The supplement file* that enclosed to this homework consists Time Versus Force data. The first column in the file stands for time (second) and the 2nd stands for force (Volt), respectively. This data were retrieved during an impact event. In this test, an impactor strikes to a sample. A force-ring sensor that attached to the impactor generates voltage during collision. A data acquisition card gathers the generated signals.

Answer:

See explaination

Explanation:

Please kindly check attachment for the step by step solution of the given problem

Answer:

  30√2 ≈ 42.426 volts

Explanation:

The RMS value of the signal is the DC equivalent. For a sine wave, the mean of the square is half the square of the peak value. Then the Root-Mean-Square is ...

  RMS = √((Vp)^2/2) = Vp/√2

For Vp = 60 volts, the equivalent DC voltage is ...

  V = (60 volts)/√2 = 30√2 volts ≈ 42.426 volts

Explanation:

This is not so much a mathematical issue as a case study, because the response will inevitably require us to test the special Columbic shuttle disaster scenario. I would suggest that you read this in detail and present the points accordingly. Here I give as many points as I think are relevant.

The failure of a space program is definitely a complex situation, more than a simple binomial distribution. It's definitely not as simple as repeating the flip of a coin. There are several coherent factors and situations that govern the overall coordination and execution of such an event. The problem is, those who are running a project like this are still making a trade off,It is never the case that they sealed the lid on any chance of failure between multiple parameters. You try to do something, but often, as is the case above, the potentially dangerous situation is impossible or uncontrollable. Since the root cause of failure, which is dried out tiles that can not withstand heat and water, it appears that owing to the constant use of the shuttle the head architects have not foreseen this.

Answer:

7.15

Explanation:

Firstly, the COP of such heat pump must be measured that is,

              [tex]COP_{HP}=\frac{T_H}{T_H-T_L}[/tex]

Therefore, the temperature relationship, [tex]T_H=1.15\;T_L[/tex]

Then, we should apply the values in the COP.

                           [tex]=\frac{1.15\;T_L}{1.15-1}[/tex]

                           [tex]=7.67[/tex]

The number of heat rejected by the heat pump must then be calculated.

                   [tex]Q_H=COP_{HP}\times W_{nst}[/tex]

                          [tex]=7.67\times5=38.35[/tex]

We must then calculate the refrigerant mass flow rate.

                   [tex]m=0.264\;kg/s[/tex]

                   [tex]q_H=\frac{Q_H}{m}[/tex]

                         [tex]=\frac{38.35}{0.264}=145.27[/tex]

The [tex]h_g[/tex] value is 145.27 and therefore the hot reservoir temperature is 64° C.

The pressure at 64 ° C is thus 1849.36 kPa by interpolation.

And, the lowest reservoir temperature must be calculated.

                   [tex]T_L=\frac{T_H}{1.15}[/tex]

                        [tex]=\frac{64+273}{1.15}=293.04[/tex]

                        [tex]=19.89\°C[/tex]

the lowest reservoir temperature = 258.703  kpa                    

So, the pressure ratio should be = 7.15

Answer:

0.05 mg / gallon

Explanation:

mass of chemecila coming in per minute = 50*10 = 500 mg/min

at a time t min , M = mass of chemical = 500*t mg

conecntartion of chemecal = 500t/10000 = 0.05 mg / gallon

Answer:

Its C. you may proceed, but only if the path is clear

Explanation:

I just gave Quiz and  its correct

Answer:

A) V(t) = 0.4e^-2t

B) i(t) = (25tsin5t+10) A for t>0

Explanation:

Formula for calculating voltage across an inductor is expressed as:

V = Ldi/dt

Given L = 100mH = 100×10^-3

If i(t) = 6 - 2e^-2t A t >= 0

di/dt = (-2)(-2)e^-2t

di/dt = 4e^-2t

If t ≥ 0

V(t) = 100×10^-3 × (4e^-2t)

V(t) = 0.1×4e^-2t

V(t) = 0.4e^-2t for t≥0

B) Applying the same formula as above

V = Ldi/dt

Vdt = Ldi

V/L dt = di

On integration

Vt/L = i + C

When t = 0, i = -10A

Substituting the values into the formula

V(0)/L = -10 + C

0 = -10+C

C = 10

To get the current i(t) through the inductor for t>0,

Since Vt/L = i + C

Given V(t) = 5sin5t Volts

L = 200mH = 200×10^-3H

C = 10

On substituting

(5sin5t)t/0.2 = i + 10

25tsin5t = i + 10

i(t) = (25tsin5t-10) A for t>0

Answer:

It wouldn't get any louder then maybe 3db more

Explanation:

There's even a equation if you wanted to check this out but, if they are the same generator same model and all and made the same precise noise it wouldn't increase more then 3db.

