The guy wires AB and AC are attached to the top of the transmission tower. The tension in cable AB is 8.7 kN. Determine the required tension T in cable AC such that the net effect of the two cables is a downward force at point A. Determine the magnitude R of this downward force.

Find the given attachments for complete explanation

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:

Overall project duration

Explanation:

Scheduling can best be defined as the process used to determine a overall project duration.

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:

Check the v(t) signal referred to in the question and the solution to each part in the files attached

Explanation:

The detailed solutions of parts a to d are clearly expressed in the second file attached.

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:

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 "".

Learn more:

brainly.com/question/15744940

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:

(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:

None of these

Explanation:

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

Therefore, the correct answer is "None of these "

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:

Yes, the forging can be completed

Explanation:

Given h = 100 mm, ε = ㏑(450/100) = 1.504

[tex]Y_f = 230 \times 1.504^{0.15} = 244.52[/tex]

V = π·D²·L/4 = π × 50²×450/4 = 883,572.93 mm³

At h = 100 mm, A = V/h = 883,572.93 /100 = 8835.73 mm²

D = √(4·A/π) = 106.07 mm

[tex]K_f[/tex] = 1 + 0.4 × 0.1 × 106.07/100 = 1.042

F = 1.042 × 244.52 × 8835.73 = 2252199.386 N =2.25 MN

Hence the required force = 2.25 MN is less than the available force = 3 MN therefore, the forging can be completed.

Answer:

Current, I = 4A

Explanation:

Since the connection is in delta, let's convert to star.

Simplify BCD:

[tex] R1 = \frac{40 * 8}{40 + 16 + 8} = \frac{320}{64} = 5 ohms [/tex]

[tex] R2 = \frac{16 * 8}{40 + 16 + 8} = \frac{128}{64} = 2 ohms [/tex]

[tex] R3 = \frac{40 * 16}{40 + 16 + 8} = \frac{640}{64} = 10 ohms [/tex]

From figure B, it can be seen that 6 ohms and 6 ohms are connected in parallel.

Simplify:

[tex] \frac{6 * 6}{6 + 6} = \frac{36}{12} = 3 \ohms [/tex]

Req = 10 ohms + 3 ohms

Req = 13 ohms

To find the current, use ohms law.

V = IR

Where, V = 52volts and I = 13 ohms

Solve for I,

[tex] I = \frac{V}{R} = \frac{52}{13} = 4A[/tex]

Current, I = 4 A

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 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:

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°

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

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:

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:

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 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:

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:

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:

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.

Answer:

See Explanation

Explanation:

Required

- Pseudocode to determine the largest of 10 numbers

- C# program to determine the largest of 10 numbers

The pseudocode and program makes use of a 1 dimensional array to accept input for the 10 numbers;

The largest of the 10 numbers is then saved in variable Largest and printed afterwards.

Pseudocode (Number lines are used for indentation to illustrate the program flow)

1. Start:

2. Declare Number as 1 dimensional array of 10 integers

3. Initialize: counter = 0

4. Do:

4.1 Display “Enter Number ”+(counter + 1)

4.2 Accept input for Number[counter]

4.3 While counter < 10

5. Initialize: Largest = Number[0]

6. Loop: i = 0 to 10

6.1 if Largest < Number[i] Then

6.2 Largest = Number[i]

6.3 End Loop:

7. Display “The largest input is “+Largest

8. Stop

C# Program (Console)

Comments are used for explanatory purpose

using System;

namespace ConsoleApplication1

{

   class Program

   {

       static void Main(string[] args)

       {

           int[] Number = new int[10];  // Declare array of 10 elements

           //Accept Input

           int counter = 0;

           while(counter<10)

           {

               Console.WriteLine("Enter Number " + (counter + 1)+": ");

               string var = Console.ReadLine();

               Number[counter] = Convert.ToInt32(var);

               counter++;                  

           }

           //Initialize largest to first element of the array

           int Largest = Number[0];

           //Determine Largest

           for(int i=0;i<10;i++)

           {

               if(Largest < Number[i])

               {

                   Largest = Number[i];

               }

           }

           //Print Largest

           Console.WriteLine("The largest input is "+ Largest);

           Console.ReadLine();

       }

   }

}

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:

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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