A part has been tested to have Sut = 530 MPa, f = 0.9, and a fully corrected Se = 210 MPa. The design requirements call for the part to be loaded at three different fully-reversed loads and cycled at each one for a set number of cycles. First, it will be loaded at ±350 MPa for 5,000 cycles. Then, it will be loaded at ±260 MPa for 50,000 cycles. Finally, it will be loaded at ±225 MPa until it fails. How many cycles do we expect the part to last at the final loading? Use Miner's method. :g

Answer:

first is the parentheses, (3+2)=5 next is the exponent 5^2=25, next is the division 5 / 5 = 1, then the multiplication 4*1=4 and then you add 4+25=29. so the answer is 29.

Answer:

The answer is below

Explanation:

[tex]\theta_1=cos^{-1}0.28=73.74^o\ lagging\\\\S_1=125\angle 73.74^o=35\ kW+j120\ kVAR\\\\S_2=10\ kW-j40\ kVAR\\\\S_3=15\ kW[/tex]

Total power = P = 35 kW + 10 kW + 15 kW = 60 kW

Total kVAR = 120 kVAR - 40 kVAR = 80 kVAR

[tex]Total\ apparent \ power =S= S_1+S_2+S_3=(35+j120)+(10-j40)+(15)\\\\S=60\ kW+j80\ kVAR=100\angle 53.13^o\\\\Current(I)=\frac{S^*}{V^*} \frac{100000\angle -53.13^o}{1400\angle0}=71.43\angle-53.13^o\\ \\Power\ factor (PF)=cos(53.13)=0.6\ lagging\\\\The \ new\ power\ factor\ is\ to \ be\ 0.8[cos^{-1}0.8=36.87^o], hence\ since\ the\ total\ \\real\ power(P)= 60\ kW, the\ capacitor\ kVAR(Q_c)\ is:\\\\Q_c=60tan(53.13)-60tan(36.87)=80-45=35\ kVAR\\\\[/tex]

[tex]C=\frac{Q_c}{wV^2} =\frac{35000\ VAR}{2\pi*60\ Hz*1400\ V}=47.38\ \mu f[/tex]

New current (I') = [tex]\frac{S'^*}{V^*}=\frac{60000-j45000}{1400}=53.57\angle-36.87^o[/tex]

Current reduce from 71.43 A to 53.57 A

Answer:

The four factors of production are land, labor, capital, and entrepreneurship. They are the inputs needed for supply. They produce all the goods and services in an economy. That's measured by gross domestic product.

please give brainlist

hope this helped........

Answer:

True

Explanation:

Principles of plasma arc cutting, Uses a constricting nozzle to create, concentrate,  and direct the high-velocity plasma. Plasma gas is always used

in plasma arc cutting When shielding gas is also used, the process is called dual flow plasma arc cutting. I hope this helps.

Answer:

Volcanic outgassing

The burning of fossil fuels (coal, oil, gasoline, natural gas), for heating, power generation and transport

Some industrial processes such as cement making.

Explanation:

Answer:

Some of the benefits are tangible for they are visible in the design and production process, while the other benefits are intangible which may not be visible directly but result in improvement in the quality of product, better control over designing and production process, reduction of stress on the designers etc.

Answer:

The angles are missing in the question.

The angles are :

45,     30,    60,     90,    -34,     -56,      20,     -42,  -65,    -15

P=10, P=5,  P=25, P=54, P=65, P=95, P=250, P=8, P=35, P=150

Explanation:

1. P = 10,   θ = 45°  rectangular coordinates

x = r cosθ  ,   y = r sinθ

So, rectangular form is x + iy

x = P cosθ = 10 cos 45°

  = 7.07

y =P sinθ = 10 sin 45°

  = 7.07

Therefore, rectangular form

x + iy = 7.07 + i (7.07)

2. P = 5 , θ = 30°

x = 5 cos  30° = 4.33

y = 5 sin  30° = 2.5

So, (x+iy) = 4.33 + i (2.5)

3. P = 25 , θ = 60°

x = 25 cos  60° = 12.5

y = 25 sin  60° = 21.65

So, (x+iy) = 12.5 + i (21.65)

4. P = 54 , θ = 90°

x = 54 cos  90° = 0

y = 54 sin  90° = 54

So, (x+iy) = 0+ i (54)