Answer:

The average heat transfer coefficient for the exhaust gas flowing inside the tube, h = 204.41 W/m^2 - K

Explanation:

The detailed solution is attached as files below.

However, the steps followed are highlighted:

1) The average temperature was calculated as 380.5 K

2) The properties of air at 380.5 K was highlighted

3) The Prandti number was calculated. Pr = 0.693

4) The Reynold number was calculated, Re = 28716.77

5) The Nusselt umber was calculated, Nu = 75.94

6) From Nu = (hD)/k , the average heat transfer coefficient, h, was calculated and a value of 204.41 W/m^2 - K was gotten.

Answer:

high speed steel I believe

Answer:

See explaination

Explanation:

Please kindly check attachment for the step by step solution of the given problem.

Answer:

0.544×10–³

Explanation:

Please see the attached file for the solution

Answer:

33.3%

Explanation:

Given that:

specific gravity (SG) = 0.89

Diameter (D) =  0.01 ft/s

Density of oil [tex]\rho= SG\rho _{h20} = 0.89 * 1.94=1.7266\frac{sl}{ft^3}[/tex]

Since the viscosity 10000 times that of water, The reynold number [tex]R_E=\frac{\rho VD}{\mu} =\frac{1.7266*0.01*0.01}{0.234}=7.38*10^{-4}[/tex]

Since RE < 1, the drag coefficient for normal flow is given as: [tex]C_{D1}=\frac{24.4}{R_E}= \frac{20.4}{7.38*10^{-4}}=2.76*10^4[/tex]

the drag coefficient for parallel flow is given as: [tex]C_{D2}=\frac{13.6}{R_E}= \frac{13.6}{7.38*10^{-4}}=1.84*10^4[/tex]

Percent reduced = [tex]\frac{D_1-D_2}{D_2} *100=\frac{2.76-1.84}{3.3}=33.3[/tex] = 33.3%

Answer:

A.) 1mv = 2000N

B.) Impulse = 60Ns

C.) Acceleration = 66.67 m/s^2

Velocity = 4 m/s

Displacement = 0.075 metre

Absorbed energy = 60 J

Explanation:

A.) Using a mathematical linear equation,

Y = MX + C

Where M = (2000 - 0)/( 898 - 0 )

M = 2000/898

M = 2.23

Let Y = 2000 and X = 898

2000 = 2.23(898) + C

2000 = 2000 + C

C = 0

We can therefore conclude that

1 mV = 2000N

B.) Impulse is the product of force and time.

Also, impulse = momentum

Given that

Mass M = 30kg

Velocity V = 2 m/s

Impulse = M × V = momentum

Impulse = 30 × 2 = 60 Ns

C.) Force = mass × acceleration

F = ma

Substitute force and mass into the formula

2000 = 30a

Make a the subject of formula

a = 2000/30

acceleration a = 66.67 m/s^2

Since impulse = 60 Ns

From Newton 2nd law,

Force = rate of change in momentum

Where

change in momentum = -MV - (- MU)

Impulse = -MV + MU

Where U = initial velocity

60 = -60 + MU

30U = 120

U = 120/30

U = 4 m/s

Force = 2000N

Impulse = Ft

Substitute force and impulse to get time

60 = 2000t

t = 60/2000

t = 0.03 second

Using third equation of motion

V^2 = U^2 + 2as

Where S = displacement

4^2 = 2^2 + 2 × 66.67S

16 = 4 + 133.4S

133.4S = 10

S = 10/133.4

S = 0.075 metre

D.) Energy = 1/2 mV^2

Energy = 0.5 × 30 × 2^2

Energy = 15 × 4 = 60J

Answer:

The idea is to protect the puddle while it cools

Explanation:

Answer:

See explaination

Explanation:

Please kindly check attachment for the step by step solution of the given problem.

Answer:

Gc(s) = [tex]\frac{0.1s + 0.28727}{s}[/tex]

Explanation:

comparing the standard approximation with the plot attached we can tune the PI gains so that the desired response is obtained. this is because the time requirement of the setting is met while the %OS requirement is not achieved instead a 12% OS is seen from the plot.

attached is the detailed solution and the plot in Matlab

Question:

The spherical pressure vessel has an inner diameter of 2 m and a thickness of 10 mm. A strain gage having a length of 20 mm is attached to it, and it is observed to increase in length by 0.012 mm when the vessel is pressurized. Determine the pressure causing this deformation, and find the maximum in-plane shear stress, and the absolute maximum shear stress at a point on the outer surface of the vessel. The material is steel, for which Eₛₜ = 200 GPa and vₛₜ = 0.3.

