how can we prevent chemical hazards in labotary

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:

A.) 0.08 kJ/kg.K

B.) 207.8 KJ/Kg

C.) 0.808

Explanation:

From the question, the use of fluids mechanic table will be required. In order to get the compressor processes, the kinetic energy and the potential energy will be negligible while applying the ideal gas model.

Since the steam is a closed system, the carbon dioxide will be compressed adiabatically.

Please find the attached files for the solution and the remaining explanation.

Answer:

partner itself

Explanation:

Answer:

limited partner

hope this answer correct :)

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:

a) cross sectional area of the core = 0.0187 m²

b) The secondary voltage on no-load = 413 V

c) The primary currency and power factor on no load is 1.21 A and 0.168 lagging respectively.

Explanation:

See attached solution.

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:

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:

Average industrial personnel door

Explanation:

Going by the description in which the door has ventilation louvers or glass panel, this shows that it can never be any type of security door. This is because with louvers or glass panel, it can be easier for any intruder to look or break into the house or space that is intended to be protected fully.

Hence, an hollow steel composite door with an 18-guage metal facing, hung on butt hinges with nonremovable pins, often with ventilation louvers or glass panels is likely an Average industrial personnel door

Answer:

mano não sei mas acho que vai dar certo porque isso aí é muito top mas é isso aí mano o cara tem que ser confiar mesmo viu que negócio é desse jeito mesmo entendeu porque sabe como é que é as coisas né nada é fácil mesmo hein mas é isso aí mano continua tentando aí mano porque Rapaz tu é louco doido agora tá difícil mesmo mas é isso aí o cara tem que ir saber se ele tá ligado eu deixei isso mesmo né mas é isso aí meu truta

Explanation:

É isso aí mano Espero que tenho ajudado aí beleza manda a tua pergunta aí beleza é isso aí mano É isso aí continua hein p

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:

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 :)

Answer and Explanation:

The coefficient of determination also called "goodness of fit" or R-squared(R²) is used in statistical measurements to understand the relationship between two variables such that changes in one variable affects the other. The level of relationship or the degree to which one affects the other is measured by 0 to 1 whereby 0 means no relationship at all and 1 means one totally affects the other while figures in between such 0.40 would mean one variable affects 40% of the other variable.

In making a decision as an engineer while using the coefficient of determination, one would try to understand the relationship between variables under consideration and make decisions based on figures obtained from calculating coefficient of determination. In other words when there is a 0 coefficient then there is no relationship between variables and an engineer would make his decisions with this in mind and vice versa.

Answer:

hello your question is incomplete attached is the complete question

A) Vc =  15 ( 1 -[tex]e^{-t/0.15s}[/tex] ) ,   ic = [tex]1.5 mAe^{-t/0.15s}[/tex]

B) attached is the relevant sketch

Explanation:

applying Thevenin's theorem to find the mathematical expression for the transient behavior of Vc and ic after closing the switch

[tex]R_{th}[/tex] = 8 k ohms || 24 k ohms = 6 k ohms

[tex]E_{th}[/tex] = [tex]\frac{20 k ohms(20 v)}{24 k ohms + 8 k ohms}[/tex]  =  15 v

t = RC = (10 k ohms( 15 uF) = 0.15 s

Also; Vc = [tex]E( 1 - e^{-t/t} )[/tex]

hence Vc = 15 ( 1 - [tex]e^{-t/0.15}[/tex] )

ic = [tex]\frac{E}{R} e^{-t/t}[/tex] = [tex]\frac{15}{10} e^{-t/t}[/tex] = [tex]1.5 mAe^{-t/0.15s}[/tex]

attached

Correct question reads;

Assume that you and your best friend each have $1000 to invest. You invest your money in a fund that pays 10% per year compound interest. Your friend invests her money at a bank that pays 10% per year simple interest. At the end of 1 year, the difference in the total amount for each of you is:

(a) You have $10 more than she does

(b) You have $100 more than she does

(c) You both have the same amount of money

(d) She has $10 more than you do

Answer:

(d) She has $10 more than you do

Explanation:

Using the compound interest formula

A= P [ (1-i)^n-1

Where P = Principal/invested amount, i = annual interest rate in percentage, and n = number of compounding periods.

My compound interest is:

= 1000 [ (1-0.1)^1-1

= $1000

$1,000 + $1,000 invested= $2,000 total amount received.

My friend's simple interest is;

To determine the total amount accrued we use the formula:

P(1 + rt) Where:

P = Invested Amount (1000)

I = Interest Amount (10,000)

r = Rate of Interest per year (10% or 0.2)

t = Time Period (1 )

= 1000 (1 + rt)

= 1000 (1 + 0.1x1)

= $1100 + $1000 invested = $2100 total amount received.

Therefore, we observe that she (my friend) has $100 more than I do.

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:

a) What is the surface temperature, in °C, after 400 s?

   T (0,400 sec) = 800°C

b) Yes, the surface temperature is greater than the ignition temperature of oak (400°C) after 400 s

c) What is the temperature, in °C, 1 mm from the surface after 400 s?

   T (1 mm, 400 sec) = 798.35°C

Explanation:

oak initial Temperature = 25°C = 298 K

oak exposed to gas of temp = 800°C = 1073 K

h = 20 W/m².K

From the book, Oak properties are e=545kg/m³   k=0.19w/m.k   Cp=2385J/kg.k

Assume: Volume = 1 m³, and from energy balance the heat transfer is an unsteady state.

From energy balance: [tex]\frac{T - T_{\infty}}{T_i - T_{\infty}} = Exp (\frac{-hA}{evCp})t[/tex]

Initial temperature wall = [tex]T_i[/tex]

Surface temperature = T

Gas exposed temperature = [tex]T_{\infty}[/tex]

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:

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

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:

a. 6 seconds

b. 180 feet

Explanation:

Images attached to show working.

a. You have the position of the truck so you integrate twice. Use the formula and plug in the time t = 7 sec. Check out uniform acceleration. The time at which the truck's velocity is zero  is when it stops.

b. Determine the initial speed. Plug in the time calculated in the previous step. From this we can observe that the truck comes to a stop before the end of the ramp.

Answer:

a. Tm = 0.3192min.

b. MRR = 396.91mm^{3}/s.

Explanation:

Given the following data;

Drill diameter, D = 12.7mm

Depth, L = 60mm

Cutting speed, V = 25m/min = 25,000m

Feed, F = 0.30mm/rev

To find the cutting time;

Cutting time, Tm =?

[tex]Tm = \frac{L}{Fr}[/tex] .......eqn 1

We would first solve for the feed rate (F);

[tex]Fr = NF[/tex] .......eqn 2

But we need to find the rotational speed (N);

[tex]N= \frac{V}{\pi *D}[/tex]

[tex]N= \frac{25000}{3.142*12.7}[/tex]

[tex]N= \frac{25000}{39.90}[/tex]

N = 626.57rev/min.

Substiting N into eqn 2;

[tex]Fr = NF[/tex]

Fr = 626.57 * 0.30

Fr = 187.97mm/min.

Substiting F into eqn 1;

[tex]Tm = \frac{L}{Fr}[/tex]

[tex]Tm = \frac{60}{187.97}[/tex]

Tm = 0.3192min.

Therefore, the cutting time is 0.3192 minutes.

For the material removal rate (MRR);

[tex]MRR = \frac{\pi *D^{2}Fr}{4}[/tex]

[tex]MRR = \frac{3.142*12.7^{2}*187.97}{4}[/tex]

[tex]MRR = \frac{3.142*161.29*187.97}{4}[/tex]

[tex]MRR = \frac{95258.16}{4}[/tex]

[tex]MRR = 23814.54mm^{3}/min[/tex]

Time in seconds, we divide by 60;

MRR = 23814.54/60 =396.91mm^{3}/s.

Therefore, the material removal rate (MRR) is 396.91mm^{3}/s.

Answer: True

Explanation:

The outer surface temperature of a glass filled with iced water will drop below dew-point temperature of the surrounding air, thereby resulting into the moisture in the vicinity of the glass to condense.

After some time, the condensate may start dripping down because of gravity. This is due to the fact that water will start condensing on glass surface below due point temperature. The adhesion force that water has with glass is low.

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:

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:

E = 2940 J

Explanation:

It is given that,

Mass, m = 12 kg

Position at which the object is placed, h = 25 m

We need to find the potential energy of the mass. It is given by the formula as follows :

E = mgh

g is acceleration due to gravity

[tex]E=12\times 9.8\times 25\\\\E=2940\ J[/tex]

So, the potential energy of the mass is 2940 J.

Answer:

A.) 0 > K > 9.6

B.) K = 9.6

C.) w = +/- 2 sqrt (3)

Explanation:

G(s)= K(s+4)/s(s+1.2)(s+2)

For a closed loop stability, we can analyse by using Routh - Horwitz analysis.

To make the pole completely imaginary, K must be equal to 9.6 Because for oscillations. Whereas, one pair of pole must lie at the imaginary axis.

Please find the attached files for the solution

Answer:

A) condenser pressure = 9.73 kPa,

B) 10242 kw

C) 36.9%

Explanation:

given data

entrance pressure of steam = 12.5 MPa

temperature of steam = 550⁰c

flow rate of steam = 7.7 kg/s

outer pressure = 2 MPa

reheated steam temperature = 450⁰c

isentropic efficiency of turbine( nt ) = 85% = 0.85

isentropic efficiency of pump = 90% = 0.90

From steam tables

at 12.5 MPa and 550⁰c ; h3 = 3476.5 kJ/kg,  S3 = 6.6317 kJ/kgK

also for an Isentropic expansion

S4s = S3 .

therefore when S4s = 6.6317 kJ/kg and P4 = 2 MPa

h4s = 2948.1 kJ/kg

nt = 0.85

nt (0.85) = [tex]\frac{h3-h4}{h3-h4s}[/tex] = [tex]\frac{3476.5 - h4}{3476.5 - 2948.1}[/tex]

making h4 subject of the equation

h4 = 3476.5 - 0.85 (3476.5 - 2948.1)

h4 = 3027.3 kj/kg

at P5 = 2 MPa and T5 = 450⁰c

h5 = 3358.2 kj/kg,  s5 = 7.2815 kj/kgk

at P6 , x6 = 0.95  and s5 = s6

using nt = 0.85 we can calculate for h6 and h6s

from the chart attached below we can see that

p6 = 9.73 kPa, h6 = 2463.3 kj/kg

B) the net power output

solution is attached below

c) thermal efficiency

thermal efficiency = 1 - [tex]\frac{qout}{qin}[/tex] = 1 - ( 2273.7/ 3603.8) = 36.9% ≈ 37%

Answer:

A. Forces that act perpendicular to the surface and pull an object apart exert a tensile stress on the object.

Explanation:

Tensile stress is due to tension forces on a material. Tensile force acts perpendicularly away from the surface of the substance. The pull on the material due to the tensile force exerts tensile stress on the material, that tends to pull the material apart. The magnitude of the tensile stress is given as

σ = [tex]\frac{P}{A}[/tex]

where σ is the tensile stress

P is the tensile force pulling the material apart

A is the cross-sectional area through which the tensile force acts perpendicularly.