Scheduling can best be defined as the process used to determine:​

Answer:

The total force resisted by the flange bolts is  163.98 N

Explanation:

Solution

The first step is to find  the pipe cross section at the inlet section

Now,

A₁ = π /4 D₁²

D₁ =  diameter of the pipe at the inlet section

Now we insert 8 cm for D₁ which gives us A₁ = π /4 D (8)²

=50.265 cm² * ( 1 m²/100² cm²)

= 5.0265 * 10^⁻³ m²

Secondly, we find cross section area of  the pipe at the inlet section

A₂ = π /4 D₂²

D₂ =  diameter of the pipe at the inlet section

Now we insert 5 cm for D₁ which gives us A₁ = π /4 D (5)²

= 19.63 cm² * ( 1 m²/100² cm²)

= 1.963 * 10^⁻³ m²

Now,

we write down the conversation mass relation which is stated as follows:

Q₁ = Q₂

Where Q₁ and Q₂ are both the flow rate at the exist and inlet.

We now insert A₁V₁ for Q₁ and A₂V₂ for Q₂

So,

V₁ and V₂ are defined as the velocities at the inlet and exit

We now insert 5.0265 * 10^⁻³ m² for A₁ 5 m/s for V₁ and 1.963 * 10^⁻³ m² for A₂

= 5.0265  * 5 = 1.963 * V₂

V₂ = 12.8 m/s

Note: Kindly find an attached copy of the part of the solution to the given question below

Answer:

Explanation:

The frequency components in the message signal are

f1 = 100Hz, f2 = 200Hz and f3 = 400Hz

When amplitude modulated with a carrier signal of frequency fc = 100kHz

Generates the following frequency components

Lower side band

[tex]100k - 100 = 99.9kHz\\\\100k - 200 = 99.8kHz\\\\100k - 400 = 99.6kHz\\\\[/tex]

Carrier frequency 100kHz

Upper side band

[tex]100k + 100 = 100.1kHz\\\\100k + 200 = 100.2kHz\\\\100k + 400 = 100.4kHz[/tex]

After passing through the SSB filter that filters the lower side band, the transmitted frequency component will be

[tex]100k, 100.1k, 100.2k\ \texttt {and}\ 100.4kHz[/tex]

At the receive these are mixed (superheterodyned) with local ocillator frequency whichh is 100.02KHz, the output frequencies will be

[tex]100.02 - 100.1k = 0.08k = 80Hz\\\\100.02 - 100.2k = 0.18k = 180Hz\\\\100.02 - 100.4 = 0.38k = 380Hz[/tex]

After passing through the SSB filter that filters the higher side band, the transmitted frequency component will be

[tex]100k, 99.9k, 99.8k\ \ and \ \99.6kHz[/tex]

At the receive these are mixed (superheterodyned) with local oscillator frequency which is 100.02KHz, and then fed to the detector whose output frequencies will be

[tex]100.02 - 99.9k = 0.12k = 120Hz\\\\100.02 - 99.8k = 0.22k = 220Hz\\\\100.02 - 99.6k = 0.42k = 420Hz[/tex]

A) The frequency Components of the Detector Output are;

80 Hz, 120 Hz and 380 Hz

B) The frequency Components if only the lower sideband is transmitted are; 120 Hz, 220 Hz and 420 Hz

A) We are given the frequency components in the message signal as;

We are told that the carrier signal has a frequency; fc = 100kHz

Thus, the frequency components generated are;

Lower side band:

Upper side band:

We are told that the local oscillator now supplies a sinusoidal wave of frequency 100.02 kHz.

Thus, the output frequencies are;

100.02 kHz - 100.1 kHz = 80 Hz

100.02 kHz - 100.2 kHz = 180 Hz

100.02 kHz - 100.4 kHz = 380 Hz

B) Repeating the analysis assuming only the lower sideband is repeated gives us the frequencies as;

100.02 kHz - 99.9 kHz = 120 Hz

100.02 kHz - 99.8 kHz = 220 Hz

100.02 kHz - 99.6 kHz = 420 Hz

Read more about Message Signals at; https://brainly.com/question/25904079

Answer:

The answer is VN =37.416 m/s

Explanation:

Recall that:

Pressure (atmospheric) = 100 kPa

So. we solve for the maximum velocity (m/s) to which water can be accelerated by the nozzles

Now,

Pabs =Patm + Pgauge = 800 KN/m²

Thus

PT/9.81 + VT²/2g =PN/9.81  + VN²/2g

Here

Acceleration due to gravity = 9.81 m/s

800/9.81 + 0

= 100/9.81 + VN²/19.62

Here,

9.81 * 2= 19.62

Thus,

VN²/19.62 = 700/9.81

So,

VN² =1400

VN =37.416 m/s

Note: (800 - 100) = 700

Answer:

[tex]V2 = 37.417ms^{-1}[/tex]

Explanation:

Given the following data;

Water enters the turbine nozzles (inlet) = 800kPa = 800000pa.

