elongation = 21.998 mm

Therefore, the elongation of the metal alloy is 21.998 mm

6 0
3 years ago
I need help on what’s and input and output to make the flowchart
e-lub [12.9K]

Input: what is put in, taken in, or operated on by any process or system.

Output: the amount of something produced by a person, machine, or industry.

3 0
3 years ago
Omplete the following program: [0.5 X 4 = 2]
borishaifa [10]
I’m confused on this one
4 0
4 years ago
Shear plane angle and shear strain: In an orthogonal cutting operation, the tool has a rake angle = 16°. The chip thickness befo
Oduvanchick [21]

Answer:

shear plane angle Ф = 26.28°

shear strain 2.20

Explanation:

given data

angle = 16°

chip thickness t1 = 0.32 mm

cut yields chip thickness t2 = 0.72 mm

solution

we get here first chip thickness ratio that is

chip thickness ratio = \frac{t1}{t2}    ................. 1

put here value

chip thickness ratio  = \frac{0.32}{0.72}  

chip thickness ratio r = 0.45

so here shear angle will be Ф

tan Ф = \frac{r*cos\alpha }{1-rsin\alpha}   ............2

tan Ф = \frac{0.45*cos16 }{1-rsin16}  

tan Ф = 0.4938

Ф = 26.28°

and

now we get shear strain that is

shear strain r = cot Ф + tan (Ф - α )   ................3

shear strain r  = cot(26.28) + tan (26.28 - 16 )

shear strain r = 2.20

6 0
4 years ago
During experiments on laser etching of microchips for IBM, a nuclear engineer places a discshaped sample 5 mm in diameter in a v
Andru [333]

Answer:

Steady state temperature is approximately 801K

Explanation:

Detailed explanation and calculation is shown in the images below.

7 0
4 years ago
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