Suppose that a wing component on an aircraft is fabricated from an aluminum alloy that has a plane-strain fracture toughness of 26.0 MPa m0.5. It has been determined that fracture results at a stress of 112 MPa when the maximum internal crack length is 8.6 mm. For this same component and alloy, compute the stress level at which fracture will occur for a critical internal crack length of 3.0 mm.

Answer:

T = 858.25 s

Explanation:

Given data:

Reheat stage for a 100-mm-thick steel plate ( 7830 kg/m3,  c 550 J/kg K, k 48 W/m K),

initial uniform temperature ( Ti ) = 200 c

Final temperature = 550 c

convection coefficient  = 250 w/m^2 k

products combustion temp = 800 c

calculate how long the plate should be left in the furnace ( to attain 550 c )

first calculate/determine the Fourier series Number ( Fo )

[tex]\frac{T_{0}-T_{x} }{T_{1}-T_{x} } = C_{1} e^{(-0.4888^{2}*Fo )}[/tex]

= 0.4167 = [tex]1.0396e^{-0.4888*Fo}[/tex]

therefore Fo =  3.8264

Now determine how long the plate should be left in the furnace

Fo = [tex](\frac{k}{pc_{p} } ) ( \frac{t}{(L/2)^2} )[/tex]

k = 48

p = 7830

L = 0.1

Input the values into the relation and make t subject of the formula

hence t = 858.25 s

Explanation:

1. increase This due to increase in the pore volume.

2.increase . This is due to the fact that more liquid phase will be present at the firing.

3. Increase. This increase is because of the fact that clay on cooling forms glass.Thus, gaining more strength as the liquid phase formed fills in pore volume.

4. Increase, Rate of corrosion depends upon the surface area exposed.Since, upon vitrification surface area would increase, therefore corrosion increases.

5. Increase , glass has higher thermal conductivity than the pores it fills.

Given:

We have given two statements.

Statement 1: Proper footwear may include both leather and steel-toed shoes.

Statement 2:  Leather-soled shoes provide slip resistance.

Find:

Which statement is true.

Solution:

A slip-resistant outsole is smoother and more slip-resistant than other outsole formulations when exposed to water and oil. A smoother outsole in rubber ensures a slip-resistant shoe can handle a slippery floor more effectively.

Slip resistant shoes have an interlocked tread pattern that does not close the water in, enabling the slip resistant sole to touch the floor to provide better slip resistance.

Leather-soled shoes don't provide slop resistance.

Therefore, both the Technicians are wrong.

From the statements made by both technician A and Technician B, we can say that; both technicians are wrong.

We are given the statements made by both technicians;

Technician A: Proper footwear may include both leather and steel-toed shoes.

Technician B: Leather-soled shoes provide slip resistance.

Now, they are talking about safety shoes to be worn in workshops.

A shoe that is Slip resistant will have rubber soles and tread patterns that can help to have better grip of wet or greasy floors.

This is the type of shoe that should be worn by technicians in the workshop.

Thus, Technician A is wrong because proper footwear does not include leather shoes.

Similarly, technician B is also wrong because leather shoes are not safety shoes.

Read more about slip resistant shoes at; https://brainly.com/question/17411739

Answer:

The liquids are TOLUENE because the equilibrum vapor above it will be flammable ( D )

Explanation:

Liquids stored at : 1 atm , 25⁰c  and they are vented with air

Determining whether the equilibrum vapor above the liquid will be flammable

We can determine this by using Antoine equation to calculate saturation vapor pressure also apply Dalton's law to determine the volume % concentration of air and finally we compare answer to flammable limits to determine which liquid will be flammable

A) For acetone

using the Antoine equation to calculate saturation vapor pressure

[tex]In(P^{out} ) = A - \frac{B}{C + T}[/tex]

values gotten appendix E ( chemical process safety (3rd edition) )

A = 16.6513

B = 2940.46

C = -35.93

T = 298 k      input values into Antoine equation

therefore ; [tex]p^{out}[/tex] = 228.4 mg

calculate volume percentage using Dalton's law

= V% = (saturation vapor pressure / pressure ) *100

         = (228.4 mmHg / 760 mmHg) * 100 = 30.1%

The liquid is not flammable because its UFL = 12.8%

B) For Benzene

using the Antoine equation to calculate saturation vapor pressure

[tex]In(P^{out} ) = A - \frac{B}{C + T}[/tex]

values gotten appendix E ( chemical process safety (3rd edition) )