Nozzle outlets = 100kPa = 100000pa.

Density of water = 1000kg/m³.

We would apply, the Bernoulli equation between the inlet and outlet;

[tex]\frac{P_{1} }{d}+\frac{V1^{2} }{2} +gz_{1} = \frac{P_{2} }{d}+\frac{V2^{2} }{2} +gz_{2}[/tex]

Where, V1 is approximately equal to zero(0).

Z[tex]z_{1} = z_{2}[/tex]

Therefore, to find the maximum velocity, V2;

[tex]V2 = \sqrt{2(\frac{P_{1} }{d}-\frac{P_{2} }{d}) }[/tex]

[tex]V2 = \sqrt{2(\frac{800000}{1000}-\frac{100000}{1000}) }[/tex]

[tex]V2 = \sqrt{2(800-100)}[/tex]

[tex]V2 = \sqrt{2(700)}[/tex]

[tex]V2 = \sqrt{1400}[/tex]

[tex]V2 = 37.417ms^{-1}[/tex]

Hence, the maximum velocity, V2 is 37.417m/s

Answer:

None of these

Explanation:

There are different types of amplifiers, and each has different characteristics.

Therefore, the correct answer is "None of these "

Answer:

(1). False, (2). True, (3). False, (4). False, (5). True.

Explanation:

The term ''contouring'' in this question does not have to do with makeup but it has to deal with the measurement of all surfaces in planes. It is a measurement in which the rough and the contours are being measured. So, let us check each questions again.

(1). In contouring, it is necessary to measure position and not velocity for feedback.

ANSWER : b =>False. IT IS NECESSARY TO MEASURE BOTH FOR FEEDBACK.

(2). In contouring during 2-axis NC machining, the two axes are moved at the same speed to achieve the desired contour.

ANSWER: a=> True.

(3). Job shop is another term for process layout.

ANSWER: b => False

JOB SHOP IS A FLEXIBLE PROCESS THAT IS BEING USED during manufacturing process and are meant for job Production. PROCESS LAYOUT is used in increasing Efficiency.

(4). Airplanes are normally produced using group technology or cellular layout.

ANSWER: b => False.

(5). In manufacturing, value-creating time is greater than takt time.

ANSWER: a => True.

Answer:

[tex]W_{net} = - 1223 kJ[/tex]

Explanation:

State 1:

[tex]P_1 = 20 bar\\T_1 = 360^{0}C\\ h_1 = 3159.3 kJ/kg\\S_1 = 6.9917 kJ/kg[/tex]

State 2:

[tex]P_2 = 20 bar\\x_2 = 1 \\ h_2 = 2799.5 kJ/kg\\u_2 = 2600.3 kJ/kg\\v_2 = 0.09963m^3/kg[/tex]

State 3:

[tex]P_2 = 5 bar\\v_2 = v_3 \\v_3 = v_f + x_3 (v_g - v_f)\\0.09963 = (1.0926 * 10^{-3}) +x_3 (0.3749 - (1.0926 * 10^{-3}))\\x_3 = 0.263[/tex]

[tex]u_{3} = u_f + x_3 ( u_g - u_f)\\u_{3} = 639.68 + 0.263 (2561.2 - 639.68)\\u_{3} = 1146.2 kJ/kg[/tex]

State 4:

[tex]P_{4} = 5 bar\\T_4 = 360^0 C\\h_4 = 3188.4 kJ/kg\\S_4 = 7.660 kJ/kg-K\\Q_{12} = h_2 - h_1 = 2799.5-3159.3 = -359 kJ/kg\\Q_{23} = u_3 - h_2 =1146.2-2006.3 = -1454.1 kJ/kg\\Q_{34} = h_4 - h_3 = 3188.4-1196.04 = 1992.36 kJ/kg\\Q_{41} = T(S_1 - S_4) = (360 + 273) (6.9917 - 7.660) = -423.04 kJ/kg[/tex]

Calculate the network done for the cycle

[tex]W_{net} = m( Q_{12} + Q_{23} + Q_{34} + Q_{41})\\W_{net} = 5( -359.8 - 1454.1 + 1992.36 - 423.04)\\W_{net} = -1223 kJ[/tex]

Answer:

Under the normal sign convention, the distributed load on a beam is equal to the: O The second derivative of the bending moment with respect to the length of the beam O Negative of the rate of change of the shear force with respect to the length of the beam.

