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.
• Take the mass of the impactor as 30 kg and strike velocity as 2.0 m/s.
• Pick the best numerical technique, justify your choice.
• All results must be in SI units.

a) Determine a factor that will be used in the conversion from V to N. The calibration data that supplied by manufacturer as below:
• Take the mass of the impactor as 30 kg and strike velocity as 2.0 m/s.
• Pick the best numerical technique, justify your choice.
• All results must be in SI units.

a) Determine a factor that will be used in the conversion from V to N. The calibration data that supplied by manufacturer as below:• Take the mass of the impactor as 30 kg and strike velocity as 2.0 m/s.
• Pick the best numerical technique, justify your choice.
• All results must be in SI units.

a) Determine a factor that will be used in the conversion from V to N. The calibration data that supplied by manufacturer as below:
b) Compute the impulse of this event.
c) Obtain acceleration, velocity and displacement histories by using Newton’s second law of motion.
d) Compute the absorbed energy during collision.

Answer:

bye hooking plugs up to it to amp it up

Answer:

Check the explanation

Explanation:

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

Answer:

See explaination

Explanation:

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

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:

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

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:

<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:

The answer is "+9.05 kw"

Explanation:

In the given question some information is missing which can be given in the following attachment.

The solution to this question can be defined as follows:

let assume that flow is from 1 to 2 then

Q= 1kw

m=0.1 kg/s

From the steady flow energy equation is:

[tex]m\{n_1+ \frac{v^2_1}{z}+ gz_1 \}+Q= m \{h_2+ \frac{v^2_2}{2}+ gz_2\}+w\\\\\ change \ energy\\\\0.1[1.005 \times 800]-1= 0.01[1.005\times 700]+w\\\\w= +9.05 \ kw\\\\[/tex]

If the sign of the work performed is positive, it means the work is done on the surrounding so, that the expected direction of the flow is right.

Answer:

See explaination

Explanation:

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

Answer:

  12 volts

Explanation:

The voltages across parallel-connected items are identical. (In fact, that's why you can measure the voltage by connecting the voltmeter in parallel with the circuit element.)

The voltage drop across each bulb is 12 volts.

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:

d[0] = 11111111

d[1] = 11011101

d[2] = 1111011

Explanation:

Assume that the number of bits is 8. The voltage range input is -8 to 7 volts. The range is thus 15V, and the resolution is 15/2^8 = 0.0586 volts. We will first add +8 to the input to convert it to a 0-15v signal. Then find the equivalent bit representation. For 7.8 volts, the binary signal will be all 1's, since the max input voltage for the ADC is 7 volts. For 4.95, we have 4.95+8 = 12.95 volts. Thus, N = 12.95/0.0586 = 221. The binary representation is 11011101. For -0.8, we have -0.8 + 8 = 7.2. Thus, N = 7.2/0.0586 = 123. The binary representation is 1111011.

Thus,

d[0] = 11111111

d[1] = 11011101

d[2] = 1111011

Answer:

high speed steel I believe

Answer:

The volume of sludge wasted each day from the secondary classifiers is Qw = 208.33 m^3 / day

Explanation:

Check the file attached for a complete solution.

The volume of the aeration tank was first calculated, V = 5000 m^3 / day.

The value of V was consequently substituted into the formula for the wasted sludge flow. The value of the wasted sludge flow was calculated to be Qw = 208.33 m^3 / day.

Answer:

Java Code was used to define classes in the abstract discount policy,The bulk discount, The buy items get one free and the combined discount

Explanation:

Solution

Code:

Main.java

public class Main {

public static void main(String[] args) {

  BulkDiscount bd=new BulkDiscount(10,5);

BuyNItemsGetOneFree bnd=new BuyNItemsGetOneFree(5);

CombinedDiscount cd=new CombinedDiscount(bd,bnd);

System.out.println("Bulk Discount :"+bd.computeDiscount(20, 20));

  System.out.println("Nth item discount :"+bnd.computeDiscount(20, 20));

 System.out.println("Combined discount :"+cd.computeDiscount(20, 20));    