A= 15.9008

B = 2788.52

C = -52.36

T = 298 k   input values into the above equation

[tex]p^{out}[/tex] = 94.5 mmHg

calculate volume percentage using Dalton's law

V% = (saturation vapor pressure / pressure ) *100

      = (94.5 / 760 ) * 100 = 12.4%

Benzene is not flammable under the given conditions because its UFL =7.1%

C) For cyclohexane

using the Antoine equation to calculate saturation vapor pressure

[tex]In(P^{out} ) = A - \frac{B}{C + T}[/tex]

values gotten appendix E ( chemical process safety (3rd edition) )

A = 15.7527

B = 2766.63

c = -50.50

T = 298 k

solving the above equation using the given values

[tex]p^{out}[/tex] = 96.9 mmHg

calculate volume percentage using Dalton's law

V% = (saturation vapor pressure / pressure ) *100

      = ( 96.9 mmHg /760 mmHg) * 100 = 12.7%

cyclohexane not flammable under the given conditions because its UFL= 8%

D) For Toluene

using the Antoine equation to calculate saturation vapor pressure

[tex]In(P^{out} ) = A - \frac{B}{C + T}[/tex]

values gotten from appendix E ( chemical process safety (3rd edition) )

A = 16.0137

B = 3096.52

C = -53.67

T = 298 k

solving the above equation using the given values

[tex]p^{out}[/tex] = 28.2 mmHg

calculate volume percentage using Dalton's law

V% = (saturation vapor pressure / pressure ) *100

     = (28.2 mmHg / 760 mmHg) * 100 = 3.7%

Toluene is flammable under the given conditions because its UFL= 7.1%

Answer:

The material will not fail

Explanation:

A rod subjected under cyclic stress will fail if the cyclic stress it is subjected to is a constant maximum value that is above the fatigue limit of the rod. but in this problem the Rod is subjected to a cyclic stress that ranges from  200 MPa(maximum stress) and 20 MPa ( minimum stress). this simply means that at all times the Rod will not experience maximum stress of 200 MPa and its Fatigue limit is also set at ~100 MPa

attached is the diagram showing the cyclic stress the rod is subjected to

Answer:

max shear = R = V = 15 kN

Explanation:

given:

load = 10 kn/m

span = 3m

max shear = R = V = wL / 2

max shear = R = V = (10 * 3) / 2

max shear = R = V = 15 kN

Answer:

partner itself

Explanation:

Answer:

limited partner

hope this answer correct :)

Answer:

56.47%

Explanation:

Determine the efficiency of the Engine

Given data :  T1 (k) = 325, P1 (kpa) = 185,

V1 (cm^3) = 410 , r = 8, T3(k) = 3420

speed ( RPM) = 4800

USING THIS FORMULA

efficiency ( n ) = [tex]1 - (\frac{1}{(rp)^{r-1} })[/tex]

= 1 -  [tex](\frac{1}{(8)^{r-1} })[/tex] = 1 - (1/8^1.4-1 )

= 0.5647 = 56.47%

Answer:

[tex]\mathbf{F_D \approx 1.071 \ lbf}[/tex]

Explanation:

Given that:

The height of a  triangular stabilizing fin on its stern is 1 ft tall

and it length is 2 ft long.

Temperature = 60 °F

The objective is to determine the drag on the fin when the submarine is traveling at a speed of 2.5 ft/s.

From these information given; we can have a diagrammatic representation describing how the  triangular stabilizing fin looks like as we resolve them into horizontal and vertical component.

The diagram can be found in the attached file below.