Sorry if the answer is wrong

Find the given attachments for complete explanation

Answer:

The goal of the software is to observe the software behavior to meet its requirement expectation. In software engineering, validating software might be harder since client's expectation may be vague or unclear.

Explanation:

Answer:

a. [tex]\dfrac{D_{1}}{ D_{2}} = \left (\dfrac{ \left{D_1} }{ {D_2}} \right )^{-3\times n}[/tex] which is constant therefore, n = constant

b. The temperature at the end of the process is 109.6°C

c. The work done by the balloon boundaries = 10.81 MJ

The work done on the surrounding atmospheric air = 10.6 MJ

Explanation:

p₁ = 100 kPa

T₁ = 27°C

D₁ = 10 m

v₂ = 1.2 × v₁

p ∝ α·D

α = Constant

[tex]v_1 = \dfrac{4}{3} \times \pi \times r^3[/tex]

[tex]\therefore v_1 = \dfrac{4}{3} \times \pi \times \left (\dfrac{10}{2} \right )^3 = 523.6 \ m^3[/tex]

v₂ = 1.2 × v₁ = 1.2 × 523.6 = 628.32 m³

Therefore, D₂ = 10.63 m

We check the following relation for a polytropic process;

[tex]\dfrac{p_{1}}{p_{2}} = \left (\dfrac{V_{2}}{V_{1}} \right )^{n} = \left (\dfrac{T_{1}}{T_{2}} \right )^{\dfrac{n}{n-1}}[/tex]

We have;

[tex]\dfrac{\alpha \times D_{1}}{\alpha \times D_{2}} = \left (\dfrac{ \dfrac{4}{3} \times \pi \times \left (\dfrac{D_2}{2} \right )^3}{\dfrac{4}{3} \times \pi \times \left (\dfrac{D_1}{2} \right )^3} \right )^{n} = \left (\dfrac{ \left{D_2} ^3}{ {D_1}^3} \right )^{n}[/tex]

[tex]\dfrac{D_{1}}{ D_{2}} = \left (\dfrac{ \left{D_2} }{ {D_1}} \right )^{3\times n} = \left (\dfrac{ \left{D_1} }{ {D_2}} \right )^{-3\times n}[/tex]

[tex]\dfrac{ D_{1}}{ D_{2}} = \left ( 1.2 \right )^{n} = \left (\dfrac{ \left{D_2} ^3}{ {D_1}^3} \right )^{n}[/tex]

[tex]log \left (\dfrac{D_{1}}{ D_{2}}\right ) = -3\times n \times log\left (\dfrac{ \left{D_1} }{ {D_2}} \right )[/tex]

n = -1/3

Therefore, the relation, pVⁿ = Constant

b. The temperature T₂ is found as follows;

[tex]\left (\dfrac{628.32 }{523.6} \right )^{-\dfrac{1}{3} } = \left (\dfrac{300.15}{T_{2}} \right )^{\dfrac{-\dfrac{1}{3}}{-\dfrac{1}{3}-1}} = \left (\dfrac{300.15}{T_{2}} \right )^{\dfrac{1}{4}}[/tex]

T₂ = 300.15/0.784 = 382.75 K = 109.6°C

c. [tex]W_{pdv} = \dfrac{p_1 \times v_1 -p_2 \times v_2 }{n-1}[/tex]

[tex]p_2 = \dfrac{p_{1}}{ \left (\dfrac{V_{2}}{V_{1}} \right )^{n} } = \dfrac{100\times 10^3}{ \left (1.2) \right ^{-\dfrac{1}{3} } }[/tex]

p₂ =  100000/0.941 = 106.265 kPa

[tex]W_{pdv} = \dfrac{100 \times 10^3 \times 523.6 -106.265 \times 10^3 \times 628.32 }{-\dfrac{1}{3} -1} = 10806697.1433 \ J[/tex]

The work done by the balloon boundaries = 10.81 MJ

Work done against atmospheric pressure, Pₐ, is given by the relation;