Answer:

See explanation below

Explanation:

Given:

d = 2m = 2*10³ = 2000

thickness, t = 10 mm

Length of strain guage = 20 mm

i) Let's calculate d/t

[tex] \frac{d}{t} = \frac{2000}{10} = 200 [/tex]

Since [tex] \frac{d}{t}[/tex] is greater than length of strain guage, the pressure vessel is thin.

For the minimum normal stress, we have:

[tex] \sigma max= \frac{pd}{4t} [/tex]

[tex] \sigma max= \frac{2000p}{4 * 20} [/tex]

= 50p

For the minimum normal strain due to pressure, we have:

[tex] E_max= \frac{change in L}{L_g} [/tex]

[tex] = \frac{0.012}{20} = 0.60*10^-^3[/tex]

The minimum normal stress for a thin pressure vessel is 0.

[tex] \sigma _min = 0 [/tex]

i) Let's use Hookes law to calculate the pressure causing this deformation.

[tex] E_max = \frac{1}{E} [\sigma _max - V(\sigma _initial + \sigma _min)] [/tex]

Substituting figures, we have:

[tex] 0.60*10^-^3 = \frac{1}{200*10^9} [50p - 0.3 (50p + 0)] [/tex]

[tex] 120 * 10^6 = 35p [/tex]

[tex] p = \frac{120*10^6}{35}[/tex]

[tex] p = 3.429 * 10^6 [/tex]

p = 3.4 MPa

ii) Calculating the maximum in-plane shear stress, we have:

[tex] \frac{\sigma _max - \sigma _int}{2}[/tex]

[tex] = \frac{50p - 50p}{2} = 0 [/tex]

Max in plane shear stress = 0

iii) To find the absolute maximum shear stress at a point on the outer surface of the vessel, we have:

[tex] \frac{\sigma _max - \sigma _min}{2}[/tex]

[tex] = \frac{50p - 0}{2} = 25p [/tex]

since p = 3.429 MPa

25p = 25 * 3.4 MPa

= 85.71 ≈ 85.7 MPa

The absolute maximum shear stress at a point on the outer surface of the vessel is 85.7 MPa

Answer:

Cause and effect

Explanation:

pls mark brainliest

The  structure that makes or turned many smiles to frowns can be regarded as compare/contrast.

The term compare/contrast  is a common terms. The act of comparing is known to be depicting the similarities, and contrasting is said to be showing differences that exist between two things.

Conclusively, from the above, we can see that it is a compare/contrast scenario as it talks about the effects of taking ice cream. It went from  smiles to frowns.

See option below

cause/effect

descriptive

compare/contrast

sequence/process

Learn more about compare/contrast from

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Answer:

<E1, E2>.

Explanation:

So, in the question above we are given that the Two Electric field vectors E1 and E2 are perpendicular to each other. Thus, we are going to have the i and the j components for the two Electric Field that is E1 and E2 respectively. That is to say the addition we give us a resultant E which is an arbitrary vector;

E = |E| cos θi + |E| sin θj. -------------------(1).

Therefore, if we make use of the components division rule we will have something like what we have below;

x = |E2|/ |E| cos θ and y = |E1|/|E| sin θ

​

Therefore, we will now have;

E = x |E2| i + y |E1| j.

The base vectors is then Given as <E1, E2>.

Answer:

Check the explanation

Explanation:

Kindly check the attached image below to see the step by step explanation to the question above.

Answer:

(a) 0.0064 kg/s

(b) 800 KPa

(c) 2.03

Explanation:

The ideal vapor compression cycle consists of following processes:

Process  1-2 Isentropic compression in a compressor

Process 2-3 Constant-pressure heat rejection in a condenser

Process 3-4 Throttling in an expansion device

Process 4-1 Constant-pressure heat absorption in an evaporator

For state 4 (while entering compressor):

x₄ = 34% = 0.34

P₄ = 120 KPa

from saturated table:

h₄ = hf + x hfg = 22.4 KJ/kg + (0.34)(214.52 KJ/kg)

h₄ = 95.34 KJ/kg

For State 1 (Entering Compressor):

h₁ = hg at 120 KPa

h₁ = 236.99 KJ/kg

s₁ = sg at 120 KPa = 0.94789 KJ/kg.k

For State 3 (Entering Expansion Valve)

Since 3 - 4 is an isenthalpic process.