5. P = 65 , θ = -34°

x = 65 cos  (-34°) = 53.88

y = 65 sin  (-34°) = -36.34

So, (x+iy) = 53.88 - i (36.34)

6. P = 95 , θ = -56°

x = 95 cos  (-56)° = 53.12

y = 95 sin  (-56)° = -78.75

So, (x+iy) = 53.12 - i (78.75)

7. P = 250 , θ = 20°

x = 250 cos  20° = 234.92

y = 250 sin 20° = 85.5

So, (x+iy) = 234.92 + i (85.5)

8. P = 8 , θ = (-42)°

x = 8 cos  (-42)° = 5.94

y = 8 sin  (-42)° = -5.353

So, (x+iy) = 5.94 - i (5.353)

9. P = 35 , θ = (-65)°

x = 35 cos  (-65)° = 14.79

y = 35 sin  (-65)° = -31.72

So, (x+iy) = 14.79 - i (31.72)

10. P = 150 , θ = (-15)°

x = 150 cos  (-15)° = 144.88

y = 150 sin  (-15)° = -38.82

So, (x+iy) = 144.88 - i (38.82)

Answer:

Hair clip: this used to hold human hairs together especially long hairs

The Design parameters to be considered are : number of teeth, durability of the material to be used and also the width ( how wide the clip can open )

Advantages : very easy to use, holds hairs tight

Disadvantages : Expensive and material used is sometimes brittle

paper clips : This used to hold papers together it is made of a thin elastic wire

The design parameters to be considered are : length of the wire, material to be used and the elasticity of the material

advantages : it is quite cheap and good strength

disadvantages :it can hold a limited amount of papers together

potato chip bag clips: These clips are used on potato chip bags to ensure that the bags are air tight

The design parameters to be considered are : Elasticity of the material to be used , and size

Advantages : it helps to prevent the potato chips from getting exposed to air

disadvantages : Not very popular and its quite expensive

Explanation:

Hair clip: this used to hold human hairs together especially long hairs

The Design parameters to be considered are : number of teeth, durability of the material to be used and also the width ( how wide the clip can open )

Advantages : very easy to use, holds hairs tight

Disadvantages : Expensive and material used is sometimes brittle

paper clips : This used to hold papers together it is made of a thin elastic wire

The design parameters to be considered are : length of the wire, material to be used and the elasticity of the material

advantages : it is quite cheap and good strength

disadvantages :it can hold a limited amount of papers together

potato chip bag clips: These clips are used on potato chip bags to ensure that the bags are air tight

The design parameters to be considered are : Elasticity of the material to be used , and size

Advantages : it helps to prevent the potato chips from getting exposed to air

disadvantages : Not very popular and its quite expensive

Answer:

To run. The machine one at a time

Explanation:

Answer: hello your question is incomplete below is the complete question

An ideal Otto Cycle has a compression ratio of 9.2 and uses air as the working fluid. At the beginning of the compression process, air is at 98kPa and 20C. The pressure is doubled during the constant volume heat addition process. Assuming constant specific heats, determine:the amount of heat transferred to the air

answer : 609.804 kj/kg

Explanation:

Given data:

compression ratio (r)= 9.2

pressure given(p1) = 98 kPa

Initial temperature =  20 + 273 = 293 k

pressure during constant volume heat addition process = 2p1

note : specific heat at constant pressure and specific heat at  constant volume varies with temperature

we use T = 300k because it is closest to T1 = 293 k

hence at T = 300 K ( ideal gas properties of air )

             [tex]u_{1}[/tex]   = 214.07 Kj/kg

            [tex]v _{r1}[/tex] = 621.2

To get  [tex]v_{r2}[/tex] = [tex]v_{r1} * \frac{v_{2} }{r}[/tex] = 621.2 * 1 / 9.2 = 67.52

ALSO    at [tex]v_{r2}[/tex] = 67.52 ( from ideal gas properties )

                [tex]u_{2}[/tex] = 518.9 kj/kg

                T2 = 708.32 k

next we apply the gas equation

[tex]\frac{p1v1}{T1} = \frac{p2v2}{T2}[/tex]

hence  p2 = (9.2) * [tex]\frac{708.32}{293} * 98[/tex] = 2179.59 kpa

to determine T3  due to the constant volume heat addition

[tex]\frac{T3}{T2} = \frac{P3}{P2}[/tex]