Pₐ × (V₂ - V₁) = 1.01×10⁵×(628.32 - 523.6) = 10576695.3 J

The work done on the surrounding atmospheric air = 10.6 MJ

Answer:

(a) 12 seconds (b) t = 8.31 seconds (c) 10µ A (d) V = 20 V (e) V =0

Explanation:

Solution

Given that:

C = 6 µ which is = 6 * 10^ ⁻6

R = 2 MΩ, which is = 2 * 10^ 6

ε = 20 V

(a) When it is at the time constant we have the following:

λ = CR

= 6 * 10^ ⁻6 * 2 * 10^ 6

λ =12 seconds

(b) We solve for the half life of the circuit which is given below:

d₀ = d₀ [ 1- e ^ ⁺t/CR

d = decay mode]

d₀/2 =  d₀  1- e ^ ⁺t/12

2^⁻1 = e ^ ⁺t/12

Thus

t/12 ln 2

t = 12 * ln 2

t = 12 * 0.693

t = 8.31 seconds

(c) We find the current at t = 0

So,

I = d₀/dt

I = d₀/dt e ^ ⁺t/CR

= CE/CR e ^ ⁺t/CR

E/R e ^ ⁺t/CR

Thus,

at t = 0

I  E/R = 20/  2 * 10^ 6

= 10µ A

(d) We find the voltage across the capacitor at t = 0 which is shown below:

V = IR

= 10 * 10^ ⁻6 * 2 * 10^ 6

V = 20 V

(e)  We solve for he voltage across the resistor.

At t = 0

I = 0

V =0

Answer:

  9.29 N . . . . weight of 0.948 kg sphere

Explanation:

The sum of torques on the link BOA is zero, so we have ...

  (right force at A)(OA) = (up force at B)(OB·sin(135°))

Solving for the force at B, we have ...

  up force at B = (10.53 N)(253 mm)/((553 mm)/√2) ≈ 6.81301 N

This force is due to the difference between the buoyant force on the float sphere and the weight of the float sphere. Dividing by the acceleration due to gravity, it translates to the difference in mass between the water displaced and the mass of the sphere.

  ∆mass = (6.81301 N)/(9.8 m/s^2) = 0.695205 kg

__

The center of the sphere of diameter 0.1553 m is below the waterline by ...

  (553 mm)cos(45°) -(353 mm) = 38.0300 mm

The volume of the spherical cap of radius 155.3/2 = 77.65 mm and height 77.65+38.0300 = 115.680 mm can be found from the formula ...

  V = (π/3)h^3(3r -h) = (π/3)(115.680^2)(3·77.65 -115.68) mm^3 ≈ 1.64336 L

So the mass of water contributing to the buoyant force is 1.64336 kg. For the net upward force to correspond to a mass of 0.695305 kg, the mass of the float sphere must be ...

  1.64336 kg -0.695205 kg ≈ 0.948 kg

The weight of the float sphere is then (9.8 m/s^2)·(0.948 kg) = 9.29 N

The weight of the 0.948 kg float sphere is about 9.29 N.

Complete Question:

Grain diameter 1 (mm) = 4.4E-02

Yield stress 1 (MPa) = 131

Grain diameter 2 (mm) = 7.7E-03

Yield Stress 2 (MPa) = 268

The yield strength for an alloy that has an average grain diameter, d1, is listed above as Yield Stress 1 . At a grain diameter of d2, the yield strength increases Yield Stress 2. At what grain diameter, in mm, will the yield strength be 217 MPa

Answer:

d = 1.3 * 10⁻² m

Explanation:

According to the Hall Petch equation:

[tex]\sigma_y = \sigma_0 + k/\sqrt{d} \\[/tex]

At [tex]d_{1} = 4.4 * 10^{-2} mm[/tex], [tex]\sigma_{y1} = 131 MPa = 131 N/ mm^2[/tex]

[tex]131 = \sigma_0 + k/\sqrt{4.4 * 10^{-2}} \\k = 27.45 - 0.2096 \sigma_0[/tex]

At [tex]d_{2} = 7.7 * 10^{-3} mm[/tex], [tex]\sigma_{y2} = 131 MPa = 268 N/ mm^2[/tex]

[tex]268 = \sigma_0 + (27.45 - 0.2096 \sigma_0)/\sqrt{7.7 * 10^{-3}} \\23.5036 = 27.47 - 0.1219 \sigma_0\\ \sigma_0 = 32.45 N/mm^2[/tex]

k = 27.45 - 0.2096(32.45)

k = 20.64

At [tex]\sigma_y = 217 MPa[/tex], reapplying Hall Petch law:

[tex]\sigma_y = \sigma_0 + k/\sqrt{d} \\[/tex]

[tex]217 =32.45 + 20.64/\sqrt{d} \\217 - 32.45 = 20.64/\sqrt{d}\\184.55 = 20.64/ \sqrt{d} \\\sqrt{d} = 20.64/184.55\\\sqrt{d} = 0.11184\\d = 0.013 mm[/tex]

d = 1.3 * 10⁻² m

Answer:

George  Floyd (BLACK  LIFES  MATTER)

C O V I D - 19

Quarantine  

no sports

wearing a mask

and a whole lot of other stuff

Explanation:

Answer:

the % recovery of aluminum product is 80.5%

the % purity of the aluminum product is 54.7%

Explanation:

feed rate to separator = 2500 kg/hr

in one hour, there will be 2500 kg/hr x 1 hr = 2500 kg of material is fed into the  machine

of this 2500 kg, the feed is known to contain 174 kg of aluminium and 2326 kg of rejects.

After the separation, 256 kg  is collected in the product stream.

of this 256 kg, 140 kg is aluminium.

% recovery of aluminium will be = mass of aluminium in material collected in the product stream ÷ mass of aluminium contained in the feed material

% recovery of aluminium = 140kg/174kg x 100% = 80.5%

% purity of the aluminium product = mass of aluminium in final product ÷ total mass of product collected in product stream

% purity of the aluminium product = 140kg/256kg

x 100% = 54.7%

Answer:

South Pole and South Pole or North Pole and North Pole.

Answer:

The average thickness of the blubber is 0.077 m

Explanation:

Here, we want to calculate the average thickness of the Walrus blubber.

We employ a mathematical formula to calculate this;

The rate of heat transfer(H) through the Walrus blubber = dQ/dT = KA(T2-T1)/L

Where dQ is the change in amount of heat transferred

dT is the temperature gradient(change in temperature) i.e T2-T1

dQ/dT = 220 W

K is the conductivity of fatty tissue without blood = 0.20 (J/s · m · °C)

A is the surface area which is 2.23 m^2

T2 = 37.0 °C

T1 = -1.0 °C

L is ?

We can rewrite the equation in terms of L as follows;

L × dQ/dT = KA(T2-T1)

L = KA(T2-T1) ÷ dQ/dT

Imputing the values listed above;

L = (0.2 * 2.23)(37-(-1))/220

L = (0.2 * 2.23 * 38)/220 = 16.948/220 = 0.077 m

Answer:

Explanation:

D ( diameter ) = 125 m

convection coefficient of  h = 700 W/m^2

Calculate THE CROSS SECTIONAL AREA

Ac = [tex]\frac{\pi }{4} * D^2[/tex]  = [tex]\frac{\pi }{4} * ( 125 )^2[/tex] = 0.79 * 15625 = 12343.75 m^2

perimeter

p = [tex]\pi * D[/tex]  = 3.14 * 125 = 392.5 m

at 300k temperature the thermal conductivity of copper and constantan from the thermodynamic property table are :

Kcu = 401 w/m.k

Kconstantan = 23 W/m.k

To calculate the length of copper wire of the thermocouple junction

L 1 = 4.6 ([tex]\frac{Kcv Ac}{h P}[/tex]) ^ 1/2 = 4.6 [tex](\frac{401 *12343.75 }{700 *392.5})^\frac{1}{2}[/tex]

L 1 = 4.6 ( 4949843.75 / 274750 )^1/2

L 1 = 9.48 m

calculate length of constantan wire

L 2 = 4.6 [tex](\frac{kcons*Ac}{hp} )^\frac{1}{2}[/tex]

     = 4.6 ( (23 * 12343.75) / ( 700 * 392.5) ) ^1/2

L 2 = 4.6 ( 283906.25 / 274750 ) ^ 1/2

L 2 = 4.68 m

I)  therefore the minimum separation distance between the two legs of the sting L = L 1 + L 2

L = 9.48 + 4.68  = 14.16 m

ii)  Evaluating the thermal conductivity of copper and constantan

Kc ( thermal conductivity of chromel) = 19 w/m.k

Ka ( thermal conductivity of alumel ) = 29 W/m.k

distance between the legs L = L 1 + L 2

THEREFORE

L = 4.6 ( (Kcn * Ac ) / ( hp ) )^1/2  +  4.6 ( (Kac * Ac)/(hp) )^1/2

L = 4.6 [tex](\frac{Ac}{hp} )^\frac{1}{2} [ (Kcn)^\frac{1}{2} + (Kal)^\frac{1}{2} ][/tex]

