Use a delta-star conversion to simplify the delta BCD (40 , 16 , and 8 ) in the
bridge network in Fig. f and find the equivalent resistance that replaces the network
between terminals A and B, and hence find the current I if the source voltage is 52 V.​

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:

The dimension of the  square cross section is = 30mm * 30mm

Explanation:

Before proceeding with the calculations convert MPa to Kpsi

Sut ( ultimate strength ) = 770 MPa * 0.145 Kpsi/MPa = 111.65 Kpsi

the fatigue strength factor of Sut at 10^3 cycles for Se = Se' = 0.5 Sut

at 10^6 cycles" for 111.65 Kpsi  = f ( fatigue strength factor) = 0.83

To calculate the endurance limit  use  Se' = 0.5 Sut      since Sut < 1400 MPa

Se'( endurance limit ) = 0.5 * 770 = 385 Mpa

The surface condition modification factor

Ka = 57.7 ( Sut )^-0.718

Ka = 57.7 ( 770 ) ^-0.718

Ka = 0.488

Assuming the size modification factor (Kb) = 0.85 and also assuming all modifiers are equal to one

The endurance limit at the critical location of a machine part can be expressed as :

Se = Ka*Kb*Se'

Se = 0.488 * 0.85 * 385 = 160 MPa

Next:

Calculating the constants to find the number of cycles

α = [tex]\frac{(fSut)^2}{Se}[/tex]

α =[tex]\frac{(0.83*770)^2}{160}[/tex]  =  2553 MPa

b = [tex]-\frac{1}{3} log(\frac{fSut}{Se} )[/tex]

b = [tex]-\frac{1}{3} log (\frac{0.83*770}{160} )[/tex]  = -0.2005

Next :

calculating the fatigue strength using the relation

Sf = αN^b

N = number of cycles

Sf = 2553 ( 10^4) ^ -0.2005

Sf = 403 MPa

Calculate the maximum moment of the beam

M = 2000 * 0.6 = 1200 N-m

calculating the maximum stress in the beam

∝ = ∝max = [tex]\frac{Mc}{I}[/tex]

Where c = b/2 and   I = b(b^3) / 12

hence ∝max = [tex]\frac{6M}{b^3}[/tex]  =  6(1200) / b^3   =  7200/ b^3   Pa

note: b is in (meters)

The expression for the factor of safety is written as

n = [tex]\frac{Sf}{\alpha max }[/tex]

Sf = 403, n = 1.5 and ∝max = 7200 / b^3

= 1.5 = [tex]\frac{403(10^6 Pa/Mpa)}{7200 / B^3}[/tex]   making b subject of the formula in other to get the value of b

b = 0.0299 m * 10^3 mm/m

b = 29.9 mm therefore b ≈ 30 mm

since  the size factor  assumed is near the calculated size factor using this relation : de = 0.808 ( hb)^1/2

the dimension = 30 mm by 30 mm

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:

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:

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]  

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

Answer:

final temperature = 26.5°

Explanation:

Initial volume of water is 1 x 1 x 1 = 1 [tex]m^{3}[/tex]

Initial temperature of water = 20° C

Density of water = 1000 kg/[tex]m^{3}[/tex]

volume of copper block = 0.46 x 0.46 x 0.46 = 0.097 [tex]m^{3}[/tex]

Initial temperature of copper block = 100° C

Density of copper = 8960 kg/[tex]m^{3}[/tex]

Final volume of water = 1 - 0.097 = 0.903 [tex]m^{3}[/tex]

Assumptions:

mass of water remaining in the tank will be density x volume = 1000 x 0.903 = 903 kg

specific heat capacity of water c = 4186 J/K-kg

heat content of water left Hw = mcT = 903 x 4186 x 20 = 75.59 Mega-joules

mass of copper will be density x volume = 8960 x 0.097 = 869.12 kg

specific heat capacity of copper is 385 J/K-kg

heat content of copper Hc = mcT = 869.12 x 385 x 100 = 33.46 Mega-joules

total heat in the system = 75.59 + 33.46 = 109.05 Mega-joules

this heat will be distributed in the entire system

heat energy of water within the system = mcT

where T is the final temperature

= 903 x 4186 x T = 3779958T

for copper, heat will be

mcT = 869.12 x 385 = 334611.2T

these component heats will sum up to the final heat of the system, i.e

3779958T + 334611.2T = 109.05 x [tex]10^{6}[/tex]

4114569.2T = 109.05 x [tex]10^{6}[/tex]

final temperature T = (109.05 x [tex]10^{6}[/tex])/4114569.2 = 26.5°

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:

Pressure difference (ΔP) = 63,519.75 kpa

Explanation:

Given:

ρ = 1,000 kg/m³

Height of cylindrical container used (h) = 7m / 2 = 3.5m

Specific gravity (sg) = 0.85

Find:

Pressure difference (ΔP).

Computation:

⇒ Pressure difference (ΔP) = h g [ ρ(sg) + ρ]                ∵ [ g = 9.81]

⇒ Pressure difference (ΔP) = (3.5)(9.81) [ 1,000(0.85) + 1,000]

⇒ Pressure difference (ΔP) = 34.335 [8,50 + 1,000]

⇒ Pressure difference (ΔP) = 34.335 [1,850]

⇒ Pressure difference (ΔP) = 63,519.75 kpa

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:

Explanation:

Consider schedules S3, S4, and S5 below. Determine whether each schedule is strict, cascadeless, recoverable, or non-recoverable. You need to explain your reason.

S3: r1(x), r2(z), r1(z), r3(x), r3(y), w1(x), c1, w3(y), c3, r2(y), w2(z),w2(y),c2

S4: r1(x), r2(z), r1(z), r3(x), r3(y),w1(x),w3(y), r2(y),w2(z),w2(y), c1,c2, c3

S5: r1(x), r2(z), r3(x), r1(z), r2(y), r3(y), w1(x), c1, w2(z), w3(y), w2(y), c3, c2

Strict schedule:

A schedule is strict if it satisfies the following conditions:

Tj reads a data item X after Ti has written to X and Ti is terminated means aborted or committed.

Tj writes a data item X after Ti has written to X and Ti is terminated means aborted or committed.

S3 is not strict because In a strict schedule T3 must read X after C1 but here T3 reads X (r3(X)) before Then T1 has written to X (w1(X)) and T3 commits after T1.

S4 is not strict because In a strict schedule T3 must read X after C1, but here T3 reads X (r3(X)) before T1 has written to X (w1(X)) and T3 commits after T1.

S5 is not strict because T3 reads X (r3(X)) before T1 has written to X (w1(X))

but T3 commits after T1. In a strict schedule T3 must read X after C1.

Cascadeless schedule:

Cascadeless schedule follows the below condition:

Tj reads X only? after Ti has written to X and terminated means aborted or committed.

S3 is not cascadeless schedule because T3 reads X (r3(X)) before T1 commits.

S4 is not cascadeless schedule because T3 reads X (r3(X)) before T1 commits.

S5 is not cascadeless schedule because T3 reads X (r3(X)) before T1 commits or T2 reads Y (r2(Y)) before T3 commits.

But while come to the definition of cascadeless schedules S3, S4, and S4 are not cascadeless, and T3 is not affected if T1 is rolled back in any of the schedules, that is,

T3 does not have to roll back if T1 is rolled back. The problem occurs because these

schedules are not serializable.

Recoverable schedule:

Schedule that follows the below condition:

-----Tj commits after Ti if Tj has?read any data item written by Ti.

Ci > Cj means that Ci happens before Cj. Ai denotes abort Ti. To test if a schedule is

recoverable one has to include abort operations. Thus in testing the recoverability abort

operations will have to used in place of commit one at a time. Also the strictest condition is

------where a transaction neither reads nor writes to a data item, which was written to by a transaction that has not committed yet.

If A1?>C3>C2, then schedule S3 is recoverable because rolling back of T1 does not affect T2 and

T3. If C1>A3>C2. schedule S3 is not recoverable because T2 read the value of Y (r2(Y)) after T3 wrote X (w3(Y)) and T2 committed but T3 rolled back. Thus, T2 used non- existent value of Y. If C1>C3>A3, then S3 is recoverable because roll back of T2 does not affect T1 and T3.

Strictest condition of schedule S3 is C3>C2.

If A1?>C2>C3, then schedule S4 is recoverable because roll back of T1 does not affect T2 and T3. If C1>A2>C3, then schedule S4 is recoverable because the roll back of T2 will restore the value of Y that was read and written to by T3 (w3(Y)). It will not affect T1. If C1>C2>A3, then schedule S4 is not recoverable because T3 will restore the value of Y which was not read by T2.