Hence T3 = p3/p2 * T2 = 2( 708.32 ) = 1416.64 k

At T3 = 1416.64 k ( from ideal gas properties )

[tex]u_{3}[/tex] = 1128.704 kj/kg

[tex]v_{r3}[/tex] = 8.592

Determine the amount of heat transferred to the air

[tex]q_{in} = ( u_{3} - u_{2} )[/tex]

     = ( 1128.704  - 518.9 )

     = 609.804 kj/kg

Answer:

  157.38° CCW from +x axis

  67.38° CCW from +y axis

  n = (12/13)i +(5/13)j

Explanation:

The reference angle is ...

  arctan(15/36) ≈ 22.62° . . . . CW from the -x axis

The signs of the component vectors indicate this is a 2nd-quadrant vector, so the angles of interest are ...

  157.38° CCW from +x axis

  67.38° CCW from +y axis

The unit vector will be ...

  n = cos(157.38°)i +sin(157.38°)j

 n = (12/13)i +(5/13)j

Answer:

B) P1 would increase to maintain the flow rate ( correct )

C) The resistance would increase (correct )

Explanation:

flow rate = 10 liters per minute

Driving pressure (p1) = 20 cm H20

Fixed downstream pressure (p2) = 5 cm H20

The correct statements when we pinch the Lumen in the middle of the tube would be : P1 would increase to maintain the flow rate and The resistance would increase.this is because when we pinch the Lumen we reduce its diameter and the reduction of its Diameter will result to increased resistance against the flow and resistance of flow is directly proportional to pressure hence P1 would increase as well

The wrong statement would be : The flow would decrease

Answer:

yes yes it is why wouldn't it be

It is true that buying shop supplies from the shop owner to work on your own car at home is an ethical practice.

The term "ethical practises" refers to deeds and conduct that uphold established moral and social norms as well as shared values.

Honesty, fairness, transparency, accountability, respect, and responsibility are the traits that define these procedures.

The precise circumstances and the relationship between the shop owner and the individual will determine if purchasing shop materials from a shop owner to work on your own car at home is an ethical practise.

The exact situation and the relationship between the shop owner and the individual will determine whether or not purchasing shop materials from a shop owner to work on your own car at home is ethical.

Before implementing such a method, it is crucial to think about the possible outcomes and any ethical issues.

Thus, the given statement can be considered as true.

For more details regarding ethical practice, visit:

https://brainly.com/question/30263183

#SPJ2

Answer: These robots study planets from space. The Cassini spacecraft is this type of robot. Cassini studies Saturn and its moons and rings. The Voyager and Pioneer spacecraft are now traveling beyond our solar system

Explanation:

Answer:

A.

Explanation:

Answer:

  110 V

Explanation:

V = IR

V = (10.0 A)(11.0 Ω) = 110 volts

Answer:

[tex]v = 250[1 - {e^{-6000t}}][/tex] mV

Explanation:

The voltage across a capacitor at a time t, is given by:

[tex]v(t) = \frac{1}{C} \int\limits^{t}_{t_0} {i(t)} \, dt + v(t_0)[/tex]                 ----------------(i)

Where;

v(t) = voltage at time t

[tex]t_{0}[/tex] = initial time

C = capacitance of the capacitor

i(t) = current through the capacitor at time t

v(t₀) = voltage at initial time.

From the question:

C = 2μF = 2 x 10⁻⁶F

i(t) = 3[tex]e^{-6000t}[/tex] mA

t₀ = 0

v(t₀ = 0) = 0

Substitute these values into equation (i) as follows;

[tex]v = \frac{1}{2*10^{-6}} \int\limits^{t}_{0} {3e^{-6000t}} \, dt + v(0)[/tex]    

[tex]v = \frac{1}{2*10^{-6}} \int\limits^{t}_{0} {3e^{-6000t}} \, dt + 0[/tex]

[tex]v = \frac{1}{2*10^{-6}} \int\limits^{t}_{0} {3e^{-6000t}} \, dt[/tex]            