If we recall ,we know that;

Kinematic viscosity v = [tex]1.2075 \times 10^{-5} \ ft^2/s[/tex]

the density of water ρ = 62.36 lb /ft³

[tex]Re_{max} = \dfrac{Ux}{v}[/tex]

[tex]Re_{max} = \dfrac{2.5 \ ft/s \times 2 \ ft }{1.2075 \times 10 ^{-5} \ ft^2/s}[/tex]

[tex]Re_{max} = 414078.6749[/tex]

[tex]Re_{max} = 4.14 \times 10^5[/tex] which is less than < 5.0 × 10⁵

Now; For laminar flow;  the drag on  the fin when the submarine is traveling at 2.5 ft/s can be determined by using the expression:

[tex]dF_D = (\dfrac{0.664 \times \rho \times U^2 (2-x) dy}{\sqrt{Re_x}})^2[/tex]

where;

[tex](2-x) dy[/tex] = strip area

[tex]Re_x = \dfrac{2.5(2-x)}{1.2075 \times 10 ^{-5}}[/tex]

Therefore;

[tex]dF_D = (\dfrac{0.664 \times 62.36 \times 2.5^2 (2-x) dy}{\sqrt{ \dfrac{2.5(2-x)}{1.2075 \times 10 ^{-5}}}})[/tex]

[tex]dF_D = 1.136 \times(2-x)^{1/2} \ dy[/tex]

Let note that y = 0.5x from what we have in the diagram,

so , x = y/0.5

By applying the rule of integration on both sides, we have:

[tex]\int\limits \ dF_D = \int\limits^1_0 \ 1.136 \times(2-\dfrac{y}{0.5})^{1/2} \ dy[/tex]

[tex]\int\limits \ dF_D = \int\limits^1_0 \ 1.136 \times(2-2y)^{1/2} \ dy[/tex]

Let U = (2-2y)

-2dy = du

dy = -du/2

[tex]F_D = \int\limits^0_2 \ 1.136 \times(U)^{1/2} \ \dfrac{du}{-2}[/tex]

[tex]F_D = - \dfrac{1.136}{2} \int\limits^0_2 \ U^{1/2} \ du[/tex]

[tex]F_D = -0.568 [ \dfrac{\frac{1}{2}U^{ \frac{1}{2}+1 } }{\frac{1}{2}+1}]^0__2[/tex]

[tex]F_D = -0.568 [ \dfrac{2}{3}U^{\frac{3}{2} } ] ^0__2[/tex]

[tex]F_D = -0.568 [0 - \dfrac{2}{3}(2)^{\frac{3}{2} } ][/tex]

[tex]F_D = -0.568 [- \dfrac{2}{3} (2.828427125)} ][/tex]

[tex]F_D = 1.071031071 \ lbf[/tex]

[tex]\mathbf{F_D \approx 1.071 \ lbf}[/tex]

Answer:

A) slip(s1) = (1800 - 1750) / 1800 = 0.0277

B) 1800 rpm, 1350 rpm,  900 rpm, 450 rpm

Explanation:

Given data

frequency = 60 Hz,  

Line - line rms = 440 - V

3 phase induction-motor drive

number of poles = 4

Full-load rated speed = 1750

rated torque = 40 Nm

A) The plot of torque-speed characteristics for the following values of the frequency f : 60 Hz, 45 Hz, 30 Hz, and 15 Hz is attached below

first we calculate the rated speed:

Ns = [tex]\frac{120f}{p}[/tex]

f = 60 Hz .  p (number of poles) = 4

Ns[tex]_{1}[/tex] = [tex]\frac{120(60)}{4}[/tex] = 1800

full loaded rated value = 1750.

slip(s1) = (1800 - 1750) / 1800 = 0.0277

considering a linearly condition the slip is low

[tex]\frac{T1}{T2} = \frac{S1}{S2} * \frac{F2}{F1}[/tex]

S1 = 0.0277

f1 = 60 Hz

hence s2 = 0.018 therefore Ns2 = 1500

B) The speeds of operation at : 60 Hz, 45 Hz, 30 Hz, 15 Hz

for 60 Hz :

Ns = [tex]\frac{120f}{p}[/tex] = (120*60) / 4 = 1800 rpm

for 45 Hz:

Ns = 120(f) / p = (120*45) / 4 = 5400 /4 = 1350 rpm

for 30 Hz:

Ns = 120(f) / p = (120*30) / 4 = 3600 / 4 = 900 rpm

for 15 Hz:

Ns = 120(f) / p = (120*15) / 4 = 1800 / 4 = 450 rpm

Answer:

a) 0.42

b) Independent

c) 30%

d) 0.88

Explanation:

Person chooses Chicken's item : 70% = 0.7

Person chooses fish's item : 30% = 0.3

Visits in which he orders Afghani Chicken = 60% = 0.6

a) Probability that he goes to KARIM and orders Afghani Chicken:

P = 0.7 * 0.6 = 0.42

b) Two events are said to be independent when occurrence of one event does not affect the probability of the other event's occurrence. Here the person orders Afghani Chicken regardless of where he visits so the events are independent.

c)  P = 0.30 because he orders Afghani Chicken regardless of where he visits.

d)  Let A be the probability that he goes to KARIM:

P(A) = 0.7 * ( 1 - 0.6 ) = 0.28

Let A be the probability that he orders Afghani Chicken:

P(B) =  0.3 * 0.6 = 0.18

Let C be the probability that he goes to KARIM and orders Afghani chicken:

= 0.7 * 0.6 = 0.42

So probability that he goes to KARIM or orders Afghani Chicken or both:

P(A) + P(B) + P(C) = 0.28 + 0.18 + 0.42 = 0.88

Answer:

1. Slope = 53.3 x 10⁻⁶

2. Deflection = -0.00016m

Explanation:

given:

let L = 4 m (span of cantilever beam)

let w = 300 N/m (distributed load)

let EI =60 MNm² (flexural stiffness)

                 dy      w * L³        300 x 4³

1. slope = ------- = --------- =  ------------------- =  53.3 x 10⁻⁶

                  dx        6EI           6 x 60x10⁶

                                  wL⁴               300 x 4⁴

2. Deflection = y = - ----------- =  - ------------------ =   -0.00016m

                                    8EI                8 x 60x10⁶

therefore the deflection is 0.16mm downwards.

Answer:

At temperature is and relative humidity is 86% therefore,  the humidity ratio is 0.0223 and the specific volume is 14.289

At temperature is and Relative humidity is 40% therefore, the humidity ratio is  0.0066 and the specific volume is 13.535.

To calculate the mass of air can be calculated as follows:

Now , we going to calculate the volume,

The time which is required to fill the cistern can be calculated as follows:

Now, putting the value in above formula we get,

Therefore, the hours required to fill the cistern is 4.65 hours.

Explanation:

Answer:

It is difficult to introduce products modification

Answer:

A) 5 MPa , 55 MPa

B) maximum stress = 55 MPa,  maximum shear stress = 25 MPa

Explanation:

using the given Data

free surface of a solid body

α[tex]_{x}[/tex] = 50 MPa,    α[tex]_{y}[/tex] = 10 MPa , t[tex]_{xy}[/tex] = -15 MPa

attached below is the detailed solution to the question

Answer:

I think its A) 5 MPa , 55 MPa

B) ms = 55 MPa,  mss= 25 MPa

α = 50 MPa,    α = 10 MPa , t = -15 MPa

Answer:

We can prevent it by:

a) By wearing GOOGLES.

b) By wearing our Lab coat.

c) Fire extinguisher should always be present in the lab.

d) Hand Gloves must be worn.

e) No playing in the lab.

f) No touching of things/equipment's e.g bottles, in the lab.

g) No eating/snacking in the lab.

h) Always pay attention, no gisting.

i) Adult/qualified person must be present in the lab with pupils/students.

Explanation:

Hope it helps.

Answer: 0.002174

Explanation:

Given that the

Yield strength rho = 450 MPa

Length = 2 m

Elastic modulus E= 207 GPa

According to Hook's law, if the elastic limits is not reached, the elastic modulus is the ratio of elastic strength to the elastic strain ə

E = rho/ə

Make ə the subject of formula

ə = rho/ E

ə = (450 × 10^6) / (207 × 10^9)

ə = 2.174 × 10^-3

Therefore, the engineering strain which depends on engineering stress and elastic modulus is 2.174 × 10^-3

Elastic Strain has no S.I Units.

Answer:

the required stress level  at which fracture will occur for a critical internal crack length of 3.0 mm is 189.66 MPa

Explanation:

From the given information; the objective is to compute the stress level at which fracture will occur for a critical internal crack length of 3.0 mm.

The Critical Stress for a maximum internal crack can be expressed by the formula:

[tex]\sigma_c = \dfrac{K_{lc}}{Y \sqrt{\pi a}}[/tex]

[tex]Y= \dfrac{K_{lc}}{\sigma_c \sqrt{\pi a}}[/tex]

where;

[tex]\sigma_c[/tex] = critical stress required for initiating crack propagation

[tex]K_{lc}[/tex] = plain stress fracture toughness = 26 Mpa

Y = dimensionless parameter

a = length of the internal crack

given that ;

the maximum internal crack length is 8.6 mm

half length of the internal  crack will be 8.6 mm/2 = 4.3mm

half length of the internal  crack a = 4.3 × 10⁻³ m

From :

[tex]Y= \dfrac{K_{lc}}{\sigma_c \sqrt{\pi a}}[/tex]

[tex]Y= \dfrac{26}{112 \times \sqrt{\pi \times 4.3 \times 10 ^{-3}}}[/tex]

[tex]Y= \dfrac{26}{112 \times0.1162275716}[/tex]

[tex]Y= \dfrac{26}{13.01748802}[/tex]

[tex]Y=1.99731315[/tex]

[tex]Y \approx 1.997[/tex]

For this same component and alloy, we are to also compute the stress level at which fracture will occur for a critical internal crack length of 3.0 mm.

when the length of the internal crack a = 3mm

half  length of the internal  crack will be 3.0 mm / 2 = 1.5 mm

half length of the internal  crack a =1.5 × 10⁻³ m

From;

[tex]\sigma_c = \dfrac{K_{lc}}{Y \sqrt{\pi a}}[/tex]

[tex]\sigma_c = \dfrac{26}{1.997 \sqrt{\pi \times 1.5 \times 10^{-3}}}[/tex]

[tex]\sigma_c = \dfrac{26}{0.1370877444}[/tex]

[tex]\sigma_c =189.6595506[/tex]

[tex]\sigma_c =[/tex] 189.66 MPa

Thus; the required stress level  at which fracture will occur for a critical internal crack length of 3.0 mm is 189.66 MPa

Answer:

A)  222.58 kJ / kg

B)  0.8897 M^3/ kg

c)  0.7737 m^3/kg

D)  746.542 k

E)  536.017 kj/kg

efficiency = 58% ( approximately )

Explanation:

Given Data :

Gas constant (R) =  0.287 kJ/ kg.K

T1 = 310 k

P1 ( Kpa ) = 100

r = 11.5 ( compression ratio )

rp = 1.95 ( pressure ratio )

A ) specific internal energy at state 1

 = Cv*T1 =  0.718 * 310 = 222.58 kJ / kg

B) Relative specific volume at state 1

= P1*V1 = R*T1 ( ideal gas equation )

V1 = R*T1 / P1 = (0.287* 10^3*310 ) / 100 * 10^3

V1 = 88.97 / 100 = 0.8897 M^3/ kg

C ) relative specific volume at state 2

Applying  r ( compression ratio) = V1 / V2

11.5 = 0.8897 / V2

V2 = 0.8897 / 11.5 = 0.7737 m^3/kg

D) The temperature (k) at state 2

since the process is an Isentropic process we will apply the p-v-t relation

[tex]\frac{T1}{T2} = (\frac{V1}{V2}^{n-1} ) = (\frac{P2}{P1} )^{\frac{n-1}{n} }[/tex]

hence T2 = [tex]9^{1.4-1} * 310[/tex] = 2.4082 * 310 = 746.542 k

e) specific internal energy at state 2

= Cv*T2 = 0.718  * 746.542 = 536.017 kj/kg

efficiency = output /input = 390.3511 / 667.5448 ≈ 58%

attached is a free hand diagram of an Otto cycle is attached below

In this exercise we have to calculate the formula that will be able to determine the length of the cantilevered, like this:

[tex]\sigma_{max}C=\frac{M_{max}C}{I}[/tex]

So to determinated the maximum tensile and compreensive stress due to bending we can describe the formula as:

[tex]\sigma_b = \frac{MC}{I}[/tex]

Where,

So making this formula for the max, we have:

[tex]\sigma_c=\frac{MC}{I} \\\sigma_T=-\sigma_c=-\frac{MC}{I}\\\sigma_{max}=M_{max}\\[/tex]

With all this information we can put the formula as:

[tex]\sigma_{max}C=\frac{M_{max}C}{I}[/tex]

See more about stress in the beam at brainly.com/question/23637191

Answer:

The volume percent of graphite is 91.906 per cent.

Explanation:

The volume percent of graphite ([tex]\% V_{Gr}[/tex]) is determined by the following expression:

[tex]\%V_{Gr} = \frac{V_{Gr}}{V_{Gr}+V_{Fe}} \times 100\,\%[/tex]

[tex]\%V_{Gr} = \frac{1}{1+\frac{V_{Gr}}{V_{Fe}} }\times 100\,\%[/tex]

Where:

[tex]V_{Gr}[/tex] - Volume occupied by the graphite phase, measured in cubic centimeters.

[tex]V_{Fe}[/tex] - Volume occupied by the ferrite phase, measured in cubic centimeters.

The volume of each phase can be calculated in terms of its density and mass. That is:

[tex]V_{Gr} = \frac{m_{Gr}}{\rho_{Gr}}[/tex]

[tex]V_{Fe} = \frac{m_{Fe}}{\rho_{Fe}}[/tex]

Where:

[tex]m_{Gr}[/tex], [tex]m_{Fe}[/tex] - Masses of the graphite and ferrite phases, measured in grams.

[tex]\rho_{Gr}[/tex], [tex]\rho_{Fe}[/tex] - Densities of the graphite and ferrite phases, measured in grams per cubic centimeter.

Let substitute each volume in the definition of the volume percent of graphite:

[tex]\%V_{Gr} = \frac{1}{1 +\frac{\frac{m_{Gr}}{\rho_{Gr}} }{\frac{m_{Fe}}{\rho_{Fe}} } } \times 100\,\%[/tex]

[tex]\%V_{Gr} = \frac{1}{1+\left(\frac{m_{Gr}}{m_{Fe}} \right)\cdot \left(\frac{\rho_{Fe}}{\rho_{Gr}} \right)}\times 100\,\%[/tex]

Let suppose that 100 grams of cast iron are available, masses of each phase are now determined:

[tex]m_{Gr} = \frac{2.5}{100}\times (100\,g)[/tex]

[tex]m_{Gr} = 2.5\,g[/tex]

[tex]m_{Fe} = 100\,g - 2.5\,g[/tex]

[tex]m_{Fe} = 97.5\,g[/tex]

If [tex]m_{Gr} = 2.5\,g[/tex], [tex]m_{Fe} = 97.5\,g[/tex], [tex]\rho_{Fe} = 7.9\,\frac{g}{cm^{3}}[/tex] and [tex]\rho_{Gr} = 2.3\,\frac{g}{cm^{3}}[/tex], the volume percent of graphite is:

[tex]\%V_{Gr} = \frac{1}{1+\left(\frac{2.5\,gr}{97.5\,gr} \right)\cdot \left(\frac{7.9\,\frac{g}{cm^{3}} }{2.3\,\frac{g}{cm^{3}} } \right)} \times 100\,\%[/tex]

[tex]\% V_{Gr} = 91.906\,\%[/tex]

The volume percent of graphite is 91.906 per cent.

Answer:

Explanation:

Known: flow rate and inlet temperature for automobile radiator.

Overall heat transfer coefficient.

Find: Area required to achieve a prescribed outlet temperature.

Assumptions: (1) Negligible heat loss to surroundings and kinetic and

potential energy changes, (2) Constant properties.

Analysis: The required heat transfer rate is

q = (m c)h (T h,i - T h,o) = 0.05 kg/s (4209J / kg.K) 70K = 14,732 W

Using the ε-NTU method,

Cmin = Ch = 210.45 W / K

Cmax = Cc = 755.25W / K

Hence, Cmin/Cmx(Th,i - Th,o) = 210.45W / K(100K) = 21,045W

and

ε=q/qmax = 14,732W / 21,045W = 0.700

NTU≅1.5, hence

A=NTU(cmin / U) = 1.5 x 210.45W / K(200W) / m² .K) = 1.58m²

1. the air outlet is..

Tc,o = Tc,i + q / Cc = 300K + (14,732W / 755.25W / K) = 319.5K

2. using the LMTD approach ΔTlm = 51.2 K,, R=0.279 and P=0.7

hence F≅0.95 and

A = q/FUΔTlm = (14,732W) / [0.95(200W / m².K) 51.2K] = 1.51m²

Answer:

An improper integral is a definite integral that has either or both limits infinite or an integrand that approaches infinity at one or more points in the range of integration

Explanation:

Answer:

Net heat transfers:  during summer = 52.253 W/m^2 during winter = 119.375 w/m^2

Explanation:

Given data :

