A specimen of a 4340 steel alloy with a plane strain fracture toughness of 54.8 MPa (50 ksi ) is exposed to a stress of 2431 MPa (352600 psi). Assume that the parameter Y has a value of 0.94. (a) If the largest surface crack is 0.2 mm (0.007874 in.) long, determine the critical stress . Enter your answer in accordance to the question statement MPa (b) Will this specimen experience fracture

Answer: (b)

Explanation:

Given

Original length of the rod is [tex]L=100\ cm[/tex]

Strain experienced is [tex]\epsilon=82\%=0.82[/tex]

Strain is the ratio of the change in length to the original length

[tex]\Rightarrow \epsilon =\dfrac{\Delta L}{L}\\\\\Rightarrow 0.82=\dfrac{\Delta L}{100}\\\\\Rightarrow \Delta L=82\ cm[/tex]

Therefore, new length is given by (Considering the load is tensile in nature)

[tex]\Rightarrow L'=\Delta L+L\\\Rightarrow L'=82+100=182\ cm[/tex]

Thus, option (b) is correct.

Answer:

[tex]U=\pm 0.382N[/tex]

Explanation:

From the question we are told that:

Resolution: 0.25 N

Range: 0 to 100 N

Linearity error: within 0.20 N over range

Hysteresis error: within 0.30 N over range

Generally the equation for Stage Uncertainty is mathematically given by

[tex]U=\sqrt{u_0^2+u_T^2}[/tex]

Where

[tex]u_0=Zero\ order\ uncertainty[/tex]

[tex]u_0=\pm 0.5*0.25[/tex]

[tex]u_0=\pm=0.125[/tex]

And

u_T=Total instrumental Uncertainty

[tex]u_T=\sqrt{l_e^2+h_e^2}[/tex]

Where

l_e=Error of linearity

h_e=Error due to hysteresis

Hence

[tex]u_T=\sqrt{0.20^2+0.30^2}[/tex]

[tex]u_T=\pm 0.36[/tex]

Therefore

[tex]U=\sqrt{(0.125)^2+0.36^2}[/tex]

[tex]U=\pm 0.382N[/tex]

Answer:

The correct answer is "99.9%".

Explanation:

According to the information given in the question,

[tex]P(1 \ fail) = 0.1[/tex]

The probability of all fail will be:

[tex]P(all \ fail) = (0.1)^3[/tex]

                  [tex]=0.001[/tex]

hence,

[tex]P(not \ all \ fail)= 1-P(all \ fail)[/tex]

                        [tex]=0.999[/tex]

                        [tex]=99.9[/tex] (%)

Thus the above is the right answer.

Answer:

320 m

Explanation:

To find the side length of the current park, x, we solve the quadratic equation for the area of the park

x² + 200x = 166400

x² + 200x - 166400 = 0

We multiply -166400 by x² to get -166400x². We now find the factors of 166400x² that will add up to 200x. These factors are -320x and 520x

So, we re-write the expression as

x² + 200x - 166400 = 0

x² + 520x - 320x - 166400 = 0

We write out the factors of the expression,

x² + 520x - 320x - 320 × 520 = 0

Factorizing the expression, we have

x(x + 520) - 320(x + 520) = 0

(x + 520)(x - 320) = 0

x + 520 = 0 or x - 320 = 0

x = -520 or x = 320

Since x is not negative, we take the positive answer.

So, x = 320 m

Answer:

Newton per square meter (N/m2)

Explanation:

Required

Unit of ultimate tensile strength

Ultimate tensile strength (U) is calculated using:

[tex]U = \frac{Ultimate\ Force}{Area}[/tex]

The units of force is N (Newton) and the unit of Area is m^2

So, we have:

[tex]U = \frac{N}{m^2}[/tex]

or

[tex]U = N/m^2[/tex]

Hence: (c) is correct

Answer:

[tex]V_z=9.1v[/tex]

[tex]V_{zo}=8.74V[/tex]

[tex]I=10mA[/tex]

[tex]R=589 ohms[/tex]

Explanation:

From the question we are told that:

Zener diode Voltage [tex]V_z=9.1-V[/tex]

Zener diode Current [tex]I_z=9 .A[/tex]

Note

[tex]rz = 40\\\\IZK= 0.5 mA[/tex]

Supply Voltage [tex]V_s=15[/tex]

Reduction Percentage [tex]P_r= 50 \%[/tex]

Generally the equation for Kirchhoff's Voltage Law is mathematically given by

[tex]V_z=V_{zo}+I_zr_z[/tex]

[tex]9.1=V_{z0}+9*10^{-3}(40)[/tex]

