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dedylja [7]
3 years ago
13

Please help .. .... . ........​

Engineering
1 answer:
Helen [10]3 years ago
7 0

Answer:

how am i supposed to answer when there is no question?

Explanation:

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For some metal alloy, the following engineering stresses produce the corresponding engineering plastic strains prior to necking.
kirza4 [7]

Answer:

203.0160

Explanation:

Because you add then subtract then multiply buy 7 the subtract then divide then you add that to the other numbers you got than boom

7 0
2 years ago
An Ideal gas is being heated in a circular duct as while flowing over an electric heater of 130 kW. The diameter of duct is 500
Assoli18 [71]

Answer: The exit temperature of the gas in deg C is 32^{o}C.

Explanation:

The given data is as follows.

C_{p} = 1000 J/kg K,   R = 500 J/kg K = 0.5 kJ/kg K (as 1 kJ = 1000 J)

P_{1} = 100 kPa,     V_{1} = 15 m^{3}/s

T_{1} = 27^{o}C = (27 + 273) K = 300 K

We know that for an ideal gas the mass flow rate will be calculated as follows.

     P_{1}V_{1} = mRT_{1}

or,         m = \frac{P_{1}V_{1}}{RT_{1}}

                = \frac{100 \times 15}{0.5 \times 300}  

                = 10 kg/s

Now, according to the steady flow energy equation:

mh_{1} + Q = mh_{2} + W

h_{1} + \frac{Q}{m} = h_{2} + \frac{W}{m}

C_{p}T_{1} - \frac{80}{10} = C_{p}T_{2} - \frac{130}{10}

(T_{2} - T_{1})C_{p} = \frac{130 - 80}{10}

(T_{2} - T_{1}) = 5 K

T_{2} = 5 K + 300 K

T_{2} = 305 K

           = (305 K - 273 K)

           = 32^{o}C

Therefore, we can conclude that the exit temperature of the gas in deg C is 32^{o}C.

8 0
4 years ago
Discuss the organizational system that you believe would be the most effective for the safety officer in a medium-sized (100-200
marin [14]

Answer:

A safety manager is a person who designs and maintains the safety elements at workplace. A balance should be required for production and the job in providing work environment. As a safety officer in a medium sized manufacturing facility the following organizational system can be designed and maintained:

  • Maintaining a workplace as per the guidelines by Occupational safety and health association. The rules and regulation should be such that maintains the manufacturing facilities.  
  • For warning to workers proper labelling, floor mapping, signs, posters should be used.  
  • Procurement and usage of safe tools.  
  • A guideline that describes safety standard and precautionary measures should be available to the workers. They should be aware about all the steps that needs to be taken in crisis.  
  • Ensuring that the workers have enough training safety and health or accident prevention.  
  • Identify and eliminate the hazardous elements from the workplace.  
  • A strict action should be taken against the worker in case of violation of rules and not adhering with guidelines.

3 0
3 years ago
All the fnaf UNC charecters
astra-53 [7]

Answer:

T.Freddy.

T.Bonnie.

T.Chica.

G.Freddy.

N. Freddy.

F.Foxy.

Mr.Hippo.

R.Freddy.

Explanation:

8 0
3 years ago
Read 2 more answers
If we have silicon at 300K with 10 microns of p-type doping of 4.48*10^18/cc and 10 microns of n-type doping at 1000 times less
liq [111]

Answer:

The resistance is 24.9 Ω

Explanation:

The resistivity is equal to:

R=\frac{1}{N_{o}*u*V } =\frac{1}{4.48x10^{15}*1500*106x10^{-19}  } =0.93ohm*cm

The area is:

A = 60 * 60 = 3600 um² = 0.36x10⁻⁴cm²

w=\sqrt{\frac{2E(V_{o}-V) }{p}(\frac{1}{N_{A} }+\frac{1}{N_{D} })

If NA is greater, then, the term 1/NA can be neglected, thus the equation:

w=\sqrt{\frac{2E(V_{o}-V) }{p}(\frac{1}{N_{D} })

Where

V = 0.44 V

E = 11.68*8.85x10¹⁴ f/cm

V_{o} =\frac{KT}{p} ln(\frac{N_{A}*N_{D}}{n_{i}^{2}  } , if n_{i}=1.5x10^{10}cm^{-3}  \\V_{o}=0.02585ln(\frac{4.48x10^{18}*4.48x10^{15}  }{(1.5x10^{10})^{2}  } )=0.83V

w=\sqrt{\frac{2*11.68*8.85x10^{-14}*(0.83-0.44) }{1.6x10^{-19}*4.48x10^{15}  } } =3.35x10^{-5} cm=0.335um

The length is:

L = 10 - 0.335 = 9.665 um

The resistance is:

Re=\frac{pL}{A} =\frac{0.93*9.665x10^{-4} }{0.36x10^{-4} } =24.9ohm

7 0
3 years ago
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