a. Name the major strengthening mechanisms in metals and explain the working principle under each mechanism.Give the relevant equations corresponding to the mechanisms.
b. Briefly explain why small-angle grain boundaries are not as effective in interfering with the slip process as are high-angle grain boundaries.

Answer: True

Explanation:

The statement that "a masonry chimney should be braced with horizontal metal straps every few feet against the structure and into reinforced points such as wall studs to stabilize the chimney from the shaking force of an earthquake" is true.

In a scenario whereby the chimney isn't braced with the horizontal metal straps every few feet, this can lead to its collapse in case of an earthquake. Therefore, the correct option is "true".

Answer:

25 mins into hours = 0.416667 hours

25 inches as mm = 635

Explanation:

Answer:

a. True

b. False

Explanation:

a. Since true stress, σ' = σ(1 + ε) where σ = engineering stress and ε = engineering strain.

Also under tension ε > 0, so, (1 + ε) > 1

Since (1 + ε) = σ'/σ > 1, ⇒ σ' > σ

So, the true stress is greater than the engineering stress.

So, the statement is true

b. Creep is a time-dependent deformation under certain applied load.

Creep occurs at high temperatures under different types of stress.

But, Creep test is carried out at constant high temperature and constant stress.

This statement is false.

Solution :

Given data :

p = 315612 Pa

[tex]$V_1=7.07 \ m/sec$[/tex]

At exit of B,

p = [tex]$P_{atm}$[/tex]

[tex]$V_B = 26.1 \ m/sec$[/tex]

At exit of A,

[tex]p=P_{atm}[/tex]

[tex]$V_{A} = 26.1 \ m/s$[/tex]

We need to determine X component of force ([tex]$R_x$[/tex]) to hold in its place.

From figure,

[tex]$\sum F_x = m_0'V_{0x} - m_iV_{ix} $[/tex]

[tex]$=F_x+P_1A_1\sin 30=-mVA-mV_1 \sin 30$[/tex]

[tex]$=F_x=-pA_1\sin 30-m_AV_AA-m_B \sin30$[/tex]

Substitute all the values,

[tex]$=F_x=[-315612 \times \frac{\pi}{4}(0.3)^2 \sin 30]-[26.1 \times 1000 \times 26.1 \frac{\pi}{4}(0.1)^2]-[7.07 \times 1000\times 0.5 \sin 30]$[/tex][tex]$=F_x = -11154.64-5350.21-1767.28$[/tex]

[tex]$F_x = -18.2733 \ kN$[/tex]

Therefore, the force required to hold the nozzle in its place along horizontal direction.

[tex]$F_x = -18.2733 \ kN$[/tex]

Answer:

Đường dây siêu cao áp 500kV: Những chuyện giờ mới kể ... ​Ngày 27/5/1994, hệ thống đường dây điện siêu cao áp 500kV Bắc - Nam chính thức đưa ... Tại thời điểm đó, các nước như Pháp, Úc, Mỹ khi xây dựng đường dây dài nhất ... và chế ra các máy kéo dây theo đặc thù công việc của từng đơn vị.

Explanation:

This question is incomplete, the complete question is;

To convert from the U.S. Customary (FPS) system of units to the SI system of units. A first-year engineering student records three separate measurements as 653 lb, 69.0 mi/h, and 293 × 10⁶ ft². Suppose this engineering student has to turn in the results, but the professor only accepts results given in SI units.  

Required:

What is the area measurement, 293 × 10⁶ ft², in SI units?

293 × 10⁶ ft² = ?km²

Answer:

the area measurement is  27.221 km²

Explanation:

Given the data in the question;

What is the area measurement, 293 × 10⁶ ft², in SI units

we are to the result of the measured area from ft² to km²

we know that;