[tex]V_{zo}=8.74V[/tex]

Therefore

[tex]At I_z-10mA[/tex]

[tex]V_z=V_{z0}+I_zr_z[/tex]

[tex]V_z=8.74+(10*10^{-3}) (40)[/tex]

[tex]V_z=9.1v[/tex]

Generally the equation for Kirchhoff's Current Law is mathematically given by

[tex]-I+I_z+I_l=0[/tex]

[tex]I=10mA+\frac{V_z}{R_l}[/tex]

[tex]I=10mA+\frac{9.1}{0}[/tex]

[tex]I=10mA[/tex]

Therefore

[tex]R=\frac{15V-V_z}{I}[/tex]

[tex]R=\frac{15-9.1}{10*10^{-3}}[/tex]

[tex]R=589 ohms[/tex]

Answer:

i) when circuit is unloaded : R1 + R2 = 2kΩ.

ii) when 5V output voltage is applied :  R1 = 1 kΩ , R2 =  1 kΩ

iii) when 2.5 v output voltage is applied : R1 = 1500 Ω,  R2 = 500 Ω

iv) when: R1 = 1 kΩ , R2 =  1 kΩ  is connected in parallel output voltage < 5 V

When : R1 = 1500 Ω, R2 = 500 Ω is connected in parallel output voltage > 2.5V

Explanation:

Current drawn from source under loaded condition ≤ 5 mA

source voltage = 10 v ,  required output = 5 v , 2.5 v

attached below is the sketch of the circuit

Resistance values

i) when the circuit is unloaded

Req = R1 + R2 = 2 kΩ  ( Req = Vs / I  = 10 / 5*10^-3 = 2 kΩ )

ii) when  output voltage = 5 v

we will apply voltage divider rule

R1 = 1 kΩ ,

R2 =  1 kΩ

iii) When the output voltage = 2.5 v

applying voltage divider rule

R1 = 1500 Ω

R2 = 500 Ω

iv) when the load is connected to each tap one at a time

i.e. when the resistance are in parallel

when: R1 = 1 kΩ , R2 =  1 kΩ  is connected in parallel output voltage < 5 V

When : R1 = 1500 Ω, R2 = 500 Ω is connected in parallel output voltage > 2.5V

attached below is the detailed solution to the given problem

Answer:

mark me as a brainleast

Explanation:

209781

Answer:

Digestion functions become more active

Explanation:

I just took the text!

In heavy traffic areas, you should wave pedestrians across the street if there is no crosswalk: False.

A crosswalk can be defined as the marked or specially paved part of a road that is characterized by heavy traffic, so as to enable pedestrians have right of way to cross the street because drivers are required by traffic law to stop for them.

However, a driver or other road users in heavy traffic areas shouldn't wave pedestrians across the street if there is no crosswalk

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Answer:

The answer is below.

Explanation:

Some of the ways, how I think elliptic cryptography is more advanced than RSA are the following:

1. ECC - Elliptic Curve Cryptography uses smaller keys for the same level of security, particularly at greater levels of security.

2. ECC can work well and at a faster rate on a small-capacity device compared to RSA

3. It uses offer speedier SSL handshakes that enhance security

4. It offers fast signatures

5. It allows signatures to be computed in two stages, which enables lower latency than inverse throughput.

6. Relatively quick encryption and decryption

Answer:

Technician B only.

Explanation:

It is not necessary that all electric motors will use Direct current, some may also use Alternative current. Some electric motors that use DC may use brushes. There are two types of brushless motors that use alternative current.

Answer: check answers in pictures (2 p)

The resistance of the shunt required for a full-scale deflection of 20 A is 2.528 Ohm.

Resistance is a measure of an electrical circuit's resistance to current flow. Resistance is measured in ohms, which is represented by the Greek letter omega.

Full-scale deflection current = 0.015 A

Full-scale deflection current for 20 A = 20/0.015 = 1333.33 times full-scale deflection current

The current through the meter coil is given by:

Ic = Im * (Rm / (Rm + Rs))

At full scale, Ic = 1333.33 * 0.015 A = 20 A, Rm = 1.5 Ohm, and Rs = 3.5 Ohm.

Therefore:

20 A = Im * (1.5 Ohm / (1.5 Ohm + 3.5 Ohm))

Im = 20 A * (3.5 Ohm / 5 Ohm)

Im = 14 A

The current through the shunt resistor is therefore:

Ish = Im - Ic

Ish = 14 A - 20 A

Ish = -6 A

Since the shunt resistor is in parallel with the meter, its resistance can be calculated using the following formula:

Rs = Rm * (Im / Ish - 1)

Substituting the values, we get:

3.5 Ohm = 1.5 Ohm * (14 A / (-6 A) - 1)

3.5 Ohm = 1.5 Ohm * (-1.333 - 1)

3.5 Ohm = 1.5 Ohm * (-2.333)

Rs = 1.5 Ohm * (14 A / (-6 A) - 1) / (-2.333)

Rs = 2.528 Ohm

Therefore, the resistance of the shunt required for a full-scale deflection of 20 A is 2.528 Ohm.

