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lawyer [7]
3 years ago
15

Is EPA is the organization that was formed to ensure safe and health working conditions for workers by setting and enforcing sta

ndards
Engineering
1 answer:
mr_godi [17]3 years ago
3 0
Let me search this up
You might be interested in
How could the location of tests affect the performance of a catapult ?
Luba_88 [7]

Answer:

It could affect how far the projectile travels

Explanation:

Facing Uphill: Moves less far

Downhill: Moves further

3 0
3 years ago
Read 2 more answers
A heat exchanger takes compressed liquid water and converts it into steam with a temperature and pressure of 20 Mpa and 480 C; r
belka [17]

Answer:

See explaination

Explanation:

Lets first consider the term Isentropic efficiency. The isentropic efficiency of a compressor or pump is defined as the ratio of the work input to an isentropic process, to the work input to the actual process between the same inlet and exit pressures. IN practice, compressors are intentionally cooled to minimize the work input.

Please kindly check attachment for the step by step solution of the given problem.

6 0
3 years ago
The electron concentration in silicon at T = 300 K is given by
puteri [66]

Answer:

E=1.44*10^-7-2.6exp(\frac{-x}{18} )v/m

Explanation:

From the question we are told that:

Temperature of silicon T=300k

Electron concentration n(x)=10^{16}\exp (\frac{-x}{18})

                                        \frac{dn}{dx}=(10^{16} *(\frac{-1}{16})\exp\frac{-x}{16})

Electron diffusion coefficient is Dn = 25cm^2/s \approx 2.5*10^{-3}

Electron mobility is \mu n = 960 cm^2/V-s \approx0.096m/V

Electron current density Jn = -40 A/cm^2 \approx -40*10^{4}A/m^2

Generally the equation for the semiconductor is mathematically given by

Jn=qb_n\frac{dn}{dx}+nq \mu E

Therefore

-40*10^{4}=1.6*10^{-19} *(2.5*10^{-3})*(10^{16} *(\frac{-1}{16})\exp\frac{-x}{16})+(10^{16}\exp (\frac{-x}{18}))*1.6*10^{-19}*0.096* E

E=\frac{-2.5*10^-^7 exp(\frac{-x}{18})+40*10^{4}}{1.536*10^-4exp(\frac{-x}{18} )}

E=1.44*10^-7-2.6exp(\frac{-x}{18} )v/m

7 0
3 years ago
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
saveliy_v [14]

Complete Question

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 this material elongate when a true stress of 411 MPa (59610 psi) is applied if the original length is 470 mm (18.50 in.)?Assume a value of 0.22 for the strain-hardening exponent, n.

Answer:

The elongation is =21.29mm

Explanation:

In order to gain a good understanding of this solution let define some terms

True Stress

       A true stress can be defined as the quotient obtained when instantaneous applied load is divided by instantaneous cross-sectional area of a material it can be denoted as \sigma_T.

True Strain

     A true strain can be defined as the value obtained when the natural logarithm quotient of instantaneous gauge length divided by original gauge length of a material is being bend out of shape by a uni-axial force. it can be denoted as \epsilon_T.

The mathematical relation between stress to strain on the plastic region of deformation is

              \sigma _T =K\epsilon^n_T

Where K is a constant

          n is known as the strain hardening exponent

           This constant K can be obtained as follows

                        K = \frac{\sigma_T}{(\epsilon_T)^n}

No substituting  345MPa \ for  \ \sigma_T, \ 0.02 \ for \ \epsilon_T , \ and  \ 0.22 \ for  \ n from the question we have

                     K = \frac{345}{(0.02)^{0.22}}

                          = 815.82MPa

Making \epsilon_T the subject from the equation above

              \epsilon_T = (\frac{\sigma_T}{K} )^{\frac{1}{n} }

Substituting \ 411MPa \ for \ \sigma_T \ 815.82MPa \ for \ K  \ and  \  0.22 \ for \ n

       \epsilon_T = (\frac{411MPa}{815.82MPa} )^{\frac{1}{0.22} }

            =0.0443

       

From the definition we mentioned instantaneous length and this can be  obtained mathematically as follows

           l_i = l_o e^{\epsilon_T}

Where

       l_i is the instantaneous length

      l_o is the original length

Substituting  \ 470mm \ for \ l_o \ and \ 0.0443 \ for  \ \epsilon_T

             l_i = 470 * e^{0.0443}

                =491.28mm

We can also obtain the elongated length mathematically as follows

            Elongated \ Length =l_i - l_o

Substituting \ 470mm \ for l_o and \ 491.28 \ for \ l_i

          Elongated \ Length = 491.28 - 470

                                       =21.29mm

4 0
4 years ago
A study of online dating found that when including emoticons in their profiles, response rates for female users _______ by _____
Dvinal [7]

Answer:

Increased, 5%

Explanation:

Recent studies conducted on online dating sites established that the response of female users increased by 5% when emotions are in their profiles even as for male users' response also increased by 8%. Another study also revealed that those who have never used online dating sites and/or mobile dating apps believe that people who use dating apps are desperate.

8 0
3 years ago
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