1 meter = 3.2808 ft

1 km = 1000 m

1 ft = (1 / 3.2808)m

1 m = ( 1/1000 ) km

since our measured are is 293 × 10⁶ ft²

hence

A = 293 × 10⁶ × [ (1 / 3.2808)m ]²

A = 27221252.74 m²

A = 27221252.74 × [ ( 1/1000 ) km ]²

A = 27.221 km²

Therefore, the area measurement is  27.221 km²

Answer: hello attached below is the question properly written

a) 2670 Kw

b) product will be made up of vapor and liquid

c) Product will be a super cooled liquid

Explanation:

mass Flow rate ( m ) = 175 kg/min

pressure = 1 atm

molecular weight of ethylene glycol ( mw ) = 62.07 g/mol

enthalpy of vaporization ( ΔHv ) = 56.9 KJ/mol

Using values from the table 8.1 related to the question

a) Determine the rate at which heat must be transferred from condenser

Using values from the table 8.1 related to the question

ΔH = 2670 Kw

b) If heat is transferred  at a lower temperature the product will be made up of vapor and liquid

c) If heat was transferred at a higher temperature the product will be a super cooled liquid

Answer:

1 NAND gate

Explanation:

The minimum number of 2 input NAND gates that can be used to implement the combinational circuit = 1

The only true combinations conditions that can produce a false result ( i.e. condition/result different from the expected result as stated in the question )

Sensor 2 activated + Emergency switch pressed = False ( Line will keep moving )

Answer:

bood ekogcd gcaerh is an American fbnuxc

78950W the answer

Explanation:

A 75- kw, 3-, Y- connected, 50-Hz 440- V cylindrical synchronous motor operates at rated condition with 0.8 p.f leading. the motor efficiency excluding field and stator losses, is 95%and X=2.5ohms. calculate the mechanical power developed, the Armature current, back e.m.f, power angle and maximum or pull out torque of the motor

A 75- kw, 3-, Y- connected, 50-Hz 440- V cylindrical synchronous motor operates at rated condition with 0.8 p.f leading. the motor efficiency excluding field and stator losses, is 95%and X=2.5ohms. calculate the mechanical power developed, the Armature current, back e.m.f, power angle and maximum or pull out torque of the motor

False, Good housekeeping eliminates accident and fire hazards. It also maintains safe, healthy work conditions; saves time, money, materials, space, and effort; improves productivity and quality; boosts morale; and reflects an image of a well-run, successful organization.

Hope it helps you❤️

Answer:

For small temperature difference between a body and its surrounding, the rate of cooling of the body is directly proportional to the temperature difference and the surface area exposed. qf = final temperature of object

Explanation:

hope this helps you sorry if it doesn’t help you

Answer:

The basic difference between Total cost and total revenue is that the total cost includes the total expenditure incurred on the production of a commodity whereas total revenue refers to the money received from selling that commodity.

Explanation:

Answer:

here you go.

screenshot 2 should give you some basic idea

Answer:

The answer is below

Explanation:

a) The work done is equal to the loss in kinetic energy (KE)

Change in kinetic energy (ΔKE) = Final kinetic energy - initial kinetic energy

Final KE = [tex]\frac{1}{2}mv_f^2[/tex]

But the final velocity is 0 (at rest). Hence:

Final KE = [tex]\frac{1}{2}mv_f^2=\frac{1}{2}m(0)^2=0[/tex]

ΔKE = 0 - K = -K

W = ΔKE = -K

Also, the work done (W) = charge (q) * distance (d) * electric field intensity (E)

W = qEd

but q = -e, hence:

W = -e * E * d

Using:

W = ΔKE

-e * E * d = -K

E= K / (e * d)

b) The electric field is in the direction of the electrons motion

Answer:

The volume for this is 29.7

Explanation:

Trust me on this I'm an expert

Answer:

Video game designers typically have a bachelor's degree in game design, computer engineering, or computer science which takes four to five years to complete. Courses in a game design degree program may include project management, integrated video design and technology, game prototyping and level design.

Explanation:

sory sorry sorry sorrysorrysorry

Solution :

Given :

A six order Butterworth high pass filter.

∴ n = 6, [tex]w_c=1 \ rad/s[/tex]

a). The location at poles :

    [tex]$s^6-(w_c)^6=0$[/tex]

   [tex]$s^6=(w_c)^6=1^6$[/tex]

  ∴ [tex]$s^6 = 1$[/tex]

Therefore, it has 6 repeated poles at s = 1.

b). The transfer function H(S) :

    Transfer function H(S) [tex]$=\frac{1}{1+j\left(\frac{w_c}{s}\right)^6}$[/tex]

                                         [tex]$=\frac{1}{1-\left(\frac{w_c}{s}\right)^6}$[/tex]

  ∴    H(S) [tex]$=\frac{s^6}{s^6-(w_c)^6}=\frac{s^6}{s^6-1}$[/tex]

   H(S) [tex]$=\frac{Y(s)}{X(s)}=\frac{s^6}{s^6-1}$[/tex]

c). The corresponding LCCDE description :

  [tex]$=\frac{Y(s)}{X(s)}=\frac{s^6}{s^6-1}$[/tex]

   [tex]$Y(s)(s^6-1) = s^6 \times (s)$[/tex]

   [tex]$Y(s)s^6-y(s).1 = s^6 \times (s)$[/tex]

By taking inverse Laplace transformation on BS

   [tex]$L^{-1}[Y(s)s^6-Y(s)1]=L^{-1}[s^6 \times (s)]$[/tex]

   [tex]$\frac{d^6y(t)}{dt^6}-y(t)=\frac{d^6 \times (t)}{dt^6}$[/tex]

  Hence solved.

Answer:

Explanation:

From the information given:

The instantaneous expression of the electric field in the wave is:

[tex]E(r,t)= (i_x \sqrt{3} -i_z) 2 \ sin (2 \pi*10^8t + 2 \pi x/3+2 \pi z /\sqrt{3} + 30 ^0) , \ V/m[/tex]

To determine the unit vector in line with the wave electric field, we take the first term in E(r,t) for [tex]I_E^\to[/tex] as:

[tex]I_E^\to = i_x \sqrt{3}-i_z \\ \\ I_E^\to = \dfrac{i_x \sqrt{3}-i_z}{\sqrt{3 +1}} \\ \\ \mathbf{ I_E = \dfrac{i_x\sqrt{3} -i_z}{2}}[/tex]

The amplitude is denoted by the numerical value after the first term, which is:

[tex]\mathbf{E_o = 2}[/tex]

The wavelength can be determined by using the expression:

[tex]\beta =\dfrac{2 \pi}{\lambda }[/tex]

from the given instantaneous expression:

[tex]\beta = \dfrac{2 \pi}{3}x+\dfrac{2 \pi}{\sqrt{3}}z[/tex]

[tex]\beta = \sqrt{\dfrac{2 \pi}{(3)^2}+\dfrac{(2 \pi}{(\sqrt{3})^2}}[/tex]

[tex]\beta = \sqrt{\dfrac{2 \pi}{9}+\dfrac{2 \pi}{{3}}}[/tex]

Factorizing 2π

[tex]\beta =2 \pi \sqrt{\dfrac{1}{9}+\dfrac{1}{{3}}}[/tex]

[tex]\beta =2 \pi \sqrt{\dfrac{9+3}{9*3}}}[/tex]

[tex]\beta =2 \pi \sqrt{\dfrac{12}{27}}}[/tex]

[tex]\beta =2 \pi \sqrt{\dfrac{4*3}{9*3}}}[/tex]

[tex]\beta =2 \pi \sqrt{\dfrac{4}{9}}}[/tex]

[tex]\beta =2 \pi\times {\dfrac{2}{3}}}[/tex]

recall from the expression using in calculating wavelength:

[tex]\beta =\dfrac{2 \pi}{\lambda }[/tex]

∴

equating both together, we have:

[tex]\dfrac{2 \pi}{\lambda }= 2 \pi\times {\dfrac{2}{3}}}[/tex]

[tex]\lambda = \dfrac{3}{2}[/tex]

λ = 1.5 m

In line with the wave direction; unit vector [tex]i_k[/tex] can be computed as follows:

[tex]i_k = - [ \beta_1x +\beta_2z]/\beta[/tex]

where;

[tex]\beta_1 = \dfrac{2 \pi }{3} \ ; \ \beta_2 = \dfrac{2 \pi }{\sqrt{3}} \ ; \ \beta = \dfrac{2 \pi \times 2}{3} ;[/tex]

∴

[tex]i_k = - \Big[\dfrac{2 \pi}{3}x + \dfrac{2 \pi}{\sqrt{3}} z\Big]\times \dfrac{1}{\dfrac{2 \pi *2}{3}}[/tex]

[tex]i_k = - \Big[\dfrac{x}{2} + \sqrt\dfrac{{3}}{4}} z\Big][/tex]

[tex]i_k = - \Big[\dfrac{1}{2}x + \sqrt{\dfrac{3}{4} }z\Big][/tex]