[tex]v = \frac{3}{2*10^{-6}} \int\limits^{t}_{0} {e^{-6000t}} \, dt[/tex]             [Solve the integral]

[tex]v = \frac{3}{2*10^{-6}*(-6000)} {e^{-6000t}}|_0^t[/tex]

[tex]v = \frac{-3000}{12} {e^{-6000t}}|_0^t[/tex]

[tex]v = -250 {e^{-6000t}}|_0^t[/tex]

[tex]v = -250 {e^{-6000t}} - [-250 {e^{-6000(0)}][/tex]

[tex]v = -250 {e^{-6000t}} - [-250][/tex]

[tex]v = -250 {e^{-6000t}} + 250[/tex]

[tex]v = 250 -250 {e^{-6000t}}[/tex]

[tex]v = 250[1 - {e^{-6000t}}][/tex]

Therefore, the voltage across the capacitor is [tex]v = 250[1 - {e^{-6000t}}][/tex] mV

Answer: the distance in advance of the curve the sign must be located to ensure that vehicles have sufficient distance to decelerate safely is 292.07 ft

Explanation:

first we calculate the distance to place sign board in advance to curve for decelerating the speed

d = 1.47Si t + [ (Si² - Sf²) / 30(0.348 ± 0.01G)

where d is the safe stopping distance, initial speed is Si (60 mi/hr ), time reaction is t (2.5s), Sf is the final speed (40 mi/hr), G is the grade (0).

so we substitute

d = (1.47 × 60 × 2.5) + [ (60² - 40²) / 30(0.348 ± 0)

= 220.5 + ( 2000/10.44)

= 220.5 + 191.57

= 412.07 ft

Now giving that a warning sign is clearly visible at a distance of 120 ft

optimum safe distance will be

d = minimum distance - sign visible distance

d = 412.07 - 120

d = 292.07 ft

therefore the distance in advance of the curve the sign must be located to ensure that vehicles have sufficient distance to decelerate safely is 292.07 ft

Answer:

metals, composite, ceramics and polymers.

Explanation:

The four categories of engineering materials used in manufacturing are metals, composite, ceramics and polymers.

i) Metals: Metals are solids made up of atoms held by matrix of electrons. They are good conductors of heat and electricity, ductile and strong.

ii) Composite: This is a combination of two or more materials. They have high strength to weight ratio, stiff, low conductivity. E.g are wood, concrete.

iii) Ceramics: They are inorganic, non-metallic crystalline compounds with high hardness and strength as well as poor conductors of electricity and heat.

iv) Polymers: They  have low weight and are poor conductors of electricity and heat

Answer:

A ) CPI : M1 = 2.4 , M2 = 2.65

B ) MIPS : M1 = 1083, M2 = 1056

C ) The machine that has a better performance based on MIPS is M1 and this is  by 27 million number of instructions per sec

Explanation:

A) The CPI for each machine

CPI = ( Total number of execution cycles ) / ( instruction counter executed )

For Machine 1 ( M1 )

we have to make some assumptions : number of instructions = 10

number of times A was executed = 4 , Number of times B was executed = 2.5 , number of times C was executed = 2.5, Number of times D was executed = 1. and this was based on the frequency given above

hence CPI for M1 =[ ( 1 * 4 ) + ( 3 * 2.5 ) + ( 3 * 2.5 ) + ( 5 * 1 ) ] / 10

       CPI  for M1 = 2.4

For Machine 2 ( M2 )

we have to make some assumptions : number of instructions = 10

number of times A was executed = 4 , Number of times B was executed = 2.  number of times C was executed = 1.5, Number of times D was executed = 2.5 times. and this was based on the frequency given above

Hence CPI  for M1 = [ ( 2 * 4 ) + ( 2 * 2 ) + ( 3 * 1.5 ) + ( 4 * 2.5 ) ] / 10

            CPI for M2 = 2.65

B ) Calculate the native MIPS  ratings for M1 and M2

MIPS = ( instruction counts ) / ( Execution time * 10^6 )

For M1

Assumptions : number of instructions executed = 10

                        each clock cycle = 0.3846 * 10^-9.      frequency = 2.6 Ghz

first we calculate the total execution time which is equal to :