Now, let's find the resistance required for a full-scale deflection of 250 V:

Full-scale deflection current = 0.015 A

Resistance required for full-scale deflection of 250 V = 250 V / 0.015 A

Resistance required for full-scale deflection of 250 V = 16666.67 Ohm

This resistance is in parallel with the meter coil, so the total resistance in the circuit will be:

Rtotal = Rm || Rext

Rtotal = (Rm * Rext) / (Rm + Rext)

Substituting the values, we get:

Rtotal = (1.5 Ohm * 16666.67 Ohm) / (1.5 Ohm + 16666.67 Ohm)

Rtotal = 1.43 Ohm

This resistance is in series with the meter coil, so the total resistance in the circuit will be:

Rtotal = Rm + Rext + Rs

Substituting the values we get:

1.43 Ohm = 1.5 Ohm + 16666.67 Ohm + Rs

Rs = 1.43 Ohm - 1.5 Ohm - 16666.67 Ohm

Rs = -16666.74 Ohm

Thus, the calculated value of Rs is negative, it means that the shunt resistor must be used instead of a series resistor to achieve a full-scale deflection of 250 V.

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Answer:

[tex]W= 208 KJ[/tex]

Explanation:

From the question we are told that:

Take in Temperature [tex]T_1=400C=>673K[/tex]

Take Out Temperature  [tex]T_2=120+>393K[/tex]

Heat [tex]Q=500kJ[/tex]

Generally the equation for Carnot Engine Efficiency is mathematically given  by

 [tex]n_c=\frac{T_1-T_2}{T_1}[/tex]

 [tex]n_c=\frac{673-393}{673}[/tex]

 [tex]n_c=0.4[/tex]

Where

 [tex]n_c=\frac{W}{Q}[/tex]

 [tex]W = Q*n_c[/tex]

 [tex]W= 500 * 0.4[/tex]

 [tex]W= 208 KJ[/tex]

Answer:

Required pump power = - 11.205 kW

Explanation:

Pi = 600 kPa

Zi ( initial height ) = - 5m

P2 = 200 kPa

Ze ( top floor above ground level ) = 150m

Flowrate = 0.01 m^3/s = 10 kg/s

Temperature = 10°C  ( assumed value )  remains constant

calculate the value of pumping power required ( i.e. work done )

Applying the energy equation

Hi + 1/2(vi)^2 + gZi = He + 1/2(Ve)^2 + gZe + W

given negligible internal/kinetic  energy  difference in "h's = the pv terms

W = ( Pi Vi - Pe V2 ) + g( Zi - Z2 )

   = ( (600 * 0.001)  - (200 * 0.001) ) + (9.81/1000) ( - 5 - 150 )

   = - 1.1205 Kg/kJ

required pump power = flowrate * -1.1205

                                     = 10 * ( - 1.1205 ) = - 11.205 kW

Answer:

Lack of oxygen

Decompression sickness

temperature variation

lack of gravity

Cosmic radiations hazards

motion sickness

Explanation:

When an astronaut travels in the space he is aware of the challenges he might face during his journey. There are various test and rehearsals of travelling before a final travel takes place. An astronaut goes through many challenges which includes lack of oxygen supply, decompression and motion sickness. There is no gravity in space so an astronaut will have to be aware of the difficulties he might face during his travel. An hour on earth is 7 years long in space, so an astronaut should have patience and be able to deal with time variation.

Answer:

ROLAP  is a branch of OLAP that is used to contain Relational database ( RDB ). which is a very fast database ( quick process of queries )

Explanation:

ROLAP ( Relational On-line Analytical processing ) is a branch of OLAP that is used to contain Relational database ( RDB ).