[tex]\mathbf{i_k = - \Big[0.5x +0.86 z\Big]}[/tex]

Answer:

8.4627 m

Explanation:

Transmissivity( T ) = 300 m^2/day

Storativity( S )  = 0.0005

well radius ( r ) = 0.3m

Determine the drawdown in well at 100 days

Drawdown at 100 days = ∑ Drawdown at various period

We will use the equation : S = Q / U*π*T [ -0.5772 - In U ]  ----- ( 1 )

where : Q = discharge , T = transmissivity

             S = drawdown ,

U = r^2*s / 4*T*t  --- ( 2 )

r = well radius , S = Storativity, t = time period

i) During 0-20

U1 = r^2*s / u*π*t  = 1.875 * 10^-9

Input values into equation 1

S1 = 2.5885

ii) During 20-50

U2 = r^2*s / 4*π*t = 0.3^2 * 30 / u * 300 * 30 = 1.25 * 10^-9

input values into equation 1

S2 = 1.5854 m

iii) During 50 -90

U3 = r^2*s / 4*π*t = 9.375 * 10^-10

input values into equation 1

S3 = 4.2888 m

iv) During 90-100

U4 = 0

s4 = 0

Drawdown at 100 days = ∑ Drawdowns at various period

                                       = s1 + s2 + s3 + s4 = 2.5885 + 1.5854 + 4.2888 + 0

                                       = 8.4627 m

Answer:

Time required = 287.2 secs  

Explanation:

Volume  of room = 25 * 10 * 4 = 1000 m^3

power consumed by pump = 15 kW

T1 ( initial temperature ) = 12°C

P1 ( Initial pressure ) = 1 bar

COPhp = 3

Calculate time taken to raise room Temp to 27°C

average heat supplied ( ∅ ) = COPhp * power consumed by pump

                                             = 3 * 15 = 45 kW

Time required can be calculated using the relation below

∅t = P*V*Cv ( T2 - T1 )   [ p = 1.2 kg/m^3 , Cv = 0.718 KJ/kg ( air properties ) ]

45 * 10^3 ( t )  = 1.2*1000* 718 ( 27 - 12 )

∴ solving for t

t = 287.2 secs  ≈ 4.79 mins

Solution :

For R-134a, we are given :

[tex]$T_i = 25^\circ C$[/tex]

[tex]$P_i=750 \ kPa$[/tex]

[tex]$P_e=165 \ kPa$[/tex]

Now we have one inlet and one exit flow, no work and no heat transfer. The energy equation is :

[tex]$h_e+\frac{1}{2}.v_e^2= h_i+\frac{1}{2}.v_i^2 $[/tex]

We also know that the gas is throttled and there is no change in the kinetic energy.

So, [tex]$v_e=v_i$[/tex]

Now from the energy equation above, we can see that the inlet and the exit enthalpies are also the same. Therefore,

[tex]$h_i=h_e$[/tex]

From the saturated R-134a table, corresponding to [tex]P_e = 165 \ kPa[/tex], we can find the exit saturation temperature.

[tex]$T_e=-15^\circ C$[/tex]

From the saturated R-134a table, corresponding to [tex]P_e = 165 \ kPa[/tex], we can find the specific enthalpies :

[tex]$h_f = 180.19 \ kJ/kg$[/tex]

[tex]$h_{fg} = 209 \ kJ/kg$[/tex]

Calculating the exit flow quality factor,

[tex]$x_e=\frac{h_e-h_f}{h_{fg}}$[/tex]

    [tex]$=\frac{234.59-180.19}{209}$[/tex]

   = 0.26

From the saturated R-134a table, corresponding to [tex]P_e = 165 \ kPa[/tex], we can find the specific volumes :

[tex]$v_f = 0.00746 \ m^3/kg$[/tex]

[tex]$v_{fg} = 0.11932 \ m^3/kg$[/tex]

Calculating the exit specific volume :

[tex]$v_e=v_f+x_e(v_{fg})$[/tex]

   = 0.000746 + 0.26 (0.11932)

   = 0.0318 [tex]m^3/kg[/tex]

The mass flow is equal to :

[tex]$\dot{m} = A_i . \frac{v}{v_i}$[/tex]

  [tex]$=A_e . \frac{v}{v_e}$[/tex]

