A cylindrical buoy is 2m in diameter and 2.5m long and weight 22kN . The specific weight of sea water is 10.25kN/m^3 . (I) Show that buoy does not float with its axis vertical. (II). What minimum pull should be applied to a chain attached to the center of the base to keep the buoy vertical?

Answers

Answer 1

Answer:

[tex]GM<0[/tex]

So the bouy does not float with its axis vertical

Explanation:

From the question we are told that:

Diameter [tex]d=2m[/tex]

Length [tex]l=2.5m[/tex]

Weight [tex]W=22kN[/tex]

Specific weight of sea water [tex]\mu= 10.25kN/m^3[/tex]

Generally the equation for weight of cylinder is mathematically given by

Weight of cylinder = buoyancy Force

[tex]W=(pwg)Vd[/tex]

Where

[tex]V_d=\pi/4(d)^2y[/tex]

Therefore

[tex]22*10^3=10.25*10^3 *\pi/4(2)^2y\\\\\22*10^3=32201.3247y\\\\\y=1.5m[/tex]

Therefore

Center of Bouyance B

[tex]B=\frac{y}{2}=0.26m\\\\B=0.75[/tex]

Center of Gravity

[tex]G=\frac{I.B}{2}=2.6m[/tex]

Generally the equation for\BM is mathematically given by

[tex]BM=\frac{I}{vd}\\\\BM=\frac{3.142/64*2^4}{3.142/4*2^2*0.5215}\\\\BM=0.479m\\\\[/tex]

Therefore

[tex]BG=2.6-0.476\\\\BG=0.64m[/tex]

Therefore

[tex]GM=BM-BG\\\\GM=0.479m-0.64m\\\\GM=-0.161m\\\\[/tex]

Therefore

[tex]GM<0[/tex]

So the bouy does not float with its axis vertical


Related Questions

An air conditioner removes heat steadily from a house at a rate of 750 kJ/min while drawing electric power at a rate of 5.25 kW. Determine (a) the COP of this air conditioner and (b) the rate of heat transfer to the outside air.Answers:(a) 2.38, (b) 1065 kJ/min

Answers

Answer:

a) the COP of this air conditioner is 2.38

b) the rate of heat transfer to the outside air is 1065 kJ/min

Explanation:

Given the data in the question;

[ Outdoor ] ←  Q[tex]_H[/tex] [ W[tex]_{net, in[/tex] ] Q[tex]_L[/tex] ← [ House ]

Rate of heat removed from the house; Q[tex]_L[/tex]  = 750 kJ/min = ( 750 kJ/min × ( 1 kW / 60 kJ/min ) ) = 12.5 kW

Net-work input; W[tex]_{net, in[/tex] = 5.25 kW

a) The coefficient of performance of the air conditioner; COP.

COP = Q[tex]_L[/tex] / W[tex]_{net, in[/tex]

we substitute

COP = 12.5 kW / 5.25 kW

COP = 2.38

Therefore, the COP of this air conditioner is 2.38

b) the rate of heat transfer to the outside air.

Q[tex]_H[/tex] = Q[tex]_L[/tex] + W[tex]_{net[/tex]

we substitute

Q[tex]_H[/tex] = 12.5 kW + 5.25 kW

Q[tex]_H[/tex] = 17.75 kW

Q[tex]_H[/tex] = ( 17.75 × 60 ) kJ/min

Q[tex]_H[/tex] = 1065 kJ/min

Therefore, the rate of heat transfer to the outside air is 1065 kJ/min

A particle which moves in two-dimensional curvilinear motion has coordinates in millimeters which vary with time t in seconds according to X=2t^2 +3t–1 and y = 5t - 2. Determine the coordinates of the center of curvature C at time t = 1s.

Answers

Answer:

The answer is "22.501,-22.899"

Explanation:

Just as in the previous problems find the angle the velocity makes with the  x-axis and radius of curvature.

[tex]x= 2t^2 + 3t — 1\\\\y=5t-2\\\\x=4t+3\\\\y=5\\\\\tan \alpha (t = 1) =\frac{y}{x}=\frac{5}{4+3}=\frac{5}{7} \to alpha=35.54^{\circ}\\\\[/tex]

For the radius of curvature, we can use the expression from the last two  problems, but first express the position and derivatives as y(x).

[tex]y(x)=2(\frac{y+2}{5})^2+3(\frac{y+2}{5})-1=\frac{1}{25}(2y^2+23y+13)\\\\y'(x)=\frac{1}{25}(4y+23)\\\\y''(x)=\frac{4}{25}\\\\\rho(t=1)=\frac{[1+(\frac{dy}{dx})^2]^{\frac{3}{2}}}{\frac{d^2y}{dx^2}}=\frac{(1+(\frac{35}{25})^2)^{\frac{3}{2}}}{4}25=31.828[/tex]

The position for the center of the radius of curvature [tex]\vec{r}[/tex], (finding this expression is easy and is left as an exercise for the reader.)

[tex]\to \vec{r} = \hat{x}(x + \rho \sin \alpha) + \hat{y}(y- \rho \cos \alpha)\\\\= (4 + 18.501, 3-25.899)\\\\=(22.501, -22.899)[/tex]

Okay bro let’s go man yes yes

Answers

Answer:do me ti

Why not me

Why not me

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Why not me

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Do me ti

Why not me

Why not me

Explanation:

mitski

Technician A says white smoke in the exhaust of a diesel engine can be the result of a cylinder misfire in a warm engine. Technician B says blue smoke in the exhaust of a diesel engine can be caused by scored cylinder walls. Who is correct?

