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konstantin123 [22]
2 years ago
14

A tensile specimen with a 12mm initial diameter and 50mm gage length reaches maximum load at 90KN and fractures at 70KN

Engineering
1 answer:
Aleksandr-060686 [28]2 years ago
3 0

Answer:

i) 796.18 N/mm^2

ii) 1111.11 N/mm^2

Explanation:

Initial diameter ( D ) = 12 mm

Gage Length = 50 mm

maximum load ( P ) = 90 KN

Fractures at =  70 KN

minimum diameter at fracture = 10mm

<u>Calculate the engineering stress at Maximum load and the True fracture stress</u>

<em>i) Engineering stress at maximum load = P/ A </em>

= P / \pi  \frac{D^2}{4}  = 90 * 10^3 / ( 3.14 * 12^2 ) / 4

= 90,000 / 113.04 = 796.18 N/mm^2

<em>ii) True Fracture stress =  P/A </em>

= 90 * 10^3 / ( 3.24 * 10^2) / 4

= 90000 / 81  =  1111.11 N/mm^2

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Answer:

x_fat = [ 0.5*(Wsa + Wsw) -  p_muscle*V ] / V*( p_fat - p_muscle )

Explanation:

Given:

- The total volume of body = V

- The average density of the body = p_avg

- The density of muscle = p_muscle

- The density of fat = p_fat

Find:

Obtain a relation for the volume fraction of body fat x_fat

Solution:

- The volume of the fat is given by:

                          V_fat = x_fat*V

- The volume of the muscle is given by:

                          V_muscle = V - V_fat

                                            = V - x_fat*V

                                            =V*( 1 - x_fat )

- We will use the conservation of mass for the body related as:

                         mass_fat + mass_muscle = Total average mass

                         p_fat*V_fat + p_muscle*V_muscle = p_avg*V

                         p_fat*x_fat*V + p_muscle*V*( 1 - x_fat ) = p_avg*V

                         p_fat*x_fat + p_muscle*( 1 - x_fat ) = p_avg

- To determine p_1 we weigh the body in air:

                         Weight reading (Wsa) = m = p_1*V

                         p_1 = Wsa / V*g

- To determine p_2 we weigh the body in water:

                         Weight reading (Wsw) = m - p_w*V= p_1*V - p_w*V

                         Weight reading (Wsw) = V*(p_1 - p_w) = V*(p_2)

                         Where, p_2 = p_1 - p_water

                         p_2 = Wsw / V

- The average density p_avg:

                         p_avg = 0.5*(p_1 + p_2)  

                         p_avg = 0.5*(Wsa / V + Wsw / V)  

                         p_avg = 0.5*(Wsa + Wsw) / V                      

- Plug in the mass equation:

                         p_fat*x_fat + p_muscle*( 1 - x_fat ) = 0.5*(Wsa + Wsw) / V

                         x_fat*( p_fat - p_muscle ) = 0.5*(Wsa + Wsw) / V - p_muscle

                   x_fat = [ 0.5*(Wsa + Wsw) -  p_muscle*V ] / V*( p_fat - p_muscle )

                         

6 0
3 years ago
A school is playing $0.XY per kWh for electric power. To reduce its power bill, the school installs a wind turbine with a rated
Semmy [17]

Answer: Your question has some missing figures so kindly plug in the values into the solution provided to get the exact amount of money saved

answer : Electric power generated = 216 * 10^6 kJ

             money saved = $0.XY * 60000 kwh

Explanation:

<u>Calculating  the amount of electric power generated by wind turbine</u>

power generated = ( 30 * 2000 ) kWh  = 60000 kWh

Electric energy generated = 60000 kWh * 3600 kJ = 216 * 10^6 kJ

<u>Calculate money saved by school per year </u>

$0.XY * 60000 kwh

5 0
2 years ago
What is the mode of operation of a ramp digital voltimeter​
liberstina [14]

Answer:

The operating principle of a ramp type digital voltmeter is to measure the time that a linear ramp voltage takes to change from level of input voltage to zero voltage (or vice versa).

7 0
1 year ago
An aquifer has three different formations. Formation A has a thickness of 8.0 m and hydraulic conductivity of 25.0 m/d. Formatio
saveliy_v [14]

Answer:

The horizontal conductivity is 41.9 m/d.

The vertical conductivity is 37.2 m/d.

Explanation:

Given that,

Thickness of A = 8.0 m

Conductivity = 25.0 m/d

Thickness of B = 2.0 m

Conductivity = 142 m/d

Thickness of C = 34 m

Conductivity = 40 m/d

We need to calculate the horizontal conductivity

Using formula of horizontal conductivity

K_{H}=\dfrac{H_{A}K_{A}+H_{A}K_{A}+H_{A}K_{A}}{H_{A}+H_{B}+H_{C}}

Put the value into the formula

K_{H}=\dfrac{8.0\times25+2,0\times142+34\times40}{8.0+2.0+34}

K_{H}=41.9\ m/d

We need to calculate the vertical conductivity

Using formula of vertical conductivity

K_{V}=\dfrac{H_{A}+H_{B}+H_{C}}{\dfrac{H_{A}}{K_{A}}+\dfrac{H_{B}}{K_{B}}+\dfrac{H_{C}}{K_{C}}}

Put the value into the formula

K_{V}=\dfrac{8.0+2.0+34}{\dfrac{8.0}{25}+\dfrac{2.0}{142}+\dfrac{34}{40}}

K_{V}=37.2\ m/d

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The vertical conductivity is 37.2 m/d.

3 0
3 years ago
A heat engine is coupled with a dynamometer. The length of the load arm is 900 mm. The spring balance reading is 16. Applied wei
miss Akunina [59]

Answer:

P = 80.922 KW

Explanation:

Given data;

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Spring balanced  read 16 N

Applied weight is 500 N

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we know that power is given as

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P =\frac{435.6 \time 2 \pi \times 1774}{60} = 80922.65  watt

P = 80.922 KW

4 0
3 years ago
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