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schepotkina [342]
3 years ago
9

2. What is the original length of the rectangular bar if the deformation is 0.005 in with a force of 1000 lbs and an area of 0.7

5 sqin? The Modulus of Elasticity is 5,000,000 psi.
Engineering
1 answer:
Ugo [173]3 years ago
6 0

Answer:

18.75in

Explanation:

Modulus of elasticity = Stress/Strain

Since stress = Force/Area

Given

Force = 1000lb

Area = 0.75sqin

Stress = 1000/0.75

Stress = 1333.33lbsqin

Strain

Strain = Stress/Modulus of elasticity

Strain = 1333.33/5,000,000

Strain = 0.0002667

Also

Strain = extension/original length

extension = 0.005in

Original length = extension/strain

Original length = 0.005/0.0002667

Original length = 18.75in

Hence the original length of the rectangular bar is 18.75in

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Material with hardness of 220 Vickers is harder than material with a hardness of 180 Vickers. a)-True b)- False
PSYCHO15rus [73]

Answer:

Correct option a) True.

Explanation:

It is true since the Vickers hardness value refers to the force applied in a 136 ° diamond tip penetrator divided by the surface of the groove produced in the material, the lower the impression made on this greater the value will be end of the Vickers measurement and greater its hardness.

The equation to determine the Vickers hardness value will be:

Hv= ((1.854 × P)/(d²))  (kg/mm²)

Therefore a value of 220 Vickers refers to a harder material than another value of 180 Vickers.

6 0
3 years ago
Determine the total condensation rate of water vapor onto the front surface of a vertical plate that is 10 mm high and 1 m in th
castortr0y [4]

Answer:

Q =  63,827.5 W

Explanation:

Given:-

- The dimensions of plate A = ( 10 mm x 1 m )

- The fluid comes at T_sat , 1 atm.

- The surface temperature, T_s = 75°C  

Find:-

Determine the total condensation rate of water vapor onto the front surface of a vertical plate

Solution:-

- Assuming drop-wise condensation the heat transfer coefficient for water is given by Griffith's empirical relation for T_sat = 100°C.

                            h = 255,310 W /m^2.K

- The rate of condensation (Q) is given by Newton's cooling law:

                           Q = h*As*( T_sat - Ts )

                           Q = (255,310)*( 0.01*1)*( 100 - 75 )

                           Q =  63,827.5 W

8 0
3 years ago
Read 2 more answers
Fluorescent troffers are a type of _ lighting fixture
creativ13 [48]
The answer would be letter A
8 0
3 years ago
Calculate the differential pressure in kPa across the hatch of a submarine 320m below the surface of the sea. Assume the atmosph
kicyunya [14]

Answer:

The pressure difference across hatch of the submarine is 3217.68 kpa.

Explanation:

Gauge pressure is the pressure above the atmospheric pressure. If we consider gauge pressure for finding pressure differential then no need to consider atmospheric pressure as they will cancel out. According to hydrostatic law, pressure varies in the z direction only.  

Given:

Height of the hatch is 320 m

Surface gravity of the sea water is 1.025.

Density of water 1000 kg/m³.

Calculation:

Step1

Density of sea water is calculated as follows:

S.G=\frac{\rho_{sw}}{\rho_{w}}

Here, density of sea water is\rho_{sw}, surface gravity is S.G and density of water is \rho_{w}.

Substitute all the values in the above equation as follows:

S.G=\frac{\rho_{sw}}{\rho_{w}}

1.025=\frac{\rho_{sw}}{1000}

\rho_{sw}=1025 kg/m³.

Step2

Difference in pressure is calculated as follows:

\bigtriangleup p=rho_{sw}gh

\bigtriangleup p=1025\times9.81\times320

\bigtriangleup p=3217680 pa.

Or

\bigtriangleup p=(3217680pa)(\frac{1kpa}{100pa})

\bigtriangleup p=3217.68 kpa.

Thus, the pressure difference across hatch of the submarine is 3217.68 kpa.

6 0
3 years ago
Consider a fan located in a 3 ft by 3 ft square duct. Velocities at various points at the outlet are measured, and the average f
natulia [17]

Answer:

minimum electric power consumption of the fan motor is 0.1437 Btu/s

Explanation:

given data

area = 3 ft by 3 ft

air density = 0.075 lbm/ft³

to find out

minimum electric power consumption of the fan motor

solution

we know that energy balance equation that is express as

E in - E out  = \frac{dE \ system}{dt}    ......................1

and at steady state  \frac{dE \ system}{dt} = 0

so we can say from equation 1

E in = E out

so

minimum power required is

E in = W = m \frac{V^2}{2} = \rho A V \frac{V^2}{2}  

put here value

E in =  \rho A V \frac{V^2}{2}  

E in =  0.075 *3*3* 22 \frac{22^2}{2}  

E in = 0.1437 Btu/s

minimum electric power consumption of the fan motor is 0.1437 Btu/s

5 0
3 years ago
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