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Katyanochek1 [597]
3 years ago
7

Which of the following is an example of a computer simulation?

Engineering
1 answer:
nata0808 [166]3 years ago
5 0

Answer:

b)prototype

Explanation:

_÷/#^'tixgixurykgdtusurdfjczut uxdkdktxkgxkt. it

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A three-point bending test is performed on a glass specimen having a rectangular cross section of height d 5 mm (0.2 in.) and wi
Anon25 [30]

Answer:

The flexural strength of a specimen is = 78.3 M pa

Explanation:

Given data

Height = depth = 5 mm

Width = 10 mm

Length L = 45 mm

Load = 290 N

The flexural strength of a specimen is given by

\sigma = \frac{3 F L}{2 bd^{2} }

\sigma = \frac{3(290)(45)}{2 (10)(5)^{2} }

\sigma = 78.3 M pa

Therefore the flexural strength of a specimen is = 78.3 M pa

4 0
3 years ago
Fluid power is a. The technology that deals with the generation, control, and transmission of power-using pressurized fluids b.
snow_tiger [21]

Answer:  a) The technology that deals with the generation, control and transmission of power using pressurized fluids

Explanation: Fluid power is defined as the fluids which are under pressure and then are used for generation,control and transmit the power. Fluid power systems produces high forces as well as power in small amount . These systems usually tend to have better life if maintained properly. The force that are applied on this system can be monitored by gauges as well as meter.

8 0
3 years ago
What precautions should be taken to avoid the overloading of domestic electric circuits.
madam [21]

The precautions that should be taken to avoid the overloading of domestic electric circuits are:

  1. Do not put high voltage wires in one socket.
  2. Do not use many electric appliances of high power at the same time.
<h3>What are electric circuits?</h3>

Electric circuits are wires or devices that give electricity to devices that run on electricity. Running of electric devices should be done carefully because our body can come in contact with the current.

Thus, the precautions are to keep high voltage lines away from one socket. Use only a few high-power electric appliances at once.

To learn more about electric circuits, refer to the below link:

brainly.com/question/28221759

#SPJ4

4 0
1 year ago
A spherical container made of steel has 20 ft outer diameter and wal thickness of 1/2 inch. Knowing the internal pressure is 50
anastassius [24]

Answer:

maximum normal stress = 5975 psi

maximum shear stress = 2987.50 psi

Explanation:

Given data

dia = 20 ft

wall thickness = 1/2 inch

internal pressure  = 50 psi

To find out

the maximum normal stress and the maximum shearing stress

Solution

By the Mohr's circle we will find out shear stress

first we calculate inner radius

i.e. r = (diameter/2) - t

r = (20 × 12 in )/2 - ( 1/2 )

r =  120 - 0.5 = 119.5 inch

Now we find out maximum normal stress by given formula

normal stress = ( internal pressure× r ) / 2 t

normal stress = ( 50×119.5 ) / 2 × 0.5

maximum normal stress = 5975 psi

and minimum normal stress is 0, due to very small radius

and maximum shear stress will be

shear stress = ( maximum normal stress - minimum normal stress ) / 2

shear stress = ( 5975- 0 ) / 2

maximum shear stress = 2987.50 psi

5 0
3 years ago
The wall shear stress in a fully developed flow portion of a 12-in.-diameter pipe carrying water is 2.00 lb/ft^2. Determine the
soldier1979 [14.2K]

Answer:

a) -8 lb / ft^3

b) -70.4 lb / ft^3

c) 54.4 lb / ft^3

Explanation:

Given:

- Diameter of pipe D = 12 in

- Shear stress t = 2.0 lb/ft^2

- y = 62.4 lb / ft^3

Find pressure gradient dP / dx when:

a) x is in horizontal flow direction

b) Vertical flow up

c) vertical flow down

Solution:

- dP / dx as function of shear stress and radial distance r:

                      (dP - y*L*sin(Q))/ L = 2*t / r

                      dP / L - y*sin(Q) = 2*t / r

Where            dP / L = - dP/dx,

                      dP / dx = -2*t / r - y*sin(Q)

Where            r = D /2 ,

                      dP / dx = -4*t / D - y*sin(Q)

a) Horizontal Pipe Q = 0

Hence,           dP / dx = -4*2 / 1 - 62.4*sin(0)

                      dP / dx = -8 + 0

                      dP/dx   = -8 lb / ft^3

b) Vertical pipe flow up Q = pi/2

Hence,           dP / dx = -4*2 / 1 - 62.4*sin(pi/2)

                      dP / dx = 8 - 62.4

                      dP/dx   = -70.4 lb / ft^3

c) Vertical flow down Q = -pi/2

Hence,           dP / dx = -4*2 / 1 - 62.4*sin(-pi/2)

                      dP / dx = -8 + 62.4

                      dP/dx   = 54.4 lb / ft^3                      

7 0
3 years ago
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