Under normal operating conditions, the electric motor exerts a torque of 2.8 kN-m.on shaft AB. Knowing that each shaft is solid, determine the maximum shearing stress in a) shaft AB b) shaft BC c) shaft CD (25 points) Given that the torque at B

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:

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:

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

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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:

Note: The diagram referred to in this question is attached as a file below.

Answer:

The block descended a distance of 9.75m from the instant the brake is applied until it stops.

Explanation:

For clarity and easiness of expression, the calculations and the Free Body Diagram are contained in the attached file. Check the attached file below.

The block descended a distance of 9.75 m

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:

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:

Reduce sound transmission.

Explanation:

A caulking is a flexible material used to seal joints, cracks or gaps formed between building materials and pipes against leakage.

Caulking is recommended around the edges of partitions between apartments to reduce sound transmission.

Hence, in the event that an individual notices that air or sound is gaining entrance into their apartment, a caulking can be used to mitigate this noise or unwanted sound.

The caulking when applied to the gap or edges of partitions between apartments would create a tight seal and block the flow or entry of air, thereby reducing sound transmission.

Answer:

a = 0.07m or 70mm

b = 0.205m or 205mm

Explanation:

Given the following data;

Modulus of rigidity, G = 14MPa=14000000Pa.

c = 80mm = 0.08m.

P = 46kN=46000N.

Shearing stress (r) in the rubber shouldn't exceed 1.4MPa=1400000Pa.

Deflection (d) of the plate is to be at least 7mm = 0.007m.

From shearing strain;

[

[tex]Modulus Of Elasticity, E = \frac{d}{a} =\frac{r}{G}[/tex]

Making a the subject formula;

[tex]a = \frac{Gd}{r}[/tex]

Substituting into the above formula;

[tex]a = \frac{14000000*0.007}{1400000}[/tex]

[tex]a = \frac{98000}{1400000}[/tex]

[tex]a = 0.07m or 70mm[/tex]

a = 0.07m or 70mm.

Also, shearing stress;

[tex]r = \frac{P}{2bc}[/tex]

Making b the subject formula;

[tex]b = \frac{P}{2cr}[/tex]

Substituting into the above equation;

[tex]b = \frac{46000}{2*0.08*1400000}[/tex]

[tex]b = \frac{46000}{224000}[/tex]

[tex]b = 0.205m or 205mm[/tex]

b = 0.205m or 205mm

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.

Answer:

the overall elongation δ of the bar is  1.2337 mm

Explanation:

From the information given :

According to the principle of superposition being applied to the axial load P of the system; we have:

[tex]\delta = \delta_{AB} +\delta_{BC} + \delta_{CD}[/tex]    

where;

δ = overall elongation

[tex]\delta _{AB}[/tex] = elongation of bar AB

[tex]\delta _{BC}[/tex] = elongation of  bar BC

[tex]\delta _{CD} =[/tex]  elongation of bar CD]

If we replace; [tex]\dfrac{PL}{AE}[/tex] for  δ  and bt for area;

we have:

[tex]\delta = \dfrac{P_{AB}L_{AB}}{(b_{AB}t)E} +\dfrac{P_{BC}L_{BC}}{(b_{BC}t)E}+\dfrac{P_{CD}L_{CD}}{(b_{CD}t)E}[/tex]

where ;

P = load

L = length of the bar

A = area of the cross-section

E = young modulus of elasticity

Let once again replace:

P for [tex]P_{AB}, P_{BC} , P_{CD}[/tex]  (since load in all member of AB, BC and CD will remain the same )

[tex]\dfrac{L}{4}[/tex] for [tex]L_{AB}[/tex],  

[tex]\dfrac{L}{2}[/tex] for [tex]L_{BC}[/tex] and

[tex]\dfrac{L}{4}[/tex] for [tex]L_{CD}[/tex]

[tex]2\dfrac{b}{3}[/tex] for  [tex]b_{BC}[/tex]

b for  [tex]b_{CD}[/tex]

[tex]\delta = \dfrac{P (\dfrac{L}{4})}{btE}+ \dfrac{P (\dfrac{L}{2})}{2 \dfrac{b}{3}tE}+\dfrac{P (\dfrac{L}{4})}{btE}[/tex]

[tex]\delta = \dfrac{PL}{btE}[\dfrac{1}{4}+ \dfrac{1}{2}*\dfrac{3}{2}+ \dfrac{1}{4}][/tex]

[tex]\delta = \dfrac{5}{4}\dfrac{PL}{btE} --- \ (1)[/tex]

The stress in the central portion can be calculated as:

[tex]\sigma = \dfrac{P}{A}[/tex]

[tex]\sigma = \dfrac{P}{\dfrac{2}{3}bt}[/tex]

[tex]\sigma = \dfrac{3P}{2bt}[/tex]

So; Now:

[tex]\delta = \dfrac{5}{4}* \dfrac{2 * \sigma}{3}*\dfrac{L}{E}[/tex]

[tex]\delta= \dfrac{5}{4}* \dfrac{2 * 200}{3}*\dfrac{570}{77*10^3 \ MPa}[/tex]

δ = 1.2337 mm

Therefore, the overall elongation δ of the bar is  1.2337 mm

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

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:

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

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:

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:

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:

a. load factor = 0.893

shrinkage factor = 0.848

b. Number of Trucks loads = 113,585 Trucks loads

Explanation:

Here, we start by identifying the factors as given in the question.

