Talc and graphite are two of the lowest minerals on the hardness scale. They are also described by terms like greasy or soapy. Both have a crystal structure characterized by sheet-structures at the atomic level, yet they don't behave like micas. What accounts for their unusual physical properties

Image of wheel is missing, so i attached it.

Answer:

ω = 14.95 rad/s

Explanation:

We are given;

Mass of wheel; m = 20kg

T = 20 N

k_o = 0.3 m

Since the wheel starts from rest, T1 = 0.

The mass moment of inertia of the wheel about point O is;

I_o = m(k_o)²

I_o = 20 * (0.3)²

I_o = 1.8 kg.m²

So, T2 = ½•I_o•ω²

T2 = ½ × 1.8 × ω²

T2 = 0.9ω²

Looking at the image of the wheel, it's clear that only T does the work.

Thus, distance is;

s_t = θr

Since 4 revolutions,

s_t = 4(2π) × 0.4

s_t = 3.2π

So, Energy expended = Force x Distance

Wt = T x s_t = 20 × 3.2π = 64π J

Using principle of work-energy, we have;

T1 + W = T2

Plugging in the relevant values, we have;

0 + 64π = 0.9ω²

0.9ω² = 64π

ω² = 64π/0.9

ω = √64π/0.9

ω = 14.95 rad/s

Answer:

The total percentage is 3.16237%

Explanation:

Solution

Now,

We have to know what a random error is.

A random error is an error in measured caused by factors or elements which varies from one measurement to another.

The random error is shown as follows:

The average random error  is = the error found in one measurement/n^1/2

Where

n =Number ( how many times the experiment was done)

Now that we added 10 times we have that,

n → 10

Thus,

The error in one measurement = 10%

So,

The average random error = 10 %/(10)^1/2

= (10)^1/2 %

√10%

The total percentage is = 3.16237%

Answer:

Roll force, F = 5.6 MN

Explanation:

Sheet width, b = 0.5 m

Roll contact length, [tex]l_{p} = 40 mm[/tex]

Strength coefficient, [tex]\sigma_{0} = 200 MPa[/tex]

Thickness, h = 11 mm

Since the sheet of metal is cold worked by 40%, the reduction in thickness will be:

Δh = 40% * 11 = 0.4 * 11 = 4.4 mm

Strain, e = (Δh)/h

e = 4.4/11 = 0.4

The roll force is calculated by the formula:

[tex]F = \sigma_{f} l_{p} b[/tex]

[tex]\sigma_{f} = \sigma_{0} (e+1)\\\sigma_{f} = 200 (0.4+1)\\\sigma_{f} = 200 *1.4\\\sigma_{f} = 280 MPa[/tex]

Substituting the value of [tex]\sigma_{f}[/tex], [tex]l_{p}[/tex], and b into the formula for the roll force:

[tex]F = \sigma_{f} l_{p} b\\F = 280 * 0.04 * 0.5\\F = 5.6 MN[/tex]

Answer:

Blindspot Location

Explanation:

Just took the quiz

When you do a vehicle check, you do NOT need to keep an eye on Blind spot locations. The correct option is D.

A blind spot is the area of the road that can't be seen by looking forward through windscreen, or by rear-view and side-view mirrors.

While doing vehicle check, we need to check tire inflation, cleanliness of windows and mirrors along with the functioning indicator lights and headlights.

Blind spot locations does not need to be checked.

Thus, the correct option is D.

Learn more about  Blind spot location

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#SPJ2

Bank loan facilitator, and hospital emergency care specialist are the two consumer or customer services careers.

Learn more about customer:

https://brainly.com/question/13735743

Answer:

Both technician A and technician B are correct

Explanation:

A planetary gearbox consists of a gearbox with the input shaft and the output shaft that is aligned to each other. It is used to transfer the largest torque in the compact form. A planetary gearbox has a compact size and low weight and it has high power density.

One planetary gear set can provide gear reduction, overdrive, and reverse. Also, most transmissions today use compound (multiple) planetary gears set.

So, both technician A and technician B are correct.

Question:

A piston–cylinder device contains 0.85 kg of refrigerant- 134a at -10°C. The piston that is free to move has a mass of 12 kg and a diameter of 25 cm. The local atmospheric pressure is 88 kPa. Now, heat is transferred to refrigerant-134a until the temperature is 15°C. Determine (a) the final pressure, (b) the change in the volume of the cylinder, and (c) the change in the enthalpy of the refrigerant-134a.

Answer:

a) 90.4 kPa

b) 0.0205 m³

c) 17.4 kJ/kg

Explanation:

Given:

Mass, m = 0.85 kg

a) The final pressure here is equal to the initial pressure. Let's use the formula:

[tex] P_2 = P_1 = P_a_t_m + \frac{mg}{\pi D^2 / 4}[/tex]

[tex] = 88*10^3 + \frac{12kg * 9.81}{\pi (0.25)^2 / 4} [/tex]

= 90398 Pa

≈ 90.4 KPa

Final pressure = 90.4 kPa

b) Change in volume of the cylinder:

To find the initial and final volume, let's use the values from the A-13 table for refrigerant-134a, at initial values of 90.4 kPa and -10°C and final values of 90.4 kPa and 15°C

v1 = 0.2302m³/kg

h1 = 247.76 kJ/kg

v2 = 0.2544 m³/kg

h2 = 268.2 kJ/kg

Change in volume is calculated as:

Δv = m(v2 - v1)

Δv = 0.85(0.2544 - 0.2302)

= 0.0205 m³

Change in volume = 0.0205 m³

c) Change in enthalpy

Let's use the formula:

Δh = m(h2 - h1)

= 0.85(268.2 - 247.76)

= 17.4 kJ/kg

Change in enthalpy = 17.4 kJ/kg

Answer:

We can describe 15×-10 as an expression. we would describe 6×-2< 35 as an...

