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3 years ago

What are the three main areas of bioengineering?

Engineering

1 answer:

V125BC [204]3 years ago

8 0

Answer:Surgical Instruments and Medical Devices.

Tissue Engineering.

Biomaterials

Explanation:

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Three identical fatigue specimens (denoted A, B, and C) are fabricated from a nonferrous alloy. Each is subjected to one of the

Law Incorporation [45]

Answer:

B A and C

Explanation:

Given:

Specimen σ $_{max}$ σ $_{min}$

A +450 -150

B +300 -300

C +500 -200

Solution:

Compute the mean stress

σ $_{m}$ = (σ $_{max}$ + σ $_{min}$ )/2

σ $_{mA}$ = (450 + (-150)) / 2

= (450 - 150) / 2

= 300/2

σ $_{mA}$ = 150 MPa

σ $_{mB}$ = (300 + (-300))/2

= (300 - 300) / 2

= 0/2

σ $_{mB}$ = 0 MPa

σ $_{mC}$ = (500 + (-200))/2

= (500 - 200) / 2

= 300/2

σ $_{mC}$ = 150 MPa

Compute stress amplitude:

σ $_{a}$ = (σ $_{max}$ - σ $_{min}$ )/2

σ $_{aA}$ = (450 - (-150)) / 2

= (450 + 150) / 2

= 600/2

σ $_{aA}$ = 300 MPa

σ $_{aB}$ = (300- (-300)) / 2

= (300 + 300) / 2

= 600/2

σ $_{aB}$ = 300 MPa

σ $_{aC}$ = (500 - (-200))/2

= (500 + 200) / 2

= 700 / 2

σ $_{aC}$ = 350 MPa

From the above results it is concluded that the longest fatigue lifetime is of specimen B because it has the minimum mean stress.

Next, the specimen A has the fatigue lifetime which is shorter than B but longer than specimen C.

In the last comes specimen C which has the shortest fatigue lifetime because it has the higher mean stress and highest stress amplitude.

7 0

4 years ago

The net potential energy EN between two adjacent ions, is sometimes represented by the expression

Anastaziya [24]

Answer:

as answered in the attached file.

Explanation:

The detailed steps, derivation and appropriate differentiation is as shown in the attachment

3 0

3 years ago

Air modeled as an ideal gas enters a turbine operating at steady state at 1040 K, 278 kPa and exits at 120 kPa. The mass flow ra

Kipish [7]

Answer:

a. T₂ = 837.2 K

b. ηt = 91.4%

Explanation:

Assuming k = 1.4

KNOWN: Air expands adiabatically through a turbine operating at steady state. Operating data are known.

FIND: Determine the exit temperature and the isentropic turbine efficiency.

ENGINEERING MODEL (See the pics):

(1) The control volume is at state.

(2) For the control volume, Qcv = 0 and kinetic and potential energy effects can be neglected.

(3) The air is modeled as an ideal gas with constant specific heats: k = 1.4.

ANALYSIS:

(a) Mass and energy rate balances reduce to give:

0 = - Wcv + m*(h₁ – h₂).

With h₁ – h₂ = cp*(T₁ – T₂)

then

cp = k*R/(k-1) = 1.4*(8.314/28.97)/(1.4 – 1) = 1.004 kJ/kg∙K and

T₂ = T₁ - (Wcv/(m*cp))

⇒ T₂ = 1040 K – (1120 kW)/[(5.5 kg/s)(1.004 kJ/kg∙K)(1 kJ/s/ 1 kW)

⇒ T₂ = 837.2 K

b) The isentropic efficiency is

ηt = (h₁ – h₂) / (h₁ – h₂s) = cp*(T₁ – T₂) / (cp*(T₁ – T₂s)).

To get T₂s we note that for an isentropic process of an ideal gas with constant specific heats

(T₂s / T₁) = (p₂ / p₁)∧((k-1)/k) ⇒ T₂s = T₁*(p₂ / p₁)∧((k-1)/k)

⇒ T₂s = (1040 K)*(120 / 278)∧((1.4-1)/1.4) = 818.1 K

Thus, the isentropic efficiency is

ηt = (1040 – 837.2)/(1040 – 818.1) = 0.914

ηt = 91.4%

7 0

3 years ago

Write a statement that increases numPeople by 5. Ex: If numPeople is initially 10, the output is: There are 15 people.

BartSMP [9]

Answer:

import java.util.Scanner;

public class AssigningNumberToVariable {

public static void main (String [] args) {

int numPeople;

int numPeople_inc;

numPeople = 10;

/* solution */

numPeople_inc=numPeople+5;

System.out.print("There are ");

System.out.print(numPeople_inc);

System.out.println(" people.");

}

Explanation:

1. Declare a new variable called: numPeople_inc

2. Add 5 each time: numPeople_inc=numPeople+5;

3. Print the answer: numPeople_inc=numPeople+5;

Optional, just write the line in the place where says: " your solutions goes here":

numPeople = numPeople+5;

8 0

3 years ago

4) How much concrete was required to make the Hoover<br> Dam? Where did this concrete come from?

erica [24]

I’m not sure if this is right

4 0

4 years ago