L = 4.6 ( 12343.75 /( 700 * 392.5) )^1/2   * [ 19^1/2  + 29^1/2 ]

L = 4.6 ( 12343.75 / 274750 ) ^1/2  * 5.39

L = 1.14 m

Answer:

Explanation:

a ) for transformer which steps down voltage , if V₁ and V₂ be voltage of primary and secondary coil and n₁ and n₂ be the no of turns of wire in them

V₁ /V₂ = n₁ / n₂

Here V₁ = 220 V , V₂ = 5V , n₁ = 2200 n₂ = ?

220 /5 = 2200 / n₂

n₂ = 2200 x 5 / 220

= 50

b )

for 100 % efficiency

input power = output power

V₁ I₁ = V₂I₂

I₁ and I₂ are current in primary and secondary coil

220 x I₁ = 5 x .5

I₁ = .01136 A .

c )

If n₂ = 100

V₁ /V₂ = n₁ / n₂

220 / V₂ = 2200 / 100

V₂ = 10 V

V₁ I₁ = V₂I₂

220 x .01136 = 10 I₂

I₂ = .25 A.

The common" concern with using variable frequency drives (VFDs) is C. hertz modulation.

It should be noted that a variable frequency drive simply means a type of motor drive that us used in mechanical drive system.

In this case, common" concern with using variable frequency drives (VFDs) is hertz modulation

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Answer:

Technician B is correct

Explanation:

O- rings are used with oil transmission filters to avoid transmission failures but some people use  lip seals as well. either of them is  inserted onto the outer part of the transmission system i.e it is inserted/found in-between Transmission filters and the transmission systems and it main purpose is to avoid leaks and transmission failure in the short and long term.

0-rings should be lubricated before installation this is because the o-rings are usually super tight when installing and would require lubrication to ease the installation process else the rubber might get ruptured and this would lead to instant transmission failure.

Answer:

See explanation

Explanation:

See the document for the complete FBD and the introductory part of the solution.

Static Balance ( Sum of Forces = 0 ) in all three directions

                 ∑[tex]F_G_X = W - G_x = 0[/tex]

                 [tex]G_X = W = 1500 lb[/tex]

                 ∑[tex]F_G_Y = P - G_Y = 0[/tex]

                 [tex]G_Y = P = -10,800 lb[/tex]

                ∑[tex]F_G_Z = - G_Z = 0[/tex]

Where, ( [tex]G_X, G_Y, G_Z[/tex] ) are internal forces at section ( G ) along the defined coordinate axes.

Static Balance ( Sum of Moments about G = 0 ) in all three directions

              [tex]M_G = r_O_G x F_O[/tex]

Where,

              r_OG: The vector from point O to point G

              F_OG: The force vector at point O

- The vector ( r_OG ) and ( F_OG ) can be written as follows:

              [tex]r_O_G = [ -( 3 + \frac{H}{2} ) i + (\frac{r_o}{12})j - ( \frac{r_o}{12} + \frac{L}{2})k ] \\\\r_O_G = [ -( 6 ) i + (0.25)j - (6)k ] \\[/tex]

              [tex]F_O_G = [ ( W ) i + ( P ) k ]\\\\F_O_G = [ (1500) i - ( 10,800 ) k ] lb[/tex]

- Then perform the cross product of the two vectors ( r_OG ) and ( F_OG ):

     [tex]( M_G_X )i + (M_G_Y)j+ (M_G_Z)k = \left[\begin{array}{ccc}i&j&k\\-6&0.25&-6\\1500&-10,800&0\end{array}\right] \\\\\\( M_G_X )i + (M_G_Y)j+ (M_G_Z)k = -( 6*10,800 ) i - ( 6*1500 ) j + [ ( 10,800*6) - ( 0.25*1500) ] k\\\\( M_G_X )i + (M_G_Y)j+ (M_G_Z)k = - (64,800)i - (9,000)j + (64,425)k[/tex]

- The internal torque ( T ) and shear force ( V ) that act on slice ( G ) are due to pressure force ( P ) as follows:

             [tex]T = P*[\frac{L}{2}] = (10,800)*(6) = 64,800 lb.ft[/tex]

             [tex]V = P = -10,800 lb[/tex]