So, [tex]$\frac{A_e}{A_i}=\frac{v_e}{v_i}$[/tex]

Therefore, the ratio of the exit pipe and the inlet pipe diameter is equal to

[tex]$\frac{D_e}{D_i}=\sqrt{\frac{A_e}{A_i}}$[/tex]

[tex]$\frac{D_e}{D_i}=\sqrt{\frac{v_e}{v_i}}$[/tex]

[tex]$\frac{D_e}{D_i}=\sqrt{\frac{0.0318}{0.000829}}$[/tex]

[tex]$\frac{D_e}{D_i}=6.19$[/tex]

Answer: hello some data related to your question is missing attached below is the missing data

answer:

T2 = 265°C

Explanation:

First step : calculate sum of vertical forces

∑ y = 0

Fmg - 2(0.5 Fst ) = 0

∴Fmg = ( 12 * 10^6 ) ( 2 * π/4 (0.01)^2 )

          = 1884.96 N

Also determine the Compatibility equation in order to determine the change in Temperature

ΔT = 250°C

therefore Temperature at which average normal stress becomes 12.0 MPa

ΔT = T2 - T1

250°C = T2 - 15°C

T2 = 250 + 15 = 265°C

attached below is the detailed solution

Answer:

The costs to run the dryer for one year are $ 9.03.

Explanation:

Given that the clothes dryer in my home has a power rating of 2250 Watts, and to dry one typical load of clothes the dryer will run for approximately 45 minutes, and in Ontario, the cost of electricity is $ 0.11 / kWh, to calculate the costs to run the dryer for one year the following calculation must be performed:

1 watt = 0.001 kilowatt

2250/45 = 50 watts per minute

45 x 365 = 16,425 / 60 = 273.75 hours of consumption

50 x 60 = 300 watt = 0.3 kw / h

0.3 x 273.75 = 82.125

82.125 x 0.11 = 9.03

Therefore, the costs to run the dryer for one year are $ 9.03.

Answer:

Following are the program to the given question:

def calculate_storage(filesize):#definging a method calculate_storage that takes filesize as a parameter  

 block_size = 4096#definging block_size that holds value

 full_blocks = filesize//block_size#definging full_blocks that divides the value and hold integer part

 partial_block_remainder = filesize%block_size#definging partial_block_remainder that holds remainder value

 if partial_block_remainder > 0:#definging if that compare the value

   return block_size*full_blocks+block_size#return value

 return block_size*full_blocks#return value

print(calculate_storage(1))    # calling method by passing value

print(calculate_storage(4096)) # calling method by passing value

print(calculate_storage(4097)) # calling method by passing value

Output:

4096

4096

8192

Explanation:

In this code, a method "calculate_storage" is declared that holds a value "filesize" in its parameters, inside the method "block_size" is declared that holds an integer value, and defines "full_blocks and partial_block_remainder" variable that holds the quotient and remainder value and use it to check its value and return its calculated value. Outside the method, three print method is declared that calls the method and prints its return value.

Answer:

In the clevis shown in Fig. 1-11b, find the minimum bolt diameter and the minimum thickness of each yoke that will support a load P = 14 kips without exceeding a shearing stress of 12 ksi and a bearing stress of 20 ksi.

127-clevis-double-shear-bolt.gif

Solution 127

Hide Click here to show or hide the solution

127-fbd-clevis-double-shear-bolt.gifFor shearing of rivets (double shear)

P=τA

14=12[2(14πd2)]

d=0.8618in → diameter of bolt answer

For bearing of yoke:

P=σbAb

14=20[2(0.8618t)]

t=0.4061in → thickness of yoke answer

Answer:

When you come up on a steady yellow traffic light you should always yield to cross traffic if you can yield safely. The flashing yellow light is there to inform drivers to be careful and to slow down.

Explanation:

hope it helped!

Answer:

0.2063

Explanation:

Given data:

packing factor = 0.5

percentage of reduction of powders = 70%

Calculate the final porosity

after sintering Bulk specific volume = 0.9 * 0.7 = 0.63

assuming true specific volume = 1

packing factor = 0.5 , bulk specific volume = 2

packing factor after pressing and sintering

= 1 / ( 2 * 0.63 ) = 0.7937

hence : porosity = 1 - packing factor

                            = 1 - 0.7937 = 0.2063