Explanation:

We can describe 15×-10 as an expression. we would describe 6×-2< 35 as an...

Answer:

Explanation:

The image attached to the question is shown in the first diagram below.

From the diagram given ; we can deduce a free body diagram which will aid us in solving the question.

IF we take a look at the second diagram attached below ; we will have a clear understanding of what the free body diagram of the system looks like :

From the diagram; we can determine the length of BC by using pyhtagoras theorem;

SO;

[tex]L_{BC}^2 = L_{AB}^2 + L_{AC}^2[/tex]

[tex]L_{BC}^2 = (3.5+2.5)^2+ 4^2[/tex]

[tex]L_{BC}= \sqrt{(6)^2+ 4^2}[/tex]

[tex]L_{BC}= \sqrt{36+ 16}[/tex]

[tex]L_{BC}= \sqrt{52}[/tex]

[tex]L_{BC}= 7.2111 \ m[/tex]

The cross -sectional of the cable is calculated by the formula :

[tex]A = \dfrac{\pi}{4}d^2[/tex]

where d = 4mm

[tex]A = \dfrac{\pi}{4}(4 \ mm * \dfrac{1 \ m}{1000 \ mm})^2[/tex]

A = 1.26 × 10⁻⁵ m²

However, looking at the maximum deflection  in length [tex]\delta[/tex] ; we can calculate for the force [tex]F_{BC[/tex] by using the formula:

[tex]\delta = \dfrac{F_{BC}L_{BC}}{AE}[/tex]

[tex]F_{BC} = \dfrac{ AE \ \delta}{L_{BC}}[/tex]

where ;

E = modulus elasticity

[tex]L_{BC}[/tex] = length of the cable

Replacing 1.26 × 10⁻⁵ m² for A; 200 × 10⁹ Pa for E ; 7.2111 m for [tex]L_{BC}[/tex] and 0.006 m for [tex]\delta[/tex] ; we have:

[tex]F_{BC} = \dfrac{1.26*10^{-5}*200*10^9*0.006}{7.2111}[/tex]

[tex]F_{BC} = 2096.76 \ N \\ \\ F_{BC} = 2.09676 \ kN[/tex]     ---- (1)

Similarly; we can determine the force [tex]F_{BC}[/tex] using the allowable  maximum stress; we have the following relation,

[tex]\sigma = \dfrac{F_{BC}}{A}[/tex]

[tex]{F_{BC}}= {A}*\sigma[/tex]

where;

[tex]\sigma =[/tex] maximum allowable stress

Replacing 190 × 10⁶ Pa for [tex]\sigma[/tex] ; we have :

[tex]{F_{BC}}= 1.26*10^{-5} * 190*10^{6} \\ \\ {F_{BC}}=2394 \ N \\ \\ {F_{BC}}= 2.394 \ kN[/tex]     ------ (2)

Comparing (1) and  (2)

The magnitude of the force [tex]F_{BC} = 2.09676 \ kN[/tex] since the elongation of the cable should not exceed 6mm

Finally applying the moment equilibrium condition about point A

[tex]\sum M_A = 0[/tex]

[tex]3.5 P - (6) ( \dfrac{4}{7.2111}F_{BC}) = 0[/tex]

[tex]3.5 P - 3.328 F_{BC} = 0[/tex]

[tex]3.5 P = 3.328 F_{BC}[/tex]

[tex]3.5 P = 3.328 *2.09676 \ kN[/tex]

[tex]P =\dfrac{ 3.328 *2.09676 \ kN}{3.5 }[/tex]

P = 1.9937 kN

Hence; the maximum load P that can be applied is 1.9937 kN

Answer:

The ABC Corporation

a) Total Expected Revenue (in dollars) for 20XX:

Revenue from T1 = 60,000 x $165 = $26,400,000

Revenue from T2 = 40,000 x $250 = $10,000,000

Total Revenue from T1 and T2 = $36,400,000

b) Production Level (in units) for T1 and T2

                                           T1                       T2

Total Units sold             160,000           40,000

Add Closing Inventory   25,000             9,000

Units Available for sale 185,000           49,000

less opening inventory  20,000             8,000

Production Level          165,000 units 41,000 units

c) Total Direct Material Purchases (in dollars):

Cost of direct materials used    T1                T2

A:       (165,000 x 4 x $12)   $7,920,000   $2,460,000 (41,000 x 5 x $12)

B:       (165,000 x 2 x $5)       1,650,000          615,000 (41,000 x 3 x $5)

C:                                                           0          123,000 (41,000 x 1 x$3)

Total cost                            $9,570,000     $3,198,000 Total = $12,768,000

Cost of direct per unit = $58 ($9,570,000/165,000) for T1 and $78 ($3,198,000/41,000) for T2

Cost of direct materials used for production $12,768,000

Cost of closing direct materials:

                 A  (36,000 x $12)  $432,000

                 B (32,000 x $5)        160,000

                 C (7,000 x $3)            21,000             $613,000

Cost of direct materials available for prodn   $13,381,000

Less cost of beginning direct materials:

                 A  (32,000 x $12)        $384,000

                 B  (29,000 x $5)            145,000

                 C  (6,000 x $3)                18,000        $547,000

Cost of direct materials purchases               $12,834,000

d) The Total Direct Manufacturing Labor Cost (in dollars):

                                             T1                         T2

Direct labor per unit              2 hours                  3 hours

Direct labor rate per hour    $12                        $16

Direct labor cost per unit   $24                          $48

Production level              165,000 units        41,000 units

Labor Cost ($)                $3,960,000        $1,968,000

Total labor cost  $5,928,000 ($3,960,000 + $1,968,000)

e) Total Overhead cost (in dollars):

Overhead rate  = $20 per labor hour

Overhead cost per unit: T1 = $40 ($20 x 2) and T2 = $60 ($20 x 3)

T1 overhead = $20 x 2  x 165,000) = $6,600,000

T2 overhead = $20 x 3 x 41,000) =    $2,460,000

Total Overhead cost =                        $9,060,000

Cost of goods produced:

Cost of opening inventory of materials  = $547,000

Purchases of directials materials             12,834,000

less closing inventory of materials     =      $613,000

Cost of materials used for production    12,768,000

add Labor cost                                           5,928,000

add Overhead cost                                    9,060,000

Total production cost                            $27,756,000

f) Total cost of goods sold (in dollars):

Cost of opening inventory =          $3,928,000

Total Production cost             =    $27,756,000

Cost of goods available for sale  $31,684,000

Less cost of closing inventory       $4,724,000

Total cost of goods sold            $26,960,000

g) Total expected operating income (in dollars)

Sales Revenue:  T1 and T2  $36,400,000

Cost of goods sold                 26,960,000

Gross profit                             $9,440,000

less marketing & distribution      400,000

Total Expected Operating Income = $9,040,000

Explanation:

a) Cost of beginning inventory of finished goods:

T1, (Direct materials + Labor + Overhead) X inventory units =

T1 = 20,000 x ($58 + 24 + 40) = $2,440,000

T2 = 8,000 ($78 + 48 + 60) = $1,488,000

Total cost of beginning inventory = $3,928,000

b) Cost of closing Inventory of finished goods:

T1 = 25,000 x ($58 + 24 + 40) = $3,050,000

T2 = 9,000 ($78 + 48 + 60) = $1,674,000

Total cost of closing inventory = $4,724,000

Answer:

a) [tex]\ddot n = 50400\,\frac{rev}{min^{2}}[/tex], b) [tex]n = 2450\,rev[/tex]

Explanation:

a) The acceleration experimented by the grinding wheel is:

[tex]\ddot n = \frac{4200\,\frac{rev}{min} - 0 \,\frac{rev}{min} }{\frac{5}{60}\,min }[/tex]

[tex]\ddot n = 50400\,\frac{rev}{min^{2}}[/tex]

Now, the number of revolutions done by the grinding wheel in that period of time is:

[tex]n = \frac{(4200\,\frac{rev}{min} )^{2}-(0\,\frac{rev}{min} )^{2}}{2\cdot \left(50400\,\frac{rev}{min^{2}} \right)}[/tex]

[tex]n = 175\,rev[/tex]

b) The acceleration experimented by the grinding wheel is:

[tex]\ddot n = \frac{0 \,\frac{rev}{min} - 4200\,\frac{rev}{min} }{\frac{70}{60}\,min }[/tex]

[tex]\ddot n = -3600\,\frac{rev}{min^{2}}[/tex]

Now, the number of revolutions done by the grinding wheel in that period of time is:

[tex]n = \frac{(0\,\frac{rev}{min} )^{2} - (4200\,\frac{rev}{min} )^{2}}{2\cdot \left(-3600\,\frac{rev}{min^{2}} \right)}[/tex]

[tex]n = 2450\,rev[/tex]

Answer:

a. T-V and P-V diagram are included

b. State 1: Specific volume = 0.0811753 m³/kg

State 2: Specific volume = 0.0811753 m³/kg

State 3: Specific volume = 0.0804155 m³/kg

c. Z = 51.1

d. Quality for state 1 = 100%

Quality for state 2 = 63.47%

Quality for state 3 = 100%

Explanation:

a. T-V and P-V diagram are included

b. State 1: Water vapor

P₁ = 3.0 MPa = 30 bar

T₁ = 300°C = 573.15

Saturation temperature = 233.86°C Hence the steam is super heated

Specific volume = 0.0811753 m³/kg

State 2:

Constant volume formula is P₁/T₁ = P₂/T₂

Specific volume = 0.0811753 m³/kg

T₂ = 200°C = 473.15

Therefore, P₂ = P₁/T₁ × T₂ = 3×473.15/573.15 = 2.4766 MPa

At T₂ water is mixed water and steam and the [tex]v_f[/tex] = 0.00115651 m³/kg

[tex]v_g[/tex] = 0.127222 m³/kg

State 3:

P₃ = 2.5 MPa

T₃ = 200°C

Isothermal compression P₂V₂ = P₃V₃

V₃ = P₂V₂ ÷ P₃ = 2.4766 × 0.0811753/2.5 = 0.0804155 m³/kg

Specific volume = 0.0804155 m³/kg

2) Compressibility factor is given by the relation;

[tex]Z = \dfrac{PV}{RT} = \dfrac{3\times 10^6 \times 0.0811753 }{8.3145 \times 573.15} = 51.1[/tex]

Z = 51.1

3) Gas quality, x, is given by the relation

[tex]x = \dfrac{Mass_{saturated \, vapor}}{Total \, mass} = \dfrac{v - v_f}{v_g - v_f}[/tex]

Quality at state 1 = Saturated quality = 100%

State 2 Vapor + liquid Quality

Gas quality = (0.0811753 - 0.00115651)/ (0.127222-0.00115651) = 63.47%

State 3: Saturated vapor, quality = 100%.