γn = 2800 lb/cy

γloose = 2500 lb/cy

and γcompacted = 3300 lb/cy

a. Mathematically,

Load factor = γloose/γn = 2500/2800 = 0.893

Shrinkage factor = γn/γcompacted = 2800/3300 = 0.848

b. To find the number of trucks loads with a capacity of 5 lcy/truck, we use the mathematical formula as follows;

ρlcy = 5

Load factor × Shrinkage factor = ρloose/γn × γn/γcompacted = ρlcy/ρccy

0.893 × 0.848 = 5/ρccy

ρccy =5/(0.893 × 0.848) = 6.603

The number of truck loads = 750,000/6.603 = 113,584.7 which is approximately 113,585 trucks loads

Answer:

None of these

Explanation:

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

Therefore, the correct answer is "None of these "

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

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.

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:

Answer:

• it charges banks more interest

• it sells more securities

• it decreases the money supply

Explanation:

hope this help edge 21

Answer:

Explanation:

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

Answer:

Explanation:

A negative binary number is represeneted by its 2's complement value. To get 2's complement, you just need to invert the bits and add 1 to it. So the formula is:

  twos_complement = ~val + 1

So you start out with 22 and you want to make it negative.

22₁₀ = ‭0001 0110‬₂    

~22₁₀ = ‭1110 1001‬₂   inverting the bits

~22₁₀ + 1 = ‭1110 1010‬₂   adding 1 to it.

so -22₁₀ == ~22₁₀ + 1 == ‭1110 1010‬₂

Do the same process for 16-bits and 32-bits and you'll find that the most significant bits will be padded with 1's.

-22₁₀ = ‭1110 1010‬₂     8-bits

-22₁₀ = ‭1111 1111 1110 1010‬‬₂     16-bits

-22₁₀ = ‭‭1111 1111 1111 1111 1111 1111 1110 1010‬‬‬₂  32-bits

Answer:

(a) k = 0.09 s⁻¹

(b) The velocity= ± 16.97 mm/s

Explanation:

(a) Given that the acceleration = a = k(100 - x)

Therefore;

[tex]a = \dfrac{dv}{dt} = \dfrac{dv}{dx} \times \dfrac{dx}{dt} = \dfrac{dv}{dx} \times v = k(100 - x)[/tex]

When x = 40 mm, v = 0 mm/s hence;

[tex]\int\limits^v_0 {v } \, dv = \int\limits^x_{40} {k(100 - x)} \, dx[/tex]

[tex]\dfrac{1}{2} v^2 = k \cdot \left [100\cdot x-\frac{1}{2}\cdot x^{2} \right ]_{x}^{40}[/tex]

[tex]\dfrac{1}{2} v^2 = -\dfrac{ k\cdot \left (x^{2}-200\cdot x+6400 \right ) }{2}[/tex]

At x = 100 mm, v = 18 mm/s hence we have;

[tex]\dfrac{1}{2} 18^2 = -\dfrac{ k\cdot \left (100^{2}-200\times 100+6400 \right ) }{2} = 1800\cdot k[/tex]

[tex]\dfrac{1}{2} 18^2 =162 = 1800\cdot k[/tex]

k = 162/1800 = 9/100 = 0.09 s⁻¹

(b) When x = 120 mm, we have

[tex]\dfrac{1}{2} v^2 = -\dfrac{ 0.09\times \left (120^{2}-200\times 120+6400 \right ) }{2} = 144[/tex]

Therefore;

v² = 2 × 144 = 288

The velocity, v = √288 = ±12·√2 = ± 16.97 mm/s.

Answer:

Explanation:

Given that,

The area of glass [tex]A_g[/tex] = [tex]0.11m^2[/tex]

The thickness of the glass [tex]t_g=4mm=4\times10^-^3m[/tex]

The area of the styrofoam [tex]A_s=11m^2[/tex]

The thickness of the styrofoam [tex]t_s=0.20m[/tex]

The thermal conductivity of the glass [tex]k_g=0.80J(s.m.C^o)[/tex]

The thermal conductivity of the styrofoam  [tex]k_s=0.010J(s.m.C^o)[/tex]

Inside and outside temperature difference is ΔT

The heat loss due to conduction in the window is

[tex]Q_g=\frac{k_gA_g\Delta T t}{t_g} \\\\=\frac{(0.8)(0.11)(\Delta T)t}{4.0\times 10^-^3}\\\\=(22\Delta Tt)j[/tex]

The heat loss due to conduction in the wall is

[tex]Q_s=\frac{k_sA_s\Delta T t}{t_g} \\\\=\frac{(0.010)(11)(\Delta T)t}{0.20}\\\\=(0.55\Delta Tt)j[/tex]

The net heat loss of the wall and the window is

[tex]Q=Q_g+Q_s\\\\=\frac{k_gA_g\Delta T t}{t_g}+\frac{k_sA_s\Delta T t}{t_g}\\\\=(22\Delta Tt)j +(0.55\Delta Tt)j \\\\=(22.55\Delta Tt)j[/tex]

The percentage of heat lost by the window is

[tex]=\frac{Q_g}{Q}\times 100\\\\=\frac{22\Delta T t}{22.55\Delta T t}\times 100\\\\=97.6 \%[/tex]