Therefore,

h₃ = h₄ = 95.34 KJ/kg

Since this state lies at liquid side of saturation line, therefore, h₃ must be hf. Hence from saturation table we find the pressure by interpolation.

P₃ = 800 KPa

For State 2 (Leaving Compressor)

Since, process 2-3 is at constant pressure. Therefore,

P₂ = P₃ = 800 KPa

T₂ = 70°C (given)

Saturation temperature at 800 KPa is 31.31°C, which is less than T₂. Thus, this is super heated state. From super heated property table:

h₂ = 306.9 KJ/kg

(a)

Compressor Power = m(h₂ - h₁)

where,

m = mass flow rate of refrigerant.

m = Compressor Power/(h₂ - h₁)

m = (0.450 KJ/s)/(306.9 KJ/kg - 236.99 KJ/kg)

m = 0.0064 kg/s

(b)

Condenser Pressure = P₂ = P₃ = 800 KPa

(c)

The COP of ideal vapor compression cycle is given as:

COP = (h₁ - h₄)/(h₂ - h₁)

COP = (236.99 - 95.34)/(306.9 - 236.99)

COP = 2.03

The Ph diagram is attached

Answer:

The total amount of heat generated;Q' = 2067 Btu/h

Explanation:

We are given;

Water entering temperature;T1 = 70°F

Water leaving temperature;T2 = 105°F

average velocity of water;V = 40 ft/min

Diameter of tube;D = 0.25 in = 0.25/12 ft = 0.02083 ft

Water density;ρ = 62.1 lbm/ft³

cp = 1.00 Btu/lbm·°F.

Now, the mass flow rate of the water is calculated from;

m' = ρAV

Where ρ is density, A is area and V is velocity

Area = πD²/4 = π*0.02083²/4 = 0.00034077555 ft²

m' = 62.1 * 0.00034077555 * 40

m' = 0.8465 lbm/min

Converting to lbm/hr = 0.8465 * 60 = 50.79 lbm/hr

From energy balance equation, we have;

E_in = E_out

So,

Q_in,w + m'h1 = m'h2

Q_in,w = m'h2 - m'h1

Q_in,w = m'(h2 - h1)

Now, m'(h2 - h1) can be written as;

m'cp(T2 - T1).

Thus ;

Q_in,w = m'cp(T2 - T1)

Plugging in the relevant values, we have;

Q_in,w = (50.79*1)(105 - 70)

Q_in,w = 1777.65 Btu/h

We are told that remaining 86 percent of heat generaged is removed by the cooling water. Thus;

The total amount of heat generated could be defined as;

Q' = Q_in,w/0.86

Q' = 1777.65/0.86

Q' = 2067 Btu/h

Answer:

To counter the behavior of a high Mach individual in my workplace, I could put the individual in charge of a project by himself, and don't let others work with him.

Explanation:

A high Mach individual is one who exhibits a manipulative and self-centered behavior.   The personality trait is characterized by the use of manipulation and persuasion to achieve power and results.  But, such individuals are hard to be persuaded.  They do not function well in team settings and asking them to agree to terms is very difficult.  "The presence of Machiavellianism in an organisation has been positively correlated with counterproductive workplace behaviour and workplace deviance," according to wikipedia.com.

Mach is an abbreviation for Machiavellianism.  Machiavellianism is referred to in psychology as a personality trait which sees a person so focused on their own interests that they will manipulate, deceive, and exploit others to achieve their selfish goals.  It is one of the Dark Triad traits.  The others are narcissism and psychopathy, which are very dangerous behaviors.

Answer:

all this could led to the failure of the garden hose and the tear along the length

Explanation:

For the flow of water to occur in any equipment, water has to flow from a high pressure to a low pressure. considering the pipe, water is flowing at a constant pressure of 30 psi inside the pipe which is assumed to be higher than the allowable operating pressure of the pipe. but the greatest change in pressure will occur at the end of the hose because at that point the water is trying to leave the hose into the atmosphere, therefore the great change in pressure along the length of the hose closest to the end of the hose will cause a tear there. also the other factors that might lead to the failure of the garden hose includes :

hoop stress ( which acts along the circumference of the pipe):

αh = [tex]\frac{PD}{2T}[/tex]     EQUATION 1

and Longitudinal stress ( acting along the length of the pipe )

αl = [tex]\frac{PD}{4T}[/tex]       EQUATION 2

where p = water pressure inside the hose

          d = diameter of hose, T = thickness of hose

we can as well attribute the failure of the hose to the material used in making the hose .

assume for a thin cylindrical pipe material used to be

[tex]\frac{D}{T}[/tex] ≥  20

insert this value into equation 1

αh = [tex]\frac{20 *30}{2}[/tex]  = 60/2 = 30 psi

the allowable hoop stress was developed by the material which could have also led to the failure of the garden hose

Answer:

Check the explanation

Explanation:

Kindly check the attached image below to see the step by step explanation to the question above.