Answer:

(i) IR = 100.167 A Iy = 75.125∠-120 IB = 50.083 ∠+120 (ii) IN =43.374∠ -30°

Explanation:

Solution

Given that:

Three loads  24 kW, 18 kW, and 12 kW are connected between the neutral.

Voltage = 415V

Now,

(1)The current in each line conductor

Thus,

The Voltage Vpn = vL√3

Gives us, 415/√3 = 239.6 V

Then,

IR = 24 K/ Vpn ∠0°

24K/239.6 ∠0°= 100.167 A

For Iy

Iy = 18k/239. 6

= 75.125A

Thus,

Iy = 75.125∠-120 this is as a result of the 3- 0 system

Now,

IB = 12K /239.6

= 50.083 A

Thus,

IB is =50.083 ∠+120

(ii) We find the current in the neutral conductor

which is,

IN =Iy +IB +IR

= 75.125∠-120 + 50.083∠+120 +100.167

This will give us the following summation below:

-37.563 - j65.06 - 25.0415 +j 43.373 + 100.167

Thus,

IN = 37.563- j 21.687

Therefore,

IN =43.374∠ -30°

Answer:

Exit temperature = 32 °C

Explanation:

We are given;

Initial Pressure;P1 = 100 KPa

Cp =1000 J/kg.K = 1 KJ/kg.k

R = 500 J/kg.K = 0.5 Kj/Kg.k

Initial temperature;T1 = 27°C = 273 + 27K = 300 K

volume flow rate;V' = 15 m³/s

W = 130 Kw

Q = 80 Kw

Using ideal gas equation,

PV' = m'RT

Where m' is mass flow rate.

Thus;making m' the subject, we have;

m' = PV'/RT

So at inlet,

m' = P1•V1'/(R•T1)

m' = (100 × 15)/(0.5 × 300)

m' = 10 kg/s

From steady flow energy equation, we know that;

m'•h1 + Q = m'h2 + W

Dividing through by m', we have;

h1 + Q/m' = h2 + W/m'

h = Cp•T

Thus,

Cp•T1 + Q/m' = Cp•T2 + W/m'

Plugging in the relevant values, we have;

(1*300) - (80/10) = (1*T2) - (130/10)

Q and M negative because heat is being lost.

300 - 8 + 13 = T2

T2 = 305 K = 305 - 273 °C = 32 °C

13000 + 300 - 8000 = T2

Answer: $17,206.13

Explanation:

Hi, to answer this question we have to apply the next formula:  

Annual electricity cost = (P x 0.746 x Ckwh x h) /η  

P = compressor power = 78 hp  

0.746 kw/hp= constant (conversion to kw)

Ckwh = Cost per kilowatt hour = $0.11/kWh  

h = operating hours per year = 2500 h  

η = efficiency = 93% = 0.93 (decimal form)  

Replacing with the values given :  

C = ( 78 hp x 0.746 kw/hp x 0.11 $/kwh x 2500 h ) / 0.93 = $17,206.13  

Answer:

1) 2304 kPa

2) B. 200 N/m

Explanation:

The internal pressure of the of the tank  can be found from the following relations;

Resisting wall force F = p×(1/4·π·D²)

σ×A = p×(1/4·π·D²)

Where:

σ = Allowable stress of the tank

A = Area of the wall of the tank = π·D·t

t = Thickness of the tank = 15 mm. = 0.015 m

D = Diameter of the tank = 25 m

p = Maximum permissible internal pressure pressure

∴ σ×π·D·t = p×(1/4·π·D²)

p = 4×σ×t/D = 4 × 240 ×0.015/2.5 = 5.76 MPa

With a desired safety factor of 2.5, the permissible internal pressure = 5.76/2.5 = 2.304 MPa

2) The formula for average shear flow is given as follows;

[tex]q = \dfrac{T}{2 \times A_m}[/tex]

Where:

q = Average shear flow

T = Torque = 8 N·m

[tex]A_m[/tex] = Average area enclosed within tube

t = Thickness of tube = 6.35 mm = 0.00635 m

Side length of the square cross sectioned tube, s = 203 mm = 0.203 m

Average area enclosed within tube, [tex]A_m[/tex] = (s - t)² = (0.203 - 0.00635)² = 0.039 m²

[tex]\therefore q = \dfrac{8}{2 \times 0.039} = 206.9 \, N/m[/tex]

Hence the average shear flow is most nearly 200 N/m.

Following are the solution to the given question:

Calculating the allowable stress:

[tex]\to \sigma_{allow} = \frac{\sigma_y}{FS} \\\\[/tex]

              [tex]= \frac{240}{2.5} \\\\= 96\\\\[/tex]

Calculating the Thickness:

[tex]\to t =15\ mm = \frac{15\ }{1000}= 0.015\ m\\\\[/tex]

The stress in a spherical tank is defined as

[tex]\to \sigma = \frac{pD}{4t}\\\\\to 96 = \frac{p(25)}{4(0.015)}\\\\\to p = 0.2304\;\;MPa\\\\\to p = 230.4\;\;kPa\\\\\to p \approx 230\;\;kPa\\\\[/tex]

[tex]\bold{\to A= 203^2= 41209\ mm^2} \\\\[/tex]

Calculating the shear flow:

[tex]\to q=\frac{T}{2A}[/tex]

      [tex]=\frac{8}{2 \times 41209 \times 10^{-6}}\\\\=\frac{8}{0.082418}\\\\=97.066\\\\[/tex]

[tex]\to q=97 \approx 100 \ \frac{N}{m}\\[/tex]

Therefore, the final answer is "".