- For the state of stress at point "C" we need to determine the the normal stress along x direction ( σ_x ) and planar stress ( τ_xy ) as follows:-

             σ_x = [tex]-\frac{G_x}{A} - \frac{M_G_Y. z*}{I_Y_Y} + \frac{M_G_Z. y*}{I_Z_Z}[/tex]

Where,

          A: The area of pipe cross section

          [tex]A = \pi * [ ( \frac{r_o}{12})^2 - ( \frac{r_i}{12})^2 ] = \pi * [ ( \frac{3}{12})^2 - ( \frac{2.75}{12})^2 ] = 0.03136 ft^2[/tex]

          z*: The distance of point "C" along z-direction from central axis ( x )

          [tex]z*= [\frac{r_i}{12} ] = [\frac{2.75}{12} ] = 0.22916 ft[/tex]

         I_YY: The second area moment of pipe along and about "y" axis:

         [tex]I_Y_Y = \frac{\pi }{4} * [ (\frac{r_o}{12})^4 - (\frac{r_i}{12})^4 ]=\frac{\pi }{4} * [ (\frac{3}{12})^4 - (\frac{2.75}{12})^4 ] \\\\I_Y_Y = 0.00090 ft^4[/tex]

         y*: The distance of point "C" along y-direction from central axis ( x )

         [tex]y* = 0[/tex]

- The normal stress ( σ_x ) becomes:

          σ_x = [tex][-\frac{1500}{0.03136} - \frac{-9,000*0.22916}{0.00090} + \frac{64,425*0}{0.00090} ] * (\frac{1}{12})^2 = 15.5 ksi[/tex]

- The planar stress is ( τ_xy ) is a contribution of torsion ( T ) and shear force ( V ):

           τ_xy = [tex]- \frac{T.c}{J} + \frac{V.Q}{I.t}[/tex]

Where,

           c: The radial distance from central axis ( x ) and point "C".

           [tex]c = \frac{r_i}{12} = \frac{2.75}{12} = 0.22916 ft[/tex]

          J: The polar moment of inertia of the annular cross section of pipe:

          [tex]J = \frac{\pi }{2}* [ ( \frac{r_o}{12})^4 - ( \frac{r_i}{12})^4 ] = \frac{\pi }{2}* [ ( \frac{3}{12})^4 - ( \frac{2.75}{12})^4 ] = 0.00180 ft^4[/tex]

          Q: The first moment of area for point "C" = semi-circle

          [tex]Q = Y_c*A_c = \frac{4*( r_m)}{3\pi } * \frac{\pi*( r_m)^2 }{2} = \frac{2. ( r_m)^3}{3} \\\\Q = \frac{2. [ ( \frac{r_o}{12})^3 - ( \frac{r_i}{12})^3] }{3} = \frac{2. [ ( \frac{3}{12})^3 - ( \frac{2.75}{12})^3] }{3} = 0.00239ft^3[/tex]

          I: The second area moment of pipe along and about "y" axis:

         [tex]I_Y_Y = \frac{\pi }{4} * [ (\frac{r_o}{12})^4 - (\frac{r_i}{12})^4 ]=\frac{\pi }{4} * [ (\frac{3}{12})^4 - (\frac{2.75}{12})^4 ] \\\\I_Y_Y = 0.00090 ft^4[/tex]    

         t: The effective thickness of thin walled pipe:

         [tex]t = 2* [ \frac{r_o}{12} - \frac{r_i}{12} ] = 2* [ \frac{3}{12} - \frac{2.75}{12} ] = 0.04166 ft[/tex]

- The planar stress is ( τ_xy ) becomes:

        τ_xy =  [tex][ - \frac{-64,800*0.22916}{0.0018} + \frac{-10,800*0.00239}{0.0009*0.04166} ] * [ \frac{1}{12}]^2 = 52.4 ksi[/tex]

- The principal stresses at point "C" can be determined from the following formula:-

       σ_x = 15.55 ksi,  σ_y = 0 ksi , τ_xy = 52.4 ksi

       σ_1 =[tex]\frac{sigma_x+sigma_y}{2} + \sqrt{(\frac{sigma_x+sigma_y}{2})^2 + (tow_x_y)^2 }[/tex]

       σ_2 = [tex]\frac{sigma_x+sigma_y}{2} - \sqrt{(\frac{sigma_x+sigma_y}{2})^2 + (tow_x_y)^2 }[/tex]