Answer:

1) The normal reactions at the front wheel is 9909.375 N

The normal reactions at the rear wheel is 8090.625 N

2) The least coefficient of friction required between the tyres and the road is 0.625

Explanation:

1) The parameters given are as follows;

Speed, u = 90 km/h = 25 m/s

Distance, s it takes to come to rest = 50 m

Mass, m = 1.8 tonnes = 1,800 kg

From the equation of motion, we have;

v² - u² = 2·a·s

Where:

v = Final velocity = 0 m/s

a = acceleration

∴ 0² - 25² = 2 × a × 50

a = -6.25 m/s²

Force, F =  mass, m × a = 1,800 × (-6.25) = -11,250 N

The coefficient of friction, μ, is given as follows;

[tex]\mu =\dfrac{u^2}{2 \times g \times s} = \dfrac{25^2}{2 \times 10 \times 50} = 0.625[/tex]

Weight transfer is given as follows;

[tex]W_{t}=\dfrac{0.625 \times 0.9}{3}\times \dfrac{6.25}{10}\times 18000 = 2109.375 \, N[/tex]

Therefore, we have for the car at rest;

Taking moment about the Center of Gravity CG;

[tex]F_R[/tex] × 1.3 = 1.7 × [tex]F_F[/tex]

[tex]F_R[/tex] + [tex]F_F[/tex] = 18000

[tex]F_R + \dfrac{1.3 }{1.7} \times F_R = 18000[/tex]

[tex]F_R[/tex] = 18000*17/30 = 10200 N

[tex]F_F[/tex] = 18000 N - 10200 N = 7800 N

Hence with the weight transfer, we have;

The normal reactions at the rear wheel [tex]F_R[/tex]  = 10200 N - 2109.375 N = 8090.625 N

The normal reactions at the front wheel [tex]F_F[/tex] =  7800 N + 2109.375 N = 9909.375 N

2) The least coefficient of friction, μ, is given as follows;

[tex]\mu = \dfrac{F}{R} = \dfrac{11250}{18000} = 0.625[/tex]

The least coefficient of friction, μ = 0.625.

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:

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

Answer:

The minimum diameter to withstand such tensile strength is 22.568 mm.

Explanation:

The allow is experimenting an axial load, so that stress formula for cylidrical sample is:

[tex]\sigma = \frac{P}{A_{c}}[/tex]

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

Where:

[tex]\sigma[/tex] - Normal stress, measured in kilopascals.

[tex]P[/tex] - Axial load, measured in kilonewtons.

[tex]A_{c}[/tex] - Cross section area, measured in square meters.

[tex]D[/tex] - Diameter, measured in meters.

Given that [tex]\sigma = 75\times 10^{3}\,kPa[/tex] and [tex]P = 30\,kN[/tex], diameter is now cleared and computed at last:

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

[tex]D = 2\sqrt{\frac{P}{\pi \cdot \sigma} }[/tex]

[tex]D = 2 \sqrt{\frac{30\,kN}{\pi \cdot (75\times 10^{3}\,kPa)} }[/tex]

[tex]D = 0.0225\,m[/tex]

[tex]D = 22.568\,mm[/tex]

The minimum diameter to withstand such tensile strength is 22.568 mm.

Answer:

Total contraction on the bar = 1.238 mm

Explanation:

Modulus of Elasticity, E = 85 GN/m²

Diameter of the aluminium bar, [tex]d_{Al} = 40 mm = 0.04 m[/tex]

Load, P = 160 kN

Cross sectional area of the aluminium bar without hole:

[tex]A_1 = \frac{\pi d_{Al}^2 }{4} \\A_1 = \frac{\pi 0.04^2 }{4}\\A_1 = 0.00126 m^2[/tex]

Diameter of hole, [tex]d_h = 30 mm = 0.03 m[/tex]

Cross sectional area of the aluminium bar with hole:

[tex]A_2 = \frac{\pi( d_{Al}^2 - d_{h}^2)}{4} \\A_2 = \frac{\pi (0.04^2 - 0.03^2) }{4}\\A_2 = 0.00055 m^2[/tex]

Length of the aluminium bar, [tex]L_{Al} = 600 mm = 0.6 m[/tex]

Length of the hole, [tex]L_h = 100mm = 0.1 m[/tex]

Contraction in aluminium bar without hole  [tex]= \frac{P * L_{Al}}{A_1 E}[/tex]

Contraction in aluminium bar without hole  [tex]= \frac{160*10^3 * 0.6}{0.00126 * 85 * 10^9 }[/tex]

Contraction in aluminium bar without hole = 96000/107100000

Contraction in aluminium bar without hole = 0.000896

Contraction in aluminium bar with hole  [tex]= \frac{P * L_{h}}{A_2 E}[/tex]

Contraction in aluminium bar without hole  [tex]= \frac{160*10^3 * 0.1}{0.00055 * 85 * 10^9 }[/tex]

Contraction in aluminium bar without hole = 16000/46750000

Contraction in aluminium bar without hole = 0.000342

Total contraction = 0.000896 + 0.000342

Total contraction = 0.001238 m = 1.238 mm

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:

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