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Answer:

Explanation:

Given that:

Torque T = 2300 lb - ft

Bending moment M = 1500 lb - ft

axial thrust P = 2500 lb

yield points for tension  σY= 100 ksi

yield points for shear   τY = 50 ksi

Using maximum-shear-stress theory

[tex]\sigma_A = \dfrac{P}{A}+\dfrac{Mc}{I}[/tex]

where;

[tex]A = \pi c^2[/tex]

[tex]I = \dfrac{\pi}{4}c^4[/tex]

[tex]\sigma_A = \dfrac{P}{\pi c^2}+\dfrac{Mc}{ \dfrac{\pi}{4}c^4}[/tex]

[tex]\sigma_A = \dfrac{2500}{\pi c^2}+\dfrac{1500*12c}{ \dfrac{\pi}{4}c^4}[/tex]

[tex]\sigma_A = \dfrac{2500}{\pi c^2}+\dfrac{72000c}{\pi c^3}}[/tex]

[tex]\tau_A = \dfrac{T_c}{\tau}[/tex]

where;

[tex]\tau = \dfrac{\pi c^4}{2}[/tex]

[tex]\tau_A = \dfrac{T_c}{\dfrac{\pi c^4}{2}}[/tex]

[tex]\tau_A = \dfrac{2300*12 c}{\dfrac{\pi c^4}{2}}[/tex]

[tex]\tau_A = \dfrac{55200 }{\pi c^3}}[/tex]

[tex]\sigma_{1,2} = \dfrac{\sigma_x+\sigma_y}{2} \pm \sqrt{\dfrac{(\sigma_x - \sigma_y)^2}{2}+ \tau_y^2}[/tex]

[tex]\sigma_{1,2} = \dfrac{2500+72000}{2 \pi c ^3} \pm \sqrt{\dfrac{(2500 +72000)^2}{2 \pi c^3}+ \dfrac{55200}{\pi c^3}} \ \ \ \ \ ------(1)[/tex]

Let say :

[tex]|\sigma_1 - \sigma_2| = \sigma_y[/tex]

Then :

[tex]2\sqrt{( \dfrac{2500c + 72000}{2 \pi c^3})^2+ ( \dfrac{55200}{\pi c^3})^2 } = 100(10^3)[/tex]

[tex](2500 c + 72000)^2 +(110400)^2 = 10000*10^6 \pi^2 c^6[/tex]

[tex]6.25c^2 + 360c+ 17372.16-10,000\ \pi^2 c^6 =0[/tex]

According to trial and error;

c = 0.75057 in

Replacing  c into equation (1)

[tex]\sigma_{1,2} = \dfrac{2500+72000}{2 \pi (0.75057) ^3} \pm \sqrt{\dfrac{(2500 +72000)^2}{2 \pi (0.75057)^3}+ \dfrac{55200}{\pi (0.75057)^3}}[/tex]

[tex]\sigma_{1,2} = \dfrac{2500+72000}{2 \pi (0.75057) ^3} + \sqrt{\dfrac{(2500 +72000)^2}{2 \pi (0.75057)^3}+ \dfrac{55200}{\pi (0.75057)^3}} \ \ \ OR \\ \\ \\ \sigma_{1,2} = \dfrac{2500+72000}{2 \pi (0.75057) ^3} - \sqrt{\dfrac{(2500 +72000)^2}{2 \pi (0.75057)^3}+ \dfrac{55200}{\pi (0.75057)^3}}[/tex]

[tex]\sigma _1 = 22193 \ Psi[/tex]

[tex]\sigma_2 = -77807 \ Psi[/tex]

The required diameter d  = 2c

d = 1.50 in   or   0.125 ft

Answer:

The correct answer to the following question will be "1.23 mm".

Explanation:

The given values are:

Average normal stress,

[tex]\sigma=200 \ MPa[/tex]

Elastic module,

[tex]E = 77 \ GPa[/tex]

Length,

[tex]L = 570 \ mm[/tex]

To find the deformation, firstly we have to find the equation:

⇒  [tex]\delta=\Sigma\frac{N_{i}L_{i}}{E \ A_{i}}[/tex]

⇒     [tex]=\frac{P(\frac{L}{H})}{E(bt)} +\frac{P(\frac{L}{2})}{E (bt)(\frac{2}{3})}+\frac{P(\frac{L}{H})}{Ebt}[/tex]

On taking "[tex]\frac{PL}{Ebt}[/tex]" as common, we get

⇒     [tex]=\frac{\frac{PL}{Ebt}}{[\frac{1}{4}+\frac{3}{4}+\frac{1}{4}]}[/tex]

⇒     [tex]=\frac{5PL}{HEbt}[/tex]

Now,

The stress at the middle will be:

⇒  [tex]\sigma=\frac{P}{A}[/tex]

⇒     [tex]=\frac{P}{(\frac{2}{3})bt}[/tex]

⇒     [tex]=\frac{3P}{2bt}[/tex]

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

Hence,

⇒  [tex]\delta=\frac{5 \sigma \ L}{6E}[/tex]

On putting the estimated values, we get

⇒     [tex]=\frac{5\times 200\times 570}{6\times 77\times 10^3}[/tex]

⇒     [tex]=\frac{570000}{462000}[/tex]

⇒     [tex]=1.23 \ mm[/tex]  

Answer:

HB = 3.22

Explanation:

The formula to calculate the Brinell Hardness is given as follows:

[tex]HB = \frac{2P}{\pi D\sqrt{D^{2}- d^{2} } }[/tex]

where,

HB = Brinell Hardness = ?

