can you translate this me gusta el queso :sorry i would have put 300 points buut i used them all for my last question

For a LED diode that has a= 632 nm, then the A1 is equal to:

alexgriva [62]

Answer:

1.693242

Explanation:

The colors in the Light emitting diodes have been identified by wavelength which is measured in nano-meters. Wavelength is a function of LED chip material. The LED diode which has a = 632 then A1 will be 1.63242, this is calculated by 1 / 632. Wavelength are important for human eye sensitivity. The colors emitted from the LED will depend on the semiconductor material.

5 0

2 years ago

A 50 mm diameter shaft is subjected to a static axial load of 160 kN. If the yield stress of the material is 350 MPa, the ultima

zvonat [6]

In order to develop this problem it is necessary to take into account the concepts related to fatigue and compression effort and Goodman equation, i.e, an equation that can be used to quantify the interaction of mean and alternating stresses on the fatigue life of a materia.

With the given data we can proceed to calculate the compression stress:

$\sigma_c = \frac{P}{A}$

$\sigma_c = \frac{160*10^3}{\pi/4*0.05^2}$

$\sigma_c = 81.5MPa$

Through Goodman's equations the combined effort by fatigue and compression is expressed as:

$\frac{\sigma_a}{S_e}+\frac{\sigma_c}{\sigma_u}=\frac{1}{Fs}$

Where,

$\sigma_a=$ Fatigue limit for comined alternating and mean stress

$S_e =$ Fatigue Limit

$\sigma_c=$ Mean stress (due to static load)

$\sigma_u =$ Ultimate tensile stress

$Fs =$ Security Factor

We can replace the values and assume a security factor of 1, then

$\frac{\sigma_a}{320}+\frac{81.5}{400}=\frac{1}{1}$

Re-arrenge for $\sigma_a$

$\sigma_a = 254.8Mpa$

We know that the stress is representing as,

$\sigma_a = \frac{M_c}{I}$

Then,

Where $M_c$ =Max Moment

I= Intertia

The inertia for this object is

$I=\frac{\pi d^4}{64}$

Then replacing and re-arrenge for $M_c$

$M_c = \frac{\sigma_a*\pi*d^3}{32}$

$M_c = \frac{260.9*10^6*\pi*0.05^3}{32}$

$M_c = 3201.7N.m$

Thereforethe moment that can be applied to this shaft so that fatigue does not occur is 3.2kNm

5 0

3 years ago

1. Fatigue equations are based solely on theoretical assumptions. Experimental data is only used to verify the theory. a. True.b

Rainbow [258]

Answer:

1. b. False

2. b. False

3. b. False

4. b. False

5. a. True

6. a. True

7. b. False

8. b. False

9. a. True

Explanation:

1. The fatigue properties of a material are determined by series of test.

2. For most steels there is a level of fatigue limit below which a component will survive an infinite number of cycles, for aluminum and titanium a fatigue limit can not be defined, as failure will eventually occur after enough experienced cycles.

3. Although there is a cyclic stress, there are also stresses complex circumstances involving tensile to compresive and constant stress, where the solution is given into the mean stress and the stress amplitude or stress range, which is double the stress amplitude.

4. Low‐cycle fatigue is defined as few thousand cycles and high cycle fatigue is around more than 10,000 cycles.

5. The number of cycles for failure on brittle materials are less and determined compared with the ductile materials.

6. The bending fatigue could be handled with specific load requirements for uniform bending or axial fatigue of the same section size where the material near the surface is subjected to the maximum stress, as in torsional fatigue, which can be performed on axial-type specially designed machines also, using the proper fixtures if the maximum twist required is small, in which linear motion is changed to rotational motion.

7. A SN-Curve for a given material, is a plot displayed on logarithmic scales of the magnitude of an alternating stress in relation to the number of cycles to failure

8. The strain life method measures the strain resistance of local stresses and strains around stress concentration that controls the fatigue life of the material. It is more accurate than determining fatigue performance as the stress-life method is for long life millions of cycles in elastic stresses, but an it gets an effective stress concentration in fatigue loading.

9. Linear Elastic Fracture Mechanics (LEFM) states that the material is isotropic and linear elastic so, when the stresses near the crack surpasses the material fracture toughness, the crack grows.

7 0

3 years ago

A storage tank, used in a fermentation process, is to be rotationally molded from polyethylene plastic. This tank will have a co

NNADVOKAT [17]

Answer:

The volume up to cylindrical portion is approx 32355 liters.

Explanation:

The tank is shown in the attached figure below

The volume of the whole tank is is sum of the following volumes

1) Hemisphere top

Volume of hemispherical top of radius 'r' is

$V_{hem}=\frac{2}{3}\pi r^3$

2) Cylindrical Middle section

Volume of cylindrical middle portion of radius 'r' and height 'h'

$V_{cyl}=\pi r^2\cdot h$

3) Conical bottom

Volume of conical bottom of radius'r' and angle $\theta$ is

$V_{cone}=\frac{1}{3}\pi r^3\times \frac{1}{tan(\frac{\theta }{2})}$

Applying the given values we obtain the volume of the container up to cylinder is

$V=\pi 1.5^2\times 4.0+\frac{1}{3}\times \frac{\pi 1.5^{3}}{tan30}=32.355m^{3}$

Hence the capacity in liters is $V=32.355\times 1000=32355Liters$

3 0

3 years ago

What were some challenges engineers faced in designing aqueducts.

Harman [31]

Answer:

Valleys and low-lying areas, hills and mountains, were some of the challenges faced by Roman engineers who built Aqueducts. The first aqueduct was built in Rome around 312 BC. By the 3rd century AD, it became common.

8 0

2 years ago