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vodomira [7]
3 years ago
12

Unless a structural element is intended to be battered, sloped, or cambered (or some other non-conforming position), wood, steel

, and concrete should always be installed and maintained __________ , __________ , and __________. These conditions are critical in order to ensure maximum stability, structural performance and user satisfaction. Please insert the answer into the spaces provided below.
Engineering
1 answer:
SSSSS [86.1K]3 years ago
5 0

Answer:

stability; resistance; rigidity.

Explanation:

Okay, let us fill in the gap in the question above. Please note that the capitalized words are the missing words to fill in the gap.

"Unless a structural element is intended to be battered, sloped, or cambered (or some other non-conforming position), wood, steel, and concrete should always be installed and maintained STABILITY , RESISTANCE , and RIGIDITY. These conditions are critical in order to ensure maximum stability, structural performance and user satisfaction."

This question has to deal with buildings or say structural (civil) engineering. The definition to the missing words are given below:

STABILITY: Stability occurs when we have the center of gravity coinciding with the base of the structure.

RESISTANCE : Resistance simply means the 'tension' that is is how much the structure can resist an applied force.

RIGIDITY : RIGIDITY can also be associated with resistance and it is the property of a structure to resist bending.

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What is hardness and how is it generally tested?
drek231 [11]

Answer:

Hardness is understood as the property of materials in general to resist the penetration of an indenter under load, so that the hardness represents the resistance of the material to the plastic deformation located on its surface.

Explanation:

Hardness of a material is understood as the resistance that the material opposes to its permanent surface plastic deformation by scratching or penetration. It is always true that the hardness of a material is inversely proportional to the footprint that remains on its surface when a force is applied.

In this sense, the hardness of a material can also be defined as that property of the surface layer of the material to resist any elastic deformation, plastic or destruction due to the action of local contact forces caused by another body (called indenter or penetrator), harder, of certain shape and dimensions, which does not suffer residual deformations during contact.

That is, hardness is understood as the property of materials in general to resist the penetration of an indenter under load, so that the hardness represents the resistance of the material to the plastic deformation located on its surface.

The following conclusions can be drawn from the previous definition of hardness:  

  1) hardness, by definition, is a property of the surface layer of the material, and is not a property of the material itself;  

  2) the methods of hardness by indentation presuppose the presence of contact efforts, and therefore, the hardness can be quantified within a scale;

  3) In any case, the indenter or penetrator must not undergo residual deformations during the test of hardness measurement of the body being tested.

To determine the hardness of the materials, durometers with different types of tips and ranges of loads are used on the various materials. Below are the most commonly used tests to determine the hardness of the materials.

   Rockwell hardness :

It refers to the Rockwell hardness test, a method with which the hardness or resistance of a material to be penetrated is calculated. It is characterized by being a fast and simple method that can be applied to all types of materials. An optical reader is not required.

    Brinell hardness :

Brinell hardness is a scale that is used to determine the hardness of a material through the indentation method, which consists of penetrating with a hardened steel ball tip into the hard material, a load and for a certain time.  

This test is not very precise but easy to apply. It is one of the oldest and was proposed in 1900 by Johan August Brinell, a Swedish engineer.

    Vickers hardness:

Vickers hardness is a test that is used in all types of solid and thin or soft materials. In this test, a square-shaped pyramid-shaped diamond and a   136° vertex angle are placed on the penetrating equipment.

In this test the hardness measurement is performed by calculating the diagonal penetration lengths.

However, its result is not read directly on the equipment used, therefore, the following formula must be applied to determine the hardness of the material: HV = 1.8544 · F / (dv2).

3 0
3 years ago
A stationary gas-turbine power plant operates on a simple ideal Brayton cycle with air as the working fluid. The air enters the
ololo11 [35]

Answer:

A) W' = 15680 KW

B) W' = 17113.87 KW

Explanation:

We are given;

Temperature at state 1; T1 = 290 K

Temperature at state 3; T3 = 1100 K

Rate of heat transfer; Q_in = 35000 kJ/s = 35000 Kw

Pressure of air into compressor; P_c = 95 kPa

Pressure of air into turbine; P_t = 760 kPa

A) The power assuming constant specific heats at room temperature is gotten from;

W' = [1 - ((T4 - T1)/(T3 - T2))] × Q_in

Now, we don't have T4 and T2 but they can be gotten from;

T4 = [T3 × (r_p)^((1 - k)/k)]

T2 = [T1 × (r_p)^((k - 1)/k)]

r_p = P_t/P_c

r_p = 760/95

r_p = 8

Also,k which is specific heat capacity of air has a constant value of 1.4

Thus;

Plugging in the relevant values, we have;

T4 = [(1100 × (8^((1 - 1.4)/1.4)]

T4 = 607.25 K

T2 = [290 × (8^((1.4 - 1)/1.4)]