P = Applied Load in kg = 500 kg

D = Diameter of Indenter in mm = 10 mm

d = Diameter of the indentation in mm = 1.55 mm

Therefore, using these values, we get:

[tex]HB = \frac{(2)(500)}{\pi (10)\sqrt{10^{2}- 1.55^{2} } }[/tex]

HB = 3.22

Answer:

The Poisson's Ratio of the bar is 0.247

Explanation:

The Poisson's ratio is got by using the formula

Lateral strain / longitudinal strain

Lateral strain = elongation / original width (since we are given the change in width as a result of compession)

Lateral strain = 0.15mm / 40 mm =0.00375

Please note that strain is a dimensionless quantity, hence it has no unit.

The Longitudinal strain is the ratio of the elongation to the original length in the longitudinal direction.

Longitudinal strain = 4.1 mm / 270 mm = 0.015185

Hence, the Poisson's ratio of the bar is 0.00375/0.015185 = 0.247

The Poisson's Ratio of the bar is 0.247

Please note also that this quantity also does not have a dimension

Answer:

With the Breakthrough of Technology, the rate at which things are done are becoming much more easy. but without basic science, innovation towards technology cannot occur, so the both work hand in hand in the world of technology today.

Explanation:

Technological innovation and Basic science research plays a major role in the world of science and technology today, while we all want technology innovation the more, without basic science, innovation cannot come in place,

Just as we are going further in technology, breakthroughs and growth are been made which helps on the long run in science research which in turn has made things to be done much better and easily.

Answer:

The answer is 1823.9

Explanation:

Solution

Given that:

m = 5.5 kg/s

= m₁ = m₂ = m₃

The work carried out by the energy balance is given as follows:

m₁h₁ = m₂h₂ +m₃h₃ + w

Now,

By applying the steam table we have that<

p₃ = 50 kPa

T₃ = 100°C

Which is

h₃ = 2682.4 kJ/KJ

s₃ = 7.6953 kJ/kgK

Since it is an isentropic process:

Then,

p₂ =  500 kPa

s₂=s₃ = 7.6953 kJ/kgK

which is

h₂ =3207.21 kJ/KgK

p₁ = 3HP0

s₁ = s₂=s₃ = 7.6953 kJ/kgK

h₁ =3854.85 kJ/kg

Thus,

Since 5 % of this flow diverted to p₂ =  500 kPa

Then

w =m (h₁-0.05 h₂ -0.95 )h₃

5.5(3854.85 - 0.05 * 3207.21  - 0.95 * 2682.4)

5.5( 3854.83 * 3207.21 - 0.95 * 2682.4)

5.5 ( 123363249.32 -0.95 * 2682.4)

w=1823.9

Answer:

4.75m^2

Explanation:

Given:-

- Temperature of hot fluid at inlet:  [tex]T_h_i = 300[/tex] °C

- Temperature of cold fluid at outlet: [tex]T_c_o = 120[/tex] °C

- Temperature of cold fluid at inlet: [tex]T_c_i = 35[/tex] °C

- The overall heat transfer coefficient: U = 1500 W / m^2 K

- The flow rate of cold fluid: m_c = 10,00 kg/ h

- The flow rate of hot fluid: m_h = 5,000 kg/h

Solution:-

- We will evaluate water properties at median temperatures of each fluid using table A-4.

Cold fluid:   Tci = 35°C , Tco = 35°C

                            Tcm = 77.5 °C ≈ 350 K  --- > [tex]C_p_c = 4195 \frac{J}{kg.K}[/tex]

 Hot fluid:     Thi = 300°C , Tho = 150°C ( assumed )

                             Thm = 225 °C ≈ 500 K --- > [tex]C_p_h = 4660 \frac{J}{kg.K}[/tex]

- We will use logarithmic - mean temperature rate equation as follows:

                             [tex]A_s = \frac{q}{U*dT_l_m}[/tex]

Where,

                 A_s : The surface area of heat exchange

                 ΔT_lm: the logarithmic differential mean temperature

                 q: The rate of heat transfer

- Apply the energy balance on cold fluid as follows:

                   [tex]q = m_c * C_p_c * ( T_c_o - T_c_i )\\\\q = \frac{10,000}{3600} * 4195 * ( 120 - 35 )\\\\q = 9.905*10^5 W[/tex]

- Similarly, apply the heat balance on hot fluid and evaluate the outlet temperature ( Tho ) :

                   [tex]T_h_o = T_h_i - \frac{q}{m_h * C_p_h} \\\\T_h_o = 300 - \frac{9.905*10^5}{\frac{5000}{3600} * 4660} \\\\T_h_o = 147 C[/tex]

- We will use the experimental results of counter flow ( unmixed - unmixed ) plotted as figure ( Fig . 11.11 ) of the " The fundamentals to heat transfer" and determine the value of ( P , R , F ).