  }

}

discountPolicy.java

public abstract class DiscountPolicy

{    

public abstract double computeDiscount(int count, double itemCost);

}    

BulkDiscount.java  

public class BulkDiscount extends DiscountPolicy

{    

private double percent;

private double minimum;

public BulkDiscount(int minimum, double percent)

{

this.minimum = minimum;

this.percent = percent;

}

at Override

public double computeDiscount(int count, double itemCost)

{

if (count >= minimum)

{

return (percent/100)*(count*itemCost); //discount is total price * percentage discount

}

return 0;

}

}

BuyNItemsGetOneFree.java

public class BuyNItemsGetOneFree extends DiscountPolicy

{

private int itemNumberForFree;

public BuyNItemsGetOneFree(int n)

{

  itemNumberForFree = n;

}

at Override

public double computeDiscount(int count, double itemCost)

{

if(count > itemNumberForFree)

return (count/itemNumberForFree)*itemCost;

else

  return 0;

}

}

CombinedDiscount.java

public class CombinedDiscount extends DiscountPolicy

{

private DiscountPolicy first, second;

public CombinedDiscount(DiscountPolicy firstDiscount, DiscountPolicy secondDiscount)

{

first = firstDiscount;

second = secondDiscount;

}

at Override

public double computeDiscount(int count, double itemCost)

{

double firstDiscount=first.computeDiscount(count, itemCost);

double secondDiscount=second.computeDiscount(count, itemCost);

if(firstDiscount>secondDiscount){

  return firstDiscount;

}else{

  return secondDiscount;

}

}  

}

Answer:

σ = 391.2 MPa

Explanation:

The relation between true stress and true strain is given as:

σ = k εⁿ

where,

σ = true stress = 365 MPa

k = constant

ε = true strain = Change in Length/Original Length

ε = (61.8 - 54.8)/54.8 = 0.128

n = strain hardening exponent = 0.2

Therefore,

365 MPa = K (0.128)^0.2

K = 365 MPa/(0.128)^0.2

k = 550.62 MPa

Now, we have the following data:

σ = true stress = ?

k = constant = 550.62 MPa

ε = true strain = Change in Length/Original Length

ε = (64.7 - 54.8)/54.8 = 0.181

n = strain hardening exponent = 0.2

Therefore,

σ = (550.62 MPa)(0.181)^0.2

σ = 391.2 MPa

Answer:

14.5° ; THD % = 3.873 × 100 = 387.3%.

Explanation:

Okay, in this question we are given the following parameters or data or information which is going to assist us in solving the question efficiently and they are;

(1). "500 VAR reactive power from 230 Vrms 50 Hz 1.5 KVAR reactor".

(2). Consideration of up to 12th harmonic.

So, let us delve right into the solution to the question above;

Step one: Calculate the Irms and Irms(12th) by using the formula for the equation below;

Irms = reactive power /Vrms = 500/230 = 2.174 A.

Irms(12th) = 1.5 × 10^3/ 12 × 230 = 0.543 A.

Step two: Calculate the THD.

Before the Calculation of the THD, there is the need to determine the value for the dissociation factor, h.

h = Irms(12th)/Irms = 0.543/ 2.174 = 0.25.

Thus, THD = [1/ (h)^2 - 1 ] ^1/2. = 3.873.

THD % = 3.873 × 100 = 387.3%.

Step four: angle AC - Ac converter

theta = sin^-1 (1.5 × 10^3/ 12 × 500) = 14.5°.

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;

https://brainly.com/question/25953942

Answer:

See explaination

Explanation:

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

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 answer is "go to sheet M-3 and look for a detail labeled 7".

Explanation:

In the given question some information is missing, that is choices so, the correct choice can be described as follows:

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:

DEPTH:-- the distance from the top or surface to the bottom of something.

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

Answer:

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

Explanation:

I just gave Quiz and  its correct

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:

mass of chemical coming in per minute = 60 × 20 = 1200 mg/min

at a time t(min) , M = mass of chemical = 1200 × t mg

concentration of chemical = 1200t / 600 = 2t mg / gallon

Explanation:

Since it is only fluid that is leaving and not chemical.

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