A cylinder of metal that is originally 450 mm tall and 50 mm in diameter is to be open-die upset forged to a final height of 100 mm. The strength coefficient is 230 MPa and the work hardening exponent is 0.15 while the coefficient of friction of the metal against the tool is 0.1. If the maximum force that the forging hammer can deliver is 3 MN, can the forging be completed

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:

Heat rejected to cold body = 3.81 kJ

Explanation:

Temperature of hot thermal reservoir Th = 1600 K

Temperature of cold thermal reservoir Tc = 400 K

efficiency of the Carnot's engine = 1 - [tex]\frac{Tc}{Th}[/tex]

eff. of the Carnot's engine = 1 - [tex]\frac{400}{600}[/tex]

eff = 1 - 0.25 = 0.75

efficiency of the heat engine = 70% of 0.75 = 0.525

work done by heat engine = 2 kJ

eff. of heat engine is gotten as = W/Q

where W = work done by heat engine

Q = heat rejected by heat engine to lower temperature reservoir

from the equation, we can derive that

heat rejected Q = W/eff = 2/0.525 = 3.81 kJ

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:

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%.

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

Answer:

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

Explanation:

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

The block descended a distance of 9.75 m

Answer:

The final temperature and pressure in the insulated rigid tank are [tex]48\,^{\circ}C[/tex] and [tex]375\,kPa[/tex].

Explanation:

An ideal gas is represented by the following model:

[tex]P\cdot V = \frac{m}{M}\cdot R_{u} \cdot T[/tex]

Where:

[tex]P[/tex] - Pressure, measured in kilopascals.

[tex]V[/tex] - Volume, measured in cubic meters.

[tex]m[/tex] - Mass of the ideal gas, measured in kilograms.

[tex]M[/tex] - Molar mass, measured in kilograms per kilomole.

[tex]T[/tex] - Temperature, measured in Kelvin.

[tex]R_{u}[/tex] - Universal constant of ideal gases, equal to [tex]8.314\,\frac{kPa\cdot m^{3}}{kmol\cdot K}[/tex]

As tank is rigid and insulated, it means that no volume deformations in tank, heat and mass interactions with surroundings occur during expansion process. Hence, final pressure is less that initial one, volume is doubled (due to equal partitioning) and temperature remains constant. Hence, the following relationship can be derived from model for ideal gases:

[tex]\frac{P_{1}\cdot V_{1}}{T_{1}} = \frac{P_{2}\cdot V_{2}}{T_{2}}[/tex]

Now, final pressure is cleared:

[tex]P_{2} = P_{1}\cdot \frac{T_{2}}{T_{1}}\cdot \frac{V_{1}}{V_{2}}[/tex]

[tex]P_{2} = (750\,kPa)\cdot 1 \cdot \frac{1}{2}[/tex]

[tex]P_{2} = 375\,kPa[/tex]

The final temperature and pressure in the insulated rigid tank are [tex]48\,^{\circ}C[/tex] and [tex]375\,kPa[/tex].

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:

(a) 2.39 MPa (b) 3.03 kJ (c) 3.035 kJ

Explanation:

Solution

Recall that:

A 10 gr of air is compressed isentropically

The initial air is at = 27 °C, 110 kPa

After compression air is at = a450 °C

For air,  R=287 J/kg.K

cv = 716.5 J/kg.K

y = 1.4

Now,

(a) W efind the pressure on [MPa]

Thus,

T₂/T₁ = (p₂/p₁)^r-1/r

=(450 + 273)/27 + 273) =

=(p₂/110) ^0.4/1.4

p₂ becomes  2390.3 kPa

So, p₂ = 2.39 MPa

(b) For the increase in total internal energy, is given below:

ΔU = mCv (T₂ - T₁)

=(10/100) (716.5) (450 -27)

ΔU =3030 J

ΔU =3.03 kJ

(c) The next step is to find the total work needed in kJ

ΔW = mR ( (T₂ - T₁) / k- 1

(10/100) (287) (450 -27)/1.4 -1

ΔW = 3035 J

Hence, the total work required is = 3.035 kJ

Answer:

[tex]W_{net} = - 1223 kJ[/tex]

Explanation:

State 1:

[tex]P_1 = 20 bar\\T_1 = 360^{0}C\\ h_1 = 3159.3 kJ/kg\\S_1 = 6.9917 kJ/kg[/tex]

State 2:

[tex]P_2 = 20 bar\\x_2 = 1 \\ h_2 = 2799.5 kJ/kg\\u_2 = 2600.3 kJ/kg\\v_2 = 0.09963m^3/kg[/tex]

State 3:

[tex]P_2 = 5 bar\\v_2 = v_3 \\v_3 = v_f + x_3 (v_g - v_f)\\0.09963 = (1.0926 * 10^{-3}) +x_3 (0.3749 - (1.0926 * 10^{-3}))\\x_3 = 0.263[/tex]

[tex]u_{3} = u_f + x_3 ( u_g - u_f)\\u_{3} = 639.68 + 0.263 (2561.2 - 639.68)\\u_{3} = 1146.2 kJ/kg[/tex]

State 4:

[tex]P_{4} = 5 bar\\T_4 = 360^0 C\\h_4 = 3188.4 kJ/kg\\S_4 = 7.660 kJ/kg-K\\Q_{12} = h_2 - h_1 = 2799.5-3159.3 = -359 kJ/kg\\Q_{23} = u_3 - h_2 =1146.2-2006.3 = -1454.1 kJ/kg\\Q_{34} = h_4 - h_3 = 3188.4-1196.04 = 1992.36 kJ/kg\\Q_{41} = T(S_1 - S_4) = (360 + 273) (6.9917 - 7.660) = -423.04 kJ/kg[/tex]

Calculate the network done for the cycle

[tex]W_{net} = m( Q_{12} + Q_{23} + Q_{34} + Q_{41})\\W_{net} = 5( -359.8 - 1454.1 + 1992.36 - 423.04)\\W_{net} = -1223 kJ[/tex]

Answer:

Time delay increases

Explanation:

Time delay is the delay between occurance of signal. If sampling time that is time between two samples is increased, the delay in the occurance of regenerated samples is also increased.

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:

Explanation:

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

Answer:

Explanation:

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

  twos_complement = ~val + 1

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

22₁₀ = ‭0001 0110‬₂    

~22₁₀ = ‭1110 1001‬₂   inverting the bits

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

so -22₁₀ == ~22₁₀ + 1 == ‭1110 1010‬₂

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

-22₁₀ = ‭1110 1010‬₂     8-bits

-22₁₀ = ‭1111 1111 1110 1010‬‬₂     16-bits

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

Answer:

The answer is VN =37.416 m/s

Explanation:

Recall that:

Pressure (atmospheric) = 100 kPa

So. we solve for the maximum velocity (m/s) to which water can be accelerated by the nozzles

Now,

Pabs =Patm + Pgauge = 800 KN/m²

Thus

PT/9.81 + VT²/2g =PN/9.81  + VN²/2g

Here

Acceleration due to gravity = 9.81 m/s

800/9.81 + 0

= 100/9.81 + VN²/19.62

Here,

9.81 * 2= 19.62

Thus,

VN²/19.62 = 700/9.81

So,

VN² =1400

VN =37.416 m/s

Note: (800 - 100) = 700

Answer:

[tex]V2 = 37.417ms^{-1}[/tex]

Explanation:

Given the following data;

Water enters the turbine nozzles (inlet) = 800kPa = 800000pa.

Nozzle outlets = 100kPa = 100000pa.

Density of water = 1000kg/m³.

We would apply, the Bernoulli equation between the inlet and outlet;

[tex]\frac{P_{1} }{d}+\frac{V1^{2} }{2} +gz_{1} = \frac{P_{2} }{d}+\frac{V2^{2} }{2} +gz_{2}[/tex]

Where, V1 is approximately equal to zero(0).

Z[tex]z_{1} = z_{2}[/tex]

Therefore, to find the maximum velocity, V2;

[tex]V2 = \sqrt{2(\frac{P_{1} }{d}-\frac{P_{2} }{d}) }[/tex]

[tex]V2 = \sqrt{2(\frac{800000}{1000}-\frac{100000}{1000}) }[/tex]

[tex]V2 = \sqrt{2(800-100)}[/tex]

[tex]V2 = \sqrt{2(700)}[/tex]

[tex]V2 = \sqrt{1400}[/tex]

[tex]V2 = 37.417ms^{-1}[/tex]

Hence, the maximum velocity, V2 is 37.417m/s

Answer:

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:

What is the probability of selecting the 4 of spade or black diamond from a deck of 52 playing cards?

a) 2/52

b) 4/52

c) 3/52

d) 1/5

Explanation:

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:

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:

(1). False, (2). True, (3). False, (4). False, (5). True.

Explanation:

The term ''contouring'' in this question does not have to do with makeup but it has to deal with the measurement of all surfaces in planes. It is a measurement in which the rough and the contours are being measured. So, let us check each questions again.

(1). In contouring, it is necessary to measure position and not velocity for feedback.

ANSWER : b =>False. IT IS NECESSARY TO MEASURE BOTH FOR FEEDBACK.

(2). In contouring during 2-axis NC machining, the two axes are moved at the same speed to achieve the desired contour.

ANSWER: a=> True.

(3). Job shop is another term for process layout.

ANSWER: b => False

JOB SHOP IS A FLEXIBLE PROCESS THAT IS BEING USED during manufacturing process and are meant for job Production. PROCESS LAYOUT is used in increasing Efficiency.

(4). Airplanes are normally produced using group technology or cellular layout.

ANSWER: b => False.

(5). In manufacturing, value-creating time is greater than takt time.

ANSWER: a => True.

Answer:

George  Floyd (BLACK  LIFES  MATTER)

C O V I D - 19

Quarantine  

no sports

wearing a mask

and a whole lot of other stuff

Explanation:

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]

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 goal of the software is to observe the software behavior to meet its requirement expectation. In software engineering, validating software might be harder since client's expectation may be vague or unclear.

Explanation:

Answer:

positive sides:

negative sides:

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:

Explanation:

Given that,

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

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

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

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

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

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

Inside and outside temperature difference is ΔT

The heat loss due to conduction in the window is

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

The heat loss due to conduction in the wall is

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

The net heat loss of the wall and the window is

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

The percentage of heat lost by the window is

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

Answer:

South Pole and South Pole or North Pole and North Pole.

Answer:

Answer:

• it charges banks more interest

• it sells more securities

• it decreases the money supply

Explanation:

hope this help edge 21

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°