= [ ( 1 * 4 ) + ( 3 * 2.5 ) + ( 3 * 2.5 ) + ( 5 * 1 ) ] * 0.3846 * 10 ^-9

= 9.2304 * 10 ^-9 secs

therefore the MIPS for M1

= 10 / ( 9.2304 * 10^-9 ) * 10^6  = 1083

For M2

Assumptions : number of instructions executed = 10

                        each clock cycle = 0.3846 * 10^-9.      frequency = 2.8 Ghz

first we calculate the total execution time which is equal to :

= [ (2*4) + (2*2) + (3 * 1.5 ) + ( 4 * 2.5 ) ] * 0.3846 * 10^-9 = 9.4631 * 10^-9 secs

therefor the MIPS for M2

= 10 / ( 9.4631*10^-9) * 10^6 = 1056

C ) The machine that has a better performance based on MIPS is M1 and this is by 27 million number of instructions per sec

Explanation:

The Release Train Engineer (RTE) has the main work of supporting as well as coaching the Agile Release Train (ART). They are capable of steering ART successfully and to navigate the complexity in delivering  the software in large and inter-functional environments.  

They serve the scrum master and coach teams to improve on the results.

The two actions or behaviors of the RTE are :

1. They try to create an environment of the mutual influence.  

2. Listens and also supports the teams in problem identification as well as decision-making.

The behaviors or actions that the release train engineer does include:

The release train engineer refers to a servant leader who is responsible for facilitating program-level processes and executes them.

The release train engineer is also responsible for driving continuous development, managing risks, and escalating impediments. Some of their actions include operating within the lean budget and facilitating demos.

Read related link on:

https://brainly.com/question/25254146

Answer:

the answer is A

Explanation:

Answer:

A

Explanation:

Answer:

Technical drawing, also known as drafting or draughting, is the act and discipline of making detailed drawings. These drawings visually communicate how something works or how it is to be made.

Answer:

a) ZT = [ 38.786 ∠-4.28° ] ohms

b) ZT = [ 8.673 ∠4.06° ] ohms

Explanation:

Given that;

Voltage V = 480V

three loads of 1) 10Kw, 8 kVAR

2) 5 kVA, 0.9 power factor lagging

3) 12kW, 0.95 pf leading

Now for load1

Active power P = V^2 / R

Resistance R1 = V^2 / P

= (480 * 480) / 10*1000 = 23.04 OHMS

Reactive power Q = 8kVAR

let x be reactance

Q = V^2 / x

8 * 1000 = (480 * 480) / x

x1 = j28.8 ohms (inductive)

Load 2

Active power = kVA * nf = 5 * 9 = 4.5 Kw

Reactive power = sqrt(S^2 - P^2)

= Sqrt (5^2 - 4.5^2)

= 2.179 kVAR

now Resistance R2 = V^2 / P2 = (480 * 480) / 4.5*10^3

= 51.2 ohms

Reactance X2 = V^2 / Q2 = (480 * 480) / 2.178*10^3

X2 = 105.75 ohms

x2 = -j105.75 ohms (capacitive)

Load 3

Active power = 12kW AND PF = 0.95

so

kVA rating = 12/0.95 = 12.63 kVA

reactive power Q3 = sqrt (12.63^2 - 12^2 )

= 3.94 kVAR

Resistance R3 = V^2/P3 = (480*480) / 12*10^3

= 19.2 ohms

Reactance x3 = V2 / Q3 = (480 * 480) / 3.94*10^3

x3 = 58.47 ohms

x3 = -j58.47 ohms (capacitive)

NOW

a)

series combination of R and X

1/ZT = [1/(R1 + jx1)] + [1/(R2-jx2)] + [1/(R3-jx3)]

we substitute

1/ZT = [1/(23.04 + j28.8)] + [1/(51.2 - j105.74)] + [1/(19.2-58.48)]

1/ZT = 0.02571 + j1.92688*10^-3

ZT = 38.67 -j2.897 (total impedance)

so ZT = [ 38.786 ∠-4.28° ] ohms

b)

parallel combination of R AND X

1/ZT = 1/R1 + 1/jx1 + 1/R2 + 1/(-jx2) + 1/R3 + 1/(-jX3)

1/ZT = 1/23.04 + 1/j28.8 + 1/51.2 + 1/(-j105.74) + 1/19.2 + 1/(-58.48)