Advantages of ROLAP ( reasons for the use of ROLAP )

Answer:

[tex]V_{pp}=2V[/tex]

Explanation:

Source Voltage [tex]V= 120V[/tex]

Frequency [tex]f=60Hz[/tex]

Peak output voltage [tex]Vp=15V[/tex]

Peak Output Voltage with filter [tex]V_p'=14V[/tex]

Generally the equation for Peak to peak voltage is mathematically given by

[tex]V_p'=V_p-\frac{V_{pp}}{2}[/tex]

Therefore

[tex]V_{pp}=2(V_p-v_p')[/tex]

[tex]V_{pp}=2(15-14)[/tex]

[tex]V_{pp}=2V[/tex]

Answer:

0.75

Explanation:

From this question above we have the following information

A = probability of investment in mutual fund= 0.6

B = probability of investment in government fund = 0.3

C = probability of investing in both the mutual fund and the government= 0.15

Where to find the probability of this investor investing in either of these two

= Prob(a) + prob(b) - prob(c)

= 0.6 + 0.3 - 0.15

= 0.9 - 0.15

= 0.75

If the integral is

[tex]\displaystyle \int \frac{\ln(3x^2)}{x^5}\,\mathrm dx[/tex]

substitute u = ln(3x ²) and du = 6x/(3x ²) dx = 2/x dx.

Then x ² = exp(u)/3 and x ⁴ = exp(2u)/9.

The integral is transformed to

[tex]\displaystyle \int \frac{\ln(3x^2)}{x^5}\,\mathrm dx = \int \frac{\ln(3x^2)}{2x^4} \times \dfrac2x \,\mathrm dx \\\\ = \int \frac{u}{2\times\dfrac{e^{2u}}9}\,\mathrm du \\\\ = \frac92 \int ue^{-2u}\,\mathrm du[/tex]

Integrate by parts:

[tex]f = u \implies \mathrm df = \mathrm du \\\\ \mathrm dg = e^{-2u}\,\mathrm du \implies g = -\dfrac12 e^{-2u}[/tex]

[tex]\displaystyle \int ue^{-2u}\,\mathrm du = fg - \int g\,\mathrm df \\\\ = -\dfrac12 ue^{-2u} + \displaystyle \frac12 \int e^{-2u}\,\mathrm du \\\\ = -\frac12 ue^{-2u} - \frac14 e^{-2u} + C[/tex]

Then

[tex]\displaystyle \frac92 \int ue^{-2u}\,\mathrm du = -\frac94 ue^{-2u} - \frac98 e^{-2u} + C[/tex]

which in terms of x would be

[tex]\displaystyle \int \frac{\ln(3x^2)}{x^5}\,\mathrm dx = -\frac94\times\frac{\ln(3x^2)}{9x^4} - \frac98 \times \frac1{9x^4} + C \\\\ = \boxed{-\frac{\ln(3x^2)}{4x^4}-\frac1{8x^4}+C}[/tex]

i have made notes and saved it as a pdf u can take it to answer question and make ur concept good

The minimum length of vertical curve that will provide 220 m stopping sight distance is; 458.8 m

Stopping sight distance; S = 220 m

Design Speed; V = 110 km/h

Intersection grade 1; G1 = +2.7

Intersection Grade 2; G2 = -3.5

From the table, at S = 220 m and V = 110 km/h, we can see that;

Radius of vertical curvature; K = 74

A = G1 - G2

A = 2.7 - (-3.5)

A = 2.7 + 3.5

A = 6.2

L = KA

Thus;

L = 74 × 6.2

L = 458.8 m

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Answer:

t = 6179.1 s = 102.9 min = 1.7 h

Explanation:

The energy provided by the resistance heater must be equal to the energy required to boil the water:

E = ΔQ

ηPt = mH

where.

η = efficiency = 84.5 % = 0.845

P = Power = 2.61 KW = 2610 W

t = time = ?

m = mass of water = 6.03 kg

H = Latent heat of vaporization of water = 2.26 x 10⁶ J/kg

Therefore,

(0.845)(2610 W)t = (6.03 kg)(2.26 x 10⁶ J/kg)

[tex]t = \frac{1.362\ x\ 10^7\ J}{2205.45\ W}[/tex]

t = 6179.1 s = 102.9 min = 1.7 h

Answer:

The new length of the rod is 182 cm.

Explanation:

Given that a rod that was originally 100-cm-long experiences a strain of 82%, to determine what is the new length of the rod, the following calculation must be performed:

100 x 1.82 = X

182 = X

Therefore, the new length of the rod is 182 cm.