Answer: Technician A is correct.

Explanation:

Technician A is correct because temperature of a tire will affect its pressure reading.

Tires attract heat because of their dark colour and then motion on the road generates heat. A car owner or technician should know that tire pressure is most accurate when the tire is cold (especially when atmospheric temperature is cool too).

If the tires are driven three miles first, their temperature will be high (due to the rubbing of the tires on the surface of the road). This higher temperature will result in higher per square inch (psi) readings.

Temperature has a great influence on the tire pressure.

Even if the tire is driven up to or more than 3 miles, it should still be left to cool, before tire pressure is checked.

The tire manufacturer's rating should be the maximum possible tire pressure. If an abnormal reading is gotten, the gauge should be properly checked.

Answer:

See explaination

Explanation:

Please kindly check attachment for the step by step solution of the given problem.

Answer:

bye hooking plugs up to it to amp it up

Answer:

Explanation:

a) Show how two 2-to-1 multiplexers (with no added gates) could be connected to form a 3-to-1 MUX. Input selection should be as follows: If AB = 00, select I0 If AB = 01, select I1 If AB = 1− (B is a don’t-care), select I2

We are to show how Two-2-to -1 multiplexers could be connected to form 3-to-1 MUX

If AB = 00 select [tex]I_o[/tex]

If AB = 01 select [tex]I_1[/tex]

If AB = 1_(B is don't care), select [tex]I_2[/tex]

However, the truth table is attached and shown in the first file below.

Also, the free- body diagram for 2- to - 1 MUX is shown in the second diagram attached below.

b) We are show how  two 4-to-1 and one 2-to-1 multiplexers could be connected to form an 8-to-1 MUX with three control inputs.

The perfect illustration showing how they are connected in displayed in the third free-body diagram attached below.

Where ; [tex]I_o , I_1, I_2, I_3, I_4, I_5, I_6, I_7[/tex] are the inputs of the multiplexer and Z is the output.

c)  Show how four 2-to-1 and one 4-to-1 multiplexers could be connected to form an 8-to-1 MUX with three control inputs.

For  four 2-to-1 and one 4-to-1 multiplexers could be connected to form an 8-to-1 MUX with three control inputs, we have a perfect illustration of the diagram in the last( which is the fourth) diagram attached below.

Where ; [tex]I_o , I_1, I_2, I_3, I_4, I_5, I_6, I_7[/tex] are the inputs of the multiplexer and Z is the output

This question is a multiplexer which is a topic in digital circuit.

Multiplexer is a type of combination circuit that consist of a maximum of [tex]2^n[/tex] data inputs 'n' selection lines and single output line. One of these data inputs will be connected to the output based on the values of selection lines. Another name for multiplexers is MUX.

If we have 'n' selection lines, we will get [tex]2^n[/tex] possible combinations zero and ones. Each combination will select a maximum of only one data input.

a)

Two 2-to-1 multiplexers to form a 3-to-1 MUX.

If AB = 00, select [tex]I_o[/tex]

If AB = 01, select [tex]I_1[/tex]

If AB = 1- (B is don't care) select I

The truth table for the above scenario is in the attached document below.

Figure 1 and 2 represents the solution to this question.

b).

Two 4-to-1 multiplexers and one 2-to-1 multiplexers and one 2-to-1 multiplexers are used to form an 8-to-1 MUX.

In the attached diagram, figure 3 shows a comprehensive detail of how it is structured.

Where [tex]I_o[/tex] to [tex]I_7[/tex] are the inputs of the multiplexer and Z is the output.

c) Four 2-to-1 multiplexers and one 4-to -1 multiplexer are used to form 8-to-1 MUX.

In the attached diagram, figure 4 shows how it is structured.

We would see that [tex]I_o[/tex] to [tex]I_7[/tex] are the inputs of the multiplexer and Z is the output in the system.

Learn more about multiplexers here;

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