T2 = 525.32 K

Thus;

W' = [1 - ((607.25 - 290)/(1100 - 525.32))] × 35000

W' = 0.448 × 35000

W' = 15680 KW

B) The power accounting for the variation of specific heats with temperature is given by;

W' = [1 - ((h4 - h1)/(h3 - h2))] × Q_in

From the table attached, we have the following;

At temperature of 607.25 K and by interpolation; h4 = 614.64 KJ/K

At T3 = 1100 K, h3 = 1161.07 KJ/K

At T1 = 290 K, h1 = 290.16 KJ/K

At T2 = 525.32 K, and by interpolation, h2 = 526.12 KJ/K

Thus;

W' = [1 - ((614.64 - 290.16)/(1161.07 - 526.12))] × 35000

W' = 17113.87 KW

4 0
2 years ago
A second inventor was driving down the highway in her Prius one day with her hand out the window. She happened to be driving thr
Eva8 [605]

Answer:

Explanation:

It wouldn't work because the wind energy she would be collecting would actually come from the car engine.

The relative wind velocity observed from a moving vehicle is the sum of the actual wind velocity and the velovity of the vehicle.

u' = u + v

While running a car will generate a rather high wind velocity, and increase the power generated by a wind turbine, the turbine would only be able to convert part of the wind energy into electricity while adding a lot of drag. In the end, it would generate less energy that what the drag casuses the car to waste to move the turbine.

Regenerative braking uses an electric generator connected to the wheel axle to recover part of the kinetic energy eliminated when one brakes the vehicle. Normal brakes dissipate this energy as heat, a regenerative brake uses it to recharge a batttery. Note that is is a fraction of the energy that is recovered, not all of it.

A "regenerative accelerator" makes no sense. Braking is taking kinetic energy out of the vehicle, while accelerating is adding kinetic energy to it. Cars accelerate using the power from their engines.

6 0
3 years ago
At 45° latitude, the gravitational acceleration as a function of elevation z above sea level is given by g = a − bz , where a =
Ahat [919]

Answer:

8861.75 m approximately 8862 m

Explanation:

We need to remember Newton's 2nd Law which says that the force experienced by an object is proportional to his acceleration and that the constant of proportionality between those two vectors correspond to the mass of the object.

F=ma for the weight of an object (which is a force) we have that the acceleration experienced by that object is equal to the gravitational acceleration, obtaining that  W = mg

For simplicity we work with g =9.807 \frac{m}{s^{2}} despiting the effect of the height above sea level. In this problem, we've been asked by the height above sea level that makes the weight of an object 0.30% more lighter.

In accord with the formula g = a-bz the "normal" or "standard" weight of an object is given by W = mg = ma when z = 0, so we need to find the value of z that makes W = m(a-bz) = 0.997ma meaning that the original weight decrease by a 0.30%, so now we operate...

m(a-bz) = 0.997ma now we group like terms on the same sides ma(1-0.997) = mbz we cancel equal tems on both sides and obtain that z = \frac{a}{b} (0.003) = \frac{9.807 \frac{m}{s^{2} } }{3.32*10^{-6} s^{-2} } (0.003) = 8861.75 m

7 0
3 years ago
Integer to Float Conversion All labs must be done during lab time. Each labs worth 10 points The lab can be hand in next day wit
andrew-mc [135]

Answer:

Code explained below

Explanation:

.data

msg1: .asciiz "Please input a temperature in celsius: "

msg2: .asciiz "The temperature in Fahrenheit is: => "

num: .float 0.0

.text

main:

#print the msg1

li $v0, 4

la $a0, msg1

syscall

#read the float value from user

li $v0,6 #read float syscall value is $v0

syscall #read value stored in $f0

#formula for celsius to fahrenheit is

#(temperature(C)* 9/5)+32

#li.s means load immediate float

#copy value 9.0 to $f2

li.s $f2,9.0  

#copy value 5.0 to $f3

li.s $f3,5.0

# following instructions performs: 9/5

#div.s - division of two float numbers

#divide $f2 and f3.Result will stores in $f1

div.s $f1,$f2,$f3

#following instruction performs: temperature(C) * (9/5)

#multiple $f1 and $f0.Result stored in $f1

mul.s $f1,$f1,$f0

#copy value 32 to $f4

li.s $f4,32.0

#following instruction performs: (temperature(C) * (9/5))+32

#add $f1 and $f4.Result stores in $f1

add.s $f1,$f1,$f4

#store float from $f1 to num

s.s $f1,num

#print the msg2

li $v0, 4 #print string syscall value is 4

la $a0, msg2 #copy address of msg2 to $a0

#print the float

syscall

li $v0,2 #print float syscall value is 2

l.s $f12,num #load value in num to $f12

syscall

#terminate the program

li $v0, 10 #terminate the program syscall value is 10

syscall

4 0
3 years ago
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