        σ_1 = [tex]\frac{15.55+0}{2} + \sqrt{(\frac{15.55+0}{2})^2 + (52.4)^2 } = 60.75 ksi[/tex]

        σ_2 =[tex]-\sqrt{\left(\frac{15.55+0}{2}\right)^2\:+\:\left(52.4\right)^2\:}+\frac{15.55+0}{2} = -45.20 ksi[/tex]

- The angle of maximum plane stress ( θ ):

       θ = [tex]0.5*arctan ( \frac{tow_x_y}{\frac{sigma_x-sigma_y}{2} } )= 0.5*arctan*( \frac{52.4}{7.8} ) = 40.8 deg[/tex]

Note: The plane stresses at point D are evaluated using the following procedure given above. Due to 5,000 character limit at Brainly, i'm unable to post here.

Answer:

(i) The inductance of the inductor is = 43.43 mH (ii) the impedance of the circuit is = 16∠58.61° Ω (iii) the phase difference for current and the voltage applied is Q = 58.61°

Explanation:

Solution

Given that:

I= 5 A

V = 125V

Resistance R= Not known yet

Thus

To find the resistance we have the following formula which is shown below:

R = V/I

=125/15

R =8.333Ω

Now,

Voltage = 240

Frequency = 50Hz

Current (I) remain at = 15A

Z= not known (impedance)

so,

To find the impedance we have the formula which is shown below:

Z = V/I =240/15

Z= 16Ω⇒ Z = R + jXL

Z = 8.333 + jXL = 16

Thus

√8.333² + XL² = 16²

8.333² + XL² = 16²

XL² = 186.561

XL = 13.658Ω

Now

We find the inductance of the Inductor and the impedance of the circuit.

(i) In solving for the inductance of the inductor, a formula is applied here, which is shown below:

L =  XL/w

=13.658/ 2π * 50

=13.658/314.15 = 0.043 = 43.43 mH

Note: w= 2πf

(ii) For the impedance of the circuit we have the following:

z = 8.333 + j 13.658

z = 16∠58.61° Ω

(iii) The next step is to find the phase difference between the applied voltage and current.

Q =  this is the voltage across the inductor in a series of resonant circuit.

Q can also be called the applied voltage

Thus,

Q is described as an Impedance angle

Therefore, Q = 58.81°

Answer:

Explanation:

For cylinder

Diameter d = 60 inches

thickness t = 0.625 inches

circumferential (hoop) stress = 30 ksi

[tex]hoop \ \ stress =\sigma_1=\frac{P_1d}{2t}\\\\\sigma_1=30ksi\\\\30000=\frac{P_1\times 60}{2\times0.625}\\\\P_1=624psi[/tex]

[tex]longitudinal \ \ stress =\sigma_2=\frac{P_2d}{2t}\\\\\sigma_2=30ksi\\\\30000=\frac{P_2\times 60}{4\times0.625}\\\\30000=\frac{P_2\times 60}{2.5}\\\\75000=P_2\times60\\\\P_2=\frac{75000}{60} \\\\P_1=1250psi[/tex]

Therefore maximum pressure without failure is P₁ = 625 psi

ii) For Sphere

[tex]\sigma_1=\sigma_2=\frac{Pd}{4t} \\\\P=\frac{30000\times 4 \times 0.625}{60} \\\\=\frac{75000}{60}\\\\=1250\ \ psi[/tex]

Answer:

Explanation:

The solution of all the four parts is provided in the attached figures

Answer:

The reason for their unusual properties of the greasy feel and low hardness is that the chemical bonds between the sheets is so weak that very low stresses can allow slip between the sheets.

Explanation:

Talc is a monoclinic mineral with a sheet structure similar to the micas and also has perfect cleavage that follows planes between the weakly bonded sheets.

Now, these sheets are held together only by van der Waals bonds and this allows them to slip past each other easily. Thus, this unique characteristic is responsible for talc's extreme softness, its greasy, soapy feel, and its value as a high-temperature lubricant.

While for graphite, it's carbon atoms are linked in a hexagonal network which forms sheets that are one atom thick. It's sheets are poorly connected and easily cleave or slide over one another when subjected to a small amount of force. Thus, gives graphite its very low hardness, its perfect cleavage, and its slippery feel.

So, we can conclude that the reason for their unusual properties is that the chemical bonds between the sheets is so weak that very low stresses can allow slip between the sheets; hence, the greasy feel and low hardness.

Follow traffic signs , Keep distance between cars , Be patient in traffic.