Answers

Answer:

Both

Explanation:

Because of water, fuel does not burn completely. This brings about water fumes that are white in color and looks like white smoke. If engine is cold and water is heating, it leads to steam formation like water vapor. The white times are because of not firing properly in the heated engine. Technician A is right.

Blue fine is caused by this scoring. It is also caused by dirty oil. Technician b is right too

Why does the ceramic made from Thorium and Oxygen have the chemical ratio of 2 oxygen atoms to every thorium atom (ThO2)

Answers

Gf so 467 gah can-go FTC 346

The allowable tensile stress for a 6.25 mm diameter bolt with a thread length of 5.5 mm is 207 MPa. The allowable shear stress of the material is 103 MPa. Where and how will such a bolt be most likely to fail if placed in tension

Answers

Answer:

At the threads due to shear.

Explanation:

Given :

The allowable tensile stress = 207 MPa

The allowable shear stress = 103 MPa

If a tensile force is applied, the maximum shear stress occurs at the threads of  the bolt. The bolt is most likely to fail at the critical section. The critical cross section is the section having the minimum cross sectional area.

The portion of the bolt having threads has the minimum cross sectional area.

So when the bolt is applied with a tensile force, failure is most likely to take place at the threads due to the shearing force.

Resistance depends on which three properties of a wire?

Color and texture are not directly related to a wire’s resistance.

1. color, thickness, texture
2. thickness, length, temperature
3. length, texture, temperature
4. temperature, color, texture

Answers

Answer:

2

Explanation:

From the formula R=(ro)A/l resistance depends on the length of the wire, the area of the wire(thickness) and the resistivity(ro) which depends on the material and temperature.

The number of pulses per second from IGBTs is referred to as

Answers

Referred to as Carrier Frequency

Even though the content of many alcohol blends doesn’t affect engine drive ability using gasoline with alcohol in warm weather may cause

Answers

Even though the content of many alcohol blends doesn't affect engine driveability, using gasoline with alcohol in warm weather may cause: decrease in fuel economy.

Mark brainliest
s

A ceramic specimen with an elastic modulus of 300 GPa is under a tensile stress of 800 MPa. Will it fracture if its most severe flaw is an internal crack of 0.30 mm long with a tip radius of curvature in the amount of 0.0015 mm? Please justify your conclusion. (Hint: Compare the largest stress in the specimen around the crack to the theoretical strength which is roughly E/10).

Answers

Answer:

16Gpa < 30 Gpa

there would be no fracture

Explanation:

fracture can occur if the maximum strength at the top of the biggest flaw is more than the theoretical fracture

to get the theoretical strength =

e/10 = 300/10

= 30 Gpa

we get the magnitude at the buggest flaw

= 2σ√a/ρt

σ = 800

ρτ = 0.0015

a= 0.3/2

[tex]=2*800\sqrt{\frac{\frac{0.3}{2} }{0.0015} }[/tex]

= [tex]=2*800*\sqrt{100} \\=2*800*10\\=16000MPa[/tex]

= 16Gpa < 30 Gpa

the fracture is not going to happen given that the maximum strenght is smaller than the theoretical fracture strength.

Can you use isentropic efficiency for a non-adiabatic compressor?

Can you use isothermal efficiency for an adiabatic compressor?

Answers

Mark brainliest please!

Isothermal work will be less than the adiabatic work for any given compression ratio and set of suction conditions. The ratio of isothermal work to the actual work is the isothermal efficiency. Isothermal paths are not typically used in most industrial compressor calculations.

Compressors

Compressors are used to move gases and vapors in situations where large pressure differences are necessary.

Types of Compressor

Compressors are classified by the way they work: dynamic (centrifugal and axial) or reciprocating. Dynamic compressors use a set of rotating blades to add velocity and pressure to fluid. They operate at high speeds and are driven by steam or gas turbines or electric motors. They tend to be smaller and lighter for a given service than reciprocating machines, and hence have lower costs.

Reciprocating compressors use pistons to push gas to a higher pressure. They are common in natural gas gathering and transmission systems, but are less common in process applications. Reciprocating compressors may be used when very large pressure differences must be achieved; however, since they produce a pulsating flow, they may need to have a receiver vessel to dampen the pulses.

The compression ratio, pout over pin, is a key parameter in understanding compressors and blowers. When the compression ratio is below 4 or so, a blower is usually adequate. Higher ratios require a compressor, or multiple compressor stages, be used.

When the pressure of a gas is increased in an adiabatic system, the temperature of the fluid must rise. Since the temperature change is accompanied by a change in the specific volume, the work necessary to compress a unit of fluid also changes. Consequently, many compressors must be accompanied by cooling to reduce the consequences of the adiabatic temperature rise. The coolant may flow through a jacket which surrounds the housing with liquid coolant. When multiple stage compressors are used, intercooler heat exchangers are often used between the stages.

Dynamic Compressors

Gas enters a centrifugal or axial compressor through a suction nozzle and is directed into the first-stage impeller by a set of guide vanes. The blades push the gas forward and into a diffuser section where the gas velocity is slowed and the kinetic energy transferred from the blades is converted to pressure. In a multistage compressor, the gas encounters another set of guide vanes and the compression step is repeated. If necessary, the gas may pass through a cooling loop between stages.

Compressor Work

To evaluate the work requirements of a compressor, start with the mechanical energy balance. In most compressors, kinetic and potential energy changes are small, so velocity and static head terms may be neglected. As with pumps, friction can be lumped into the work term by using an efficiency. Unlike pumps, the fluid cannot be treated as incompressible, so a differential equation is required:

Compressor Work
Evaluation of the integral requires that the compression path be known - - is it adiabatic, isothermal, or polytropic?
uncooