A rectangular bar of length L has a slot in the central half of its length. The bar has width b, thickness t, and elastic modulus E. The slot has width b/3. The overall length of the bar is L = 570 mm, and the elastic modulus of the material is 77 GPa. If the average normal stress in the central portion of the bar is 200 MPa, calculate the overall elongation δ of the bar.

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:

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:

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:

Explanation:

Given that:

V = 12.5m/s

L= 2.70m

b= 0.65m

[tex]T_{ \infty} = 27^0C= 273+27 = 300K[/tex]

[tex]T_s= 127^0C = (127+273)= 400K[/tex]

P = 1atm

Film temperature

[tex]T_f = \frac{T_s + T_{\infty}}{2} \\\\=\frac{400+300}{2} \\\\=350K[/tex]

dynamic viscosity =

[tex]\mu =20.9096\times 10^{-6} m^2/sec[/tex]

density = 0.9946kg/m³

Pr = 0.708564

K= 229.7984 * 10⁻³w/mk

Reynolds number,

[tex]Re = \frac{SUD}{\mu} =\frac{\ SUl}{\mu}[/tex]

[tex]=\frac{0.9946 \times 12.5\times 2.7}{20.9096\times 10^-^6} \\\\Re=1605375.043[/tex]

we have,

[tex]Nu=\frac{hL}{k} =0.037Re^{4/5}Pr^{1/3}\\\\\frac{h\times2.7}{29.79\times 10^-63} =0.037(1605375.043)^{4/5}(0.7085)^{1/3}\\\\h=33.53w/m^2k[/tex]

we have,

heat transfer rate from top plate

[tex]\theta _1 =hA(T_s-T_{\infty})\\\\A=Lb\\\\=2.7*0.655\\\\ \theta_1=33.53*2.7*0.65(127/27)\\\\ \theta_1=5884.51w[/tex]

Answer:

A periodic function is a function that returns to its value over a certain period at regular intervals an example is the wave form of flux density (B) = sin wt

Explanation:

A periodic function is a function that returns to its value over a certain period at regular intervals an example is the wave form of flux density (B) = sin wt

attached to the answer is a free plot of the output starting with zero degree for one coil rotating in a uniform magnetic field

B ( wave flux density ) = Bm sinwt  and w = 2[tex]\pi[/tex]f = [tex]\frac{2\pi }{T}[/tex] rad/sec

Answer:

Develop social skills

Explanation:

Answer:

strengthen your college applications

Explanation:

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:

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:

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:

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:

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

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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

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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.

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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:

A.) Time = 17.13 seconds

B.) Distance = 31.9 m

C.) V = 11.18 m/s

D.) V = 7.1 m/s

Explanation:

The initial velocity U of the automobile is 15.65 m/s.

 At the instant that the two vehicles are abreast of each other, the police car starts to pursue the automobile with initial velocity U = 0 at a constant acceleration of 1.96 m/s². Because the police is starting from rest.

For the automobile, let us use first equation of motion

V = U - at.

Acceleration a is negative since it is decelerating with a = 3.05 m/s² . And

V = 0.

Substitute U and a into the formula

0 = 15.65 - 3.05t

15.65 = 3.05t

t = 15.65/3.05

t = 5.13 seconds

But the motorist noticed the police car in his rear view mirror 12 s after the police car started the pursuit and applied his brakes and decelerates at 3.05 m/s².

The total time required for the police car to overtake the automobile will be

12 + 5.13 = 17.13 seconds.

b.) Using the third equation of motion formula for the police car at V = 11.18 m/s and a = 1.96 m/s²

V^2 = U^2 + 2aS

Where S = distance travelled.

Substitute V and a into the formula

11.18^2 = 0 + 2 × 1.96 ×S

124.99 = 3.92S

S = 124.99/3.92

S = 31.88 m

c.) The speed of the police car at the time it overtakes the automobile will be in line with the speed zone which is 11.18 m/s

d.) That will be the final velocity V of the automobile car.

We will use third equation of motion to solve that.

V^2 = U^2 + 2as

V^2 = 15.65^2 - 2 × 3.05 × 31.88

V^2 = 244.9225 - 194.468

V = sqrt( 50.4545)

V = 7.1 m/s

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:

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:

(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:

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:

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 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:

s = v√[2(H - h)/g]

This formula describes the distance from the building that Prof Murthy must be when you release the egg

Explanation:

First, we need to find the time required by the egg to reach the head of Professor. For that purpose, we use 1st equation of motion in vertical form:

Vf = Vi + gt

where,

Vf = Velocity of egg at the time of hitting head of the Professor

Vi = initial velocity of egg = 0 m/s  (Since, egg is initially at rest)

g = acceleration due to gravity

t = time taken by egg to come down

Therefore,

Vf = 0 + gt

t = Vf/g   ----- equation (1)

Now, we use 3rd euation of motion for Vf:

2gs = Vf² - Vi²

where,

s = height dropped = H - h

Therefore,

2g(H - h) = Vf²

Vf = √[2g(H - h)]

Therefore, equation (1) becomes:

t = √[2g(H - h)]/g

t = √[2(H - h)/g]

Now, consider the horizontal motion of professor. So, the minimum distance of professor from building can be found out by finding the distance covered by the professor in time t. Since, the professor is running at constant speed of v m/s. Therefore:

s = vt

s = v√[2(H - h)/g]

This formula describes the distance from the building that Prof Murthy must be when you release the egg

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:

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:

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:

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