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3 years ago

15

When engineers develop a model, which step in the engineering process is taking place

1 answer:

Sonbull [250]3 years ago

7 0

Answer: Design a solution or problem

Explanation:

When engineers develop a model, the step that is taking place in the engineering process is referred to as the design a solution or problem step.

This is when engineers tackles and solved a problem by finding solutions and this is done by incorporating their analytical, and synthetic thinking and using their skills an experience.

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Discuss importance of good communication system

meriva

Answer:

Effective communication helps decision makers by gathering and providing the information to the right person on right time. Communication performs as a motivator to the employees by notifying the employees about the job task, process of carrying and how it could be done better.

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2 years ago

The design for a new cementless hip implant is to be studied using an instrumented implant and a fixed simulated femur.

OlgaM077 [116]

Answer:

a) the velocity of the implant immediately after impact is 20 m/s

b) the average resistance of the implant is 40000 N

Explanation:

a) The impulse momentum is:

mv1 + ∑Imp(1---->2) = mv2

According the exercise:

v1=0

∑Imp(1---->2) = F(t2-t1)

m=0.2 kg

Replacing:

$0+F(t_{2} -t_{1} )=0.2v_{2}$

if F=2 kN and t2-t1=2x10^-3 s. Replacing

$0+2x10^{-3} (2x10^{-3} )=0.2v_{2} \\v_{2} =\frac{4}{0.2} =20m/s$

b) Work and energy in the system is:

T2 - U(2----->3) = T3

where T2 and T3 are the kinetic energy and U(2----->3) is the work.

$T_{2} =\frac{1}{2} mv_{2}^{2} \\T_{3} =0\\U_{2---3} =-F_{res} x$

Replacing:

$\frac{1}{2} *0.2*20^{2} -F_{res} *0.001=0\\F_{res} =40000N$

3 0

3 years ago

I need solution for this question please

AlladinOne [14]

Answer:

a true b false thos is the solution

6 0

3 years ago

Read 2 more answers

Wet steam at 15 bar is throttled adiabatically in a steady-flow process to 2 bar. The resulting stream has a temperature of 130°

cricket20 [7]

Answer:

$\Delta s = 0.8708\,\frac{kJ}{kg\cdot K}$

Explanation:

The adiabatic throttling process is modelled after the First Law of Thermodynamics:

$m\cdot (h_{in} - h_{out}) = 0$

$h_{in} = h_{out}$

Properties of water at inlet and outlet are obtained from steam tables:

State 1 - Inlet (Liquid-Vapor Mixture)

$P = 1500\,kPa$

$T = 198.29\,^{\textdegree}C$

$h = 2726.9\,\frac{kJ}{kg}$

$s = 6.3068\,\frac{kJ}{kg\cdot K}$

$x = 0.967$

State 2 - Outlet (Superheated Vapor)

$P = 200\,kPa$

$T = 130\,^{\textdegree}C$

$h = 2726.9\,\frac{kJ}{kg}$

$s = 7.1776\,\frac{kJ}{kg\cdot K}$

The change of entropy of the steam is derived of the Second Law of Thermodynamics:

$\Delta s = 7.1776\,\frac{kJ}{kg\cdot K} - 6.3068\, \frac{kJ}{kg\cdot K}$

$\Delta s = 0.8708\,\frac{kJ}{kg\cdot K}$

6 0

3 years ago

Question 8(Multiple Choice Worth 2 points)

8090 [49]

Answer:

4 number answer is correct.

8 0

2 years ago