Room temperature throughout the year = 20⁰c = 293 k

Room temperature during summer = 27⁰c = 300 k

Room temperature during winter = 14⁰c = 287 k

surface temperature of a person throughout the year = 32⁰c = 305 k

coefficient of heat transfer by natural convection (h)= 2 w /m^2 k

emissivity = 0.9

An explanation to the condition of feeling chilled during the winter and comfortable during summer  can be explained with the calculation below

The heat transfer from the surface body of a person the the room is carried out by convection and this can be calculated as

q = hΔt = 2 * ( 305  -  293) = 2 * 12 = 24 w /m^2

also calculate heat transfer through radiation using this formula

[tex]q_{rad} =[/tex] εσ [ (Temperature of body)^4- (temperature of room at each season)^4 ]

ε = 0.90,(emissivity)    σ = 5.67 *10^-8 w/^2 . k ( Boltzmann's constant)

during summer :

[tex]q_{rad}[/tex] = (0.90)*(5.67*10^-8)* ( 305^4 - 300^4 ) = 28.253 W/m^2

therefore the net heat transfer during the summer = 24 w/m^3 + 28.253 W/m^3 = 52.253 W/m^2

During winter :

[tex]q_{rad}[/tex] = (0.90)*(5.67*10^-8) * ( 305^4 - 287^4 ) = 95.375 W/m^3

therefore the net heat transfer during the winter

= 24 w/m^3 + 95.375w/m^3 = 119.375 w/m^2

Answer:

D) ΔU = -153.43 kJ and ΔH = -146.00 kJ

Explanation:

Given;

heat energy released by the exothermic reaction, ΔH = -146 kJ

number of gas mol, n = 3 mol

temperature of the gas, T = 298 K

Apply first law of thermodynamic

Change in the internal energy of the system, ΔU;

ΔU = ΔH- nRT

where;

R is gas constant = 8.314 J/mol.K

ΔU = -146kJ - (3 x 8.314 x 298)

ΔU = -146kJ - 7433 J

ΔU = -146kJ - 7.433 kJ

ΔU = -153.43 kJ

Therefore, the enthalpy change of the reaction ΔH is -146 kJ and change in the internal energy of the system is -153.43 kJ

D) ΔU = -153.43 kJ and ΔH = -146.00 kJ

Answer:

Both technician A and technician B are correct.

Explanation: Vehicle manufacturers always specify the size of the tires required for a given vehiclefor optimal efficiency,this will ensure that the speedometer is accurate and the level of the vehicle is good enough to ensure the vehicle works efficiently.

It is also a known fact that an increase in a vehicle's rpm(revolution per minute) will eventually lead to increased fuel consumption which means the fuel economy of the vehicle will be reduced making the vehicle less efficient in its fuel consumption.

Answer:

c. The tension in the cord connected to the truck is greater than 1200 lb

e. The normal force between A and B is 1200 lb.

Explanation:

The correct question should be

A 1000 lb boulder B is resting on a 200 lb platform A when truck C accidentally accelerates to the right (truck in reverse). Which of the following statements are true (select two answers)?

A free body diagram is shown below.

The normal force between the the boulder and the platform will be the sum of the force, i.e 1000 lb + 200 lb = 1200 lb

For the combination of the bodies to accelerate upwards, then the tension must be greater than the normal force, i.e T > 1200 lb

Answer:

From the calculation, we can see that the invention's COP of 2.67 does not exceed the maximum theoretical COP of 14.65. Hence his claim is valid and could be possible.

Explanation:

Heat generated Q = 200 kW

power input W = 75 kW

Temperature of heated region [tex]T_{h}[/tex] = 293 K

Temperature of heat source [tex]T_{c}[/tex] = 273 K

For this engine,

coefficient of performance COP = Q/W  = 200/75 = 2.67

The maximum theoretical COP obtainable for a heat pump is given as

COP = [tex]\frac{T_{h} }{T_{h} - T_{c} }[/tex] =  [tex]\frac{293 }{293 - 273 }[/tex] = 14.65

From the calculation, we can see that the invention's COP of 2.67 does not exceed the maximum theoretical COP of 14.65. Hence his claim is valid and could be possible.

Answer:

true

Explanation:

true

Answer:

I believe this is true

Explanation:

If your looking at something and you look at something else everything is still in perfect view and clear, in focus.

hope this helps :)