1/ZT = 0.1150 - j0.008164

ZT = 8.65 +j0.6141

so ZT = [ 8.673 ∠4.06° ] ohms

Answer:

The answer is "83.98, 1889.195, and 1889.195"

Explanation:

Given value:

[tex]\bold{P_{4}=2.34 \ kPa}[/tex]

In point a:

The value of [tex]h_{f4}=83.915 \ \ \frac{Kj}{kg}\\[/tex]

[tex]V_4=0.001002 \ \ \frac{Kj}{kg}\\\\U_4= 83.98 \ \ \frac{Kj}{kg}\\\\[/tex]

In point b:

calculating heat leaves formula= [tex]h_3-h_{f4}[/tex]

                                                      [tex]= 1973.11-83.915\\\\= 1889.195 \ \ \frac{KJ}{kg}[/tex]

In point c:

calculating Heat transfer rate formula[tex]=m(h_3-h_4)[/tex]

                                                              [tex]= 1(1889.195)\\\\ = 1889.19 \ \ kw.[/tex]

Answer:

  1.39 µmol

Explanation:

(1 gal) × (44.7 µg/L)/(121.76 g/mol)(3.785411784 L/gal) = 1.39 µmol

Answer:

The answer is "[tex]\bold{e^{-t} \sin t}[/tex]"

Explanation:

Given equation:

[tex]\to x+2x+2x=0.............(a)[/tex]

Let x= e^{mt} be a solution of the equation will be:

[tex]\to m^2+2m+2=0[/tex]

compare the value with the  [tex]am^2+bm+c=0[/tex]

[tex]\to a= 1\\\to b= -2\\\to c= 2\\[/tex]

Formula:

[tex]\bold{=\frac{-b \pm \sqrt{b^2-4ac}}{2a}}\\\\=\frac{-2 \pm \sqrt{(-2)^2-4\times 1 \times 2}}{2 \times 1}\\\\=\frac{-2 \pm \sqrt{4-8}}{2}\\\\=\frac{-2 \pm \sqrt{-4}}{2}\\\\=\frac{-2 \pm 2i}{2}\\\\[/tex]

Calculate the g.s at equation (a):

[tex]\to x = e^{-t} (c_1 \cos t+ c_2 \sin t)............(b)\\\\\to x = e^{-t} (c_2 \cos t -c_1 \sin t) - e^{-t} (c_1 \cos t + c_2\sin t) ...............(c)\\\\[/tex]

[tex]_{when} \\\\\to t= 0\\\\\to x=x_0\\\\\to x= v_0 =1 \\\\ \ then \ form \ equation \ (b) \ and \ equation \ (c)[/tex]

[tex]\to a= 0 \\ \to c_2 -c_1 =1 \\ \to c_2=1[/tex]

The value of [tex]x = e^{-t} ( 0 \times \cos t + 1 \times \sin t)[/tex]

[tex]\boxed{\bold{x = e^{-t} (\sin t)}}[/tex]

Answer:

3 sec

If you are driving a 30-foot vehicle at 55 mph, how many seconds of following distance should you allow? 30ft truck. = 3 sec. Since the truck is over 40 mph.

Explanation:

Answer:

[tex]\mathbf{stress \ \sigma = 264.6 \ Mpa}[/tex]

Explanation:

From the concept of Hooke's  Law,

[tex]E =\dfrac{ stress \ \sigma}{ strain \ \varepsilon}[/tex]

where;

[tex]strain \ \varepsilon = \dfrac{change \ in \ dimension }{original \ dimension}[/tex]

[tex]strain \ \varepsilon = \dfrac{7 \ mm }{4.75 \times 10^{3} \ mm}[/tex]

[tex]strain \ \varepsilon =0.00147368[/tex]

Recall:

[tex]E =\dfrac{ stress \ \sigma}{ strain \ \varepsilon}[/tex]

[tex]stress \ \sigma = E \times { strain \ \varepsilon}[/tex]

[tex]stress \ \sigma = 180 \times 10^{3} \ Mpa \times 0.00147[/tex]

[tex]\mathbf{stress \ \sigma = 264.6 \ Mpa}[/tex]

Thus, the stress in the pipe at the maximum allowable contraction = 264.6 Mpa