- So the relations from the figure 11.11 are:

                  [tex]P = \frac{T_c_o - T_c_i}{T_h_i - T_c_i} \\\\P = \frac{120 - 35}{300 - 35} \\\\P = 0.32[/tex]    

                 [tex]R = \frac{T_h_i - T_h_o}{T_c_o - T_c_i} \\\\R = \frac{300 - 147}{120 - 35} \\\\R = 1.8[/tex]

Therefore,         P = 0.32 , R = 1.8 ---- > F ≈ 0.97

- The log-mean temperature ( ΔT_lm - cf ) for counter-flow heat exchange can be determined from the relation:

                        [tex]dT_l_m = \frac{( T_h_i - T_c_o ) - ( T_h_o - T_c_i ) }{Ln ( \frac{( T_h_i - T_c_o )}{( T_h_o - T_c_i )} ) } \\\\dT_l_m = \frac{( 300 - 120 ) - ( 147 - 35 ) }{Ln ( \frac{( 300-120 )}{( 147-35)} ) } \\\\dT_l_m = 143.3 K[/tex]

- The log - mean differential temperature for counter flow is multiplied by the factor of ( F ) to get the standardized value of log - mean differential temperature:

                       [tex]dT_l = F*dT_l_m = 0.97*143.3 = 139 K[/tex]

- The required heat exchange area ( A_s ) can now be calculated:

                     [tex]A_s = \frac{9.905*10^5 }{1500*139} \\\\A_s = 4.75 m^2[/tex]

Answer:

A. True

Explanation:

When an electromagnetic field wave strikes a conductor, say a wire, it induces an alternating current that is proportional to the wave in the conductor. This is a reversal of generating electromagnetic wave from accelerating a charged particle. This phenomenon is used in radio antena for receiving radio wave signals and also use in medicine for body scanning.

Answer:

The minimum allowable bolt diameter required to support an applied load of P = 450 kN is 45.7 milimeters.

Explanation:

The complete statement of this question is "Five bolts are used in the connection between the axial member and the support. The ultimate shear strength of the bolts is 320 MPa, and a factor of safety of 4.2 is required with respect to fracture. Determine the minimum allowable bolt diameter required to support an applied load of P = 450 kN"

Each bolt is subjected to shear forces. In this case, safety factor is the ratio of the ultimate shear strength to maximum allowable shear stress. That is to say:

[tex]n = \frac{S_{uts}}{\tau_{max}}[/tex]

Where:

[tex]n[/tex] - Safety factor, dimensionless.

[tex]S_{uts}[/tex] - Ultimate shear strength, measured in pascals.

[tex]\tau_{max}[/tex] - Maximum allowable shear stress, measured in pascals.

The maximum allowable shear stress is consequently cleared and computed: ([tex]n = 4.2[/tex], [tex]S_{uts} = 320\times 10^{6}\,Pa[/tex])

[tex]\tau_{max} = \frac{S_{uts}}{n}[/tex]

[tex]\tau_{max} = \frac{320\times 10^{6}\,Pa}{4.2}[/tex]

[tex]\tau_{max} = 76.190\times 10^{6}\,Pa[/tex]

Since each bolt has a circular cross section area and assuming the shear stress is not distributed uniformly, shear stress is calculated by:

[tex]\tau_{max} = \frac{4}{3} \cdot \frac{V}{A}[/tex]

Where:

[tex]\tau_{max}[/tex] - Maximum allowable shear stress, measured in pascals.

[tex]V[/tex] - Shear force, measured in kilonewtons.

[tex]A[/tex] - Cross section area, measured in square meters.

As connection consist on five bolts, shear force is equal to a fifth of the applied load. That is:

[tex]V = \frac{P}{5}[/tex]

[tex]V = \frac{450\,kN}{5}[/tex]

[tex]V = 90\,kN[/tex]

The minimum allowable cross section area is cleared in the shearing stress equation:

[tex]A = \frac{4}{3}\cdot \frac{V}{\tau_{max}}[/tex]

If [tex]V = 90\,kN[/tex] and [tex]\tau_{max} = 76.190\times 10^{3}\,kPa[/tex], the minimum allowable cross section area is:

[tex]A = \frac{4}{3} \cdot \frac{90\,kN}{76.190\times 10^{3}\,kPa}[/tex]

[tex]A = 1.640\times 10^{-3}\,m^{2}[/tex]

The minimum allowable cross section area can be determined in terms of minimum allowable bolt diameter by means of this expression:

[tex]A = \frac{\pi}{4}\cdot D^{2}[/tex]

The diameter is now cleared and computed:

[tex]D = \sqrt{\frac{4}{\pi}\cdot A}[/tex]

[tex]D =\sqrt{\frac{4}{\pi}\cdot (1.640\times 10^{-3}\,m^{2})[/tex]

[tex]D = 0.0457\,m[/tex]

[tex]D = 45.7\,mm[/tex]

The minimum allowable bolt diameter required to support an applied load of P = 450 kN is 45.7 milimeters.