Answer:

A) 600.8 m/s

B) (i) 22.5 cm^2   (ii) 7.11 cm^2

Explanation:

Given data :

P1 = 20 bar ,  T1 = 2808°C

P2 = 7 bar ,  T2 = 1808°C

mass flow rate = 1.5 kg/s

Using the superheated vapor region in Table A-4

h1 = 2976.4 KJ/kg , v1 = 0.1200 m^3/kg

h2 = 2799.1 KJ/kg , v2 = 0.2847 m^3/kg  

A) calculate exit velocity  ( m/s )

given that we are to neglect heat transfer and potential energy

V2 ( exit velocity ) = ( V1^2 + 2 (√h1 - h2 )

                             = [ (80)^2 + 2 ( √ 2976.4 - 2799.1 )

                             = 600.8 m/s

B) calculate the inlet and exit flow areas ( cm^2 )

i) Inlet flow area

A1 = ( m * v1 ) / V1

    = ( 1.5 * 0.1200 )/ 80  = 22.5 cm^2

ii) exit flow area

A2 = ( m * v2 ) / V2

    = ( 1.5 * 0.2847 ) / 600.8  = 7.11 cm^2

Answer:

a) [tex]\mu=3.3[/tex]

b)  [tex]Q=1440.7KJ/Kmol[/tex]

c)  [tex]W=1872.9KJ/Kmol[/tex]

Explanation:

From the question we are told that:

Initial Temperature [tex]T_1=325k[/tex]

initial Pressure [tex]P_1=2 bar[/tex]

Final Pressure [tex]P_2=4 bar[/tex]

iso-thermal expansion [tex]T_2=250k[/tex]

a)

Generally the equation for Coefficient of performance is mathematically given by

[tex]\mu=\frac{T_2}{T_1-T_2}[/tex]

[tex]\mu=\frac{250}{325-250}[/tex]

[tex]\mu=3.3[/tex]

b)

Generally the equation for Heat Expansion is mathematically given by

[tex]Q=RT_2 In(\frac{P_2}{P_1})[/tex]

Where

R=Gas constant

[tex]R=8.314462618[/tex]

Therefore

[tex]Q=8.314462618*250 In(\frac{4}{2})[/tex]

[tex]Q=1440.7KJ/Kmol[/tex]

c)

Generally the equation for work input is mathematically given by

[tex]W=RT_1 In(\frac{P_2}{P_1})[/tex]

[tex]W=8.314462618*250 In(\frac{4}{2})[/tex]

[tex]W=1872.9KJ/Kmol[/tex]

The coefficient of performance is 3.33, the heat transfer in the isothermal expansion is 1440.71kJ/K.mol and the work input is calculated as 1872.92kJ/K.mol

Given Data:

Isothermal expansion occurs at 250K.

This is calculated as

COP =[tex]\frac{T_2}{T_1-T_2}=\frac{250}{325-250} =3.33[/tex]

[tex]Q = RT_2In(\frac{p_2}{p_1})[/tex]

Therefore; In isothermal process du = 0

R = 8.314 AkJ/K.mol

Q = 8.314 * 250 In(4/2)

Q = 1440.71kJ/K.mol

W[tex]_i_n[/tex]=[tex]RT_1In(\frac{p_2}{p_1})\\W_i_n=8.314*325In(4/2)\\W_i_n=1872.92kJ/K.mol[/tex]

The work input is 1872.92kJ/K.mol

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Answer:

very good thx

Explanation:

Answer:

I think ( MOV Metal oxide varistors )

Transients (surges) on a line can cause spikes or surges of energy that can damage delicate electronic components. A SPD device contains one or more MOV Metal oxide varistors than bypass and absorb the energy of the transient.

Answer:

. 1. Type of building

2. Location of emergency

3. Extent of emergency

Explanation:

1. The kind of building in which people find themselves could be a factor that can be used to make this decision. a lot of buildings can be easily affected by disasters such as explosions or tornadoes the extent of the effect is dependent on how the building is constructed. in some situations it is better to shelter in, while in others it is best to evacuate.

2. Another factor to be considered is the location or area where this is happening it is good to consider this so that people can be safely moved given that help can be easily accessed or if best to stay in.

3. The last is the extent of what is happening. The risk involved is one way of making the choice to evacuate or to stay.

The factors that affect the decision on whether to evacuate or shelter in place include:

It should be noted that the building where an individual lives play a vital role during emergencies. Buildings that have poor foundations can easily be affected during emergencies.

Another factor that should be considered is the location where the emergency is taking place. Lastly, the extent of the emergency can determine if the person should stay or not.

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Answer:

[tex]l=24mm[/tex]

Explanation:

From the question we are told that:

Plane strain fracture toughness of [tex]T=55 MPa-m1/2[/tex]

Y value [tex]Y=1.0[/tex]

Stress level of[tex]\sigma =200 MPa[/tex]

Generally the equation for length of a surface crack is mathematically given by

[tex]l=\frac{1}{\pi}(\frac{T}{Y*\sigma})^2[/tex]

[tex]l=\frac{1}{3.142}(\frac{55}{1*200})^2[/tex]

[tex]l=0.024m[/tex]

Therefore

in mm

[tex]l=24mm[/tex]