We have that the minimum allowable bolt diameter is mathematically given as

From the question we are told

Generally the equation for the stress   is mathematically given as

[tex]\mu= 320/4.2 \\\\\mu= 76.190 N/mm^2[/tex]

Therefore

Force = Stress * area

Force = P/2

F= 425,000 N / 2 = 212,500 N

Hence area of each bolt is given as

212,500 = 76.190*( 5* area of each bolt)

area of each bolt = 557.815

Since

area of each bolt=\pi*d^2/4

\pi*d^2/4 = 557.815

For more information on diameter visit

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Answer:

Settling Velocity (Up)= 2.048*10^-5 m/s

Reynolds number Re = 2.159*10^-3

Explanation:

We proceed as follows;

Diameter of Particle = 0.09 mm = 0.09*10^-3 m

Solid Particle Density = 2002 kg/m3

Solid Fraction, θ= 0.45

Temperature = 26.7°C

Viscosity of water = 0.8509*10^-3 kg/ms

Density of water at 26.7 °C = 996.67 kg/m3

The velocity between the interface, i.e between the suspension and clear water is given by,

U = [ ((nf/ρf)/d)D^3] [18+(1/3)D^3)(1/2)]

D = d[(ρp/ρf)-1)g*(ρf/nf)^2]^(1/3)

D = 2.147

U = 0.0003m/s (n = 4.49)

Up = 0.0003 * (1-0.45)^4.49 = 2.048*10^-5 m/s

Re=0.09*10^-3*2.048*10^-5*996.67/0.0008509 = 2.159*10^-3

Answer:

a) 4160 V

b) 12 kW and 81 kVAR

c)  54 kW and 477 kVAR

Explanation:

1) The phase voltage is given as:

[tex]V_p=\frac{3810.5}{\sqrt{3} }=2200 V[/tex]

The complex power S is given as:

[tex]S=560.1(0.707 +j0.707)+132=660\angle 36.87^o \ KVA[/tex]

[tex]where\ S^*\ is \ the \ conjugate\ of \ S\\Therefore\ S^*=660\angle -36.87^oKVA[/tex]

The line current I is given as:

[tex]I=\frac{S^*}{3V}=\frac{660000\angle -36.87}{3(2200)} =100\angle -36.87^o\ A[/tex]

The phase voltage at the sending end is:

[tex]V_s=2200\angle 0+100\angle -36.87(0.4+j2.7)=2401.7\angle 4.58^oV[/tex]

The magnitude of the line voltage at the source end of the line ([tex]V_{sL}=\sqrt{3} |V_s|=\sqrt{3} *2401.7=4160V[/tex]

b) The Total real and reactive power loss in the line is:

[tex]S_l=3|I|^2(R+jX)=3|100|^2(0.4+j2.7)=12000+j81000[/tex]

The real power loss is 12000 W = 12 kW

The reactive power loss is 81000 kVAR = 81 kVAR

c) The sending power is:

[tex]S_s=3V_sI^*=3(2401.7\angle 4.58)(100\angle 36.87)=54000+j477000[/tex]

The Real power delivered by the supply = 54000 W = 54 kW

The Reactive power delivered by the supply = 477000 VAR = 477 kVAR

Answer:

positive sides:

negative sides:

Answer:

[tex]\mathbf{h_2 =1021.9 \ mm}[/tex]

Explanation:

Given that :

The initial volume of water [tex]V_1[/tex] = 1000.00 mL = 1000000 mm³

The initial temperature of the water  [tex]T_1[/tex] = 10° C

The height of the water column h = 1000.00 mm

The final temperature of the water [tex]T_2[/tex] = 70° C

The coefficient of thermal expansion for the glass is  ∝ = [tex]3.8*10^{-6 } mm/mm \ per ^oC[/tex]

The objective is to determine the the depth of the water column

In order to do that we will need to determine the volume of the water.

We obtain the data for physical properties of water at standard sea level atmospheric from pressure tables; So:

At temperature [tex]T_1 = 10 ^ 0C[/tex]  the density of the water is [tex]\rho = 999.7 \ kg/m^3[/tex]

At temperature [tex]T_2 = 70^0 C[/tex]  the density of the water is [tex]\rho = 977.8 \ kg/m^3[/tex]

The mass of the water is  [tex]\rho V = \rho _1 V_1 = \rho _2 V_2[/tex]

Thus; we can say [tex]\rho _1 V_1 = \rho _2 V_2[/tex];

⇒ [tex]999.7 \ kg/m^3*1000 \ mL = 977.8 \ kg/m^3 *V_2[/tex]

[tex]V_2 = \dfrac{999.7 \ kg/m^3*1000 \ mL}{977.8 \ kg/m^3 }[/tex]

[tex]V_2 = 1022.40 \ mL[/tex]

[tex]v_2 = 1022400 \ mm^3[/tex]

Thus, the volume of the water after heating to a required temperature of  [tex]70^0C[/tex] is 1022400 mm³

However; taking an integral look at this process; the volume of the water before heating can be deduced by the relation:

[tex]V_1 = A_1 *h_1[/tex]

The area of the water before heating is:

[tex]A_1 = \dfrac{V_1}{h_1}[/tex]

[tex]A_1 = \dfrac{1000000}{1000}[/tex]

[tex]A_1 = 1000 \ mm^2[/tex]

The area of the heated water is :

[tex]A_2 = A_1 (1 + \Delta t \alpha )^2[/tex]

[tex]A_2 = A_1 (1 + (T_2-T_1) \alpha )^2[/tex]

[tex]A_2 = 1000 (1 + (70-10) 3.8*10^{-6} )^2[/tex]

[tex]A_2 = 1000.5 \ mm^2[/tex]

Finally, the depth of the heated hot water is:

[tex]h_2 = \dfrac{V_2}{A_2}[/tex]

[tex]h_2 = \dfrac{1022400}{1000.5}[/tex]

[tex]\mathbf{h_2 =1021.9 \ mm}[/tex]

Hence the depth of the heated hot  water is [tex]\mathbf{h_2 =1021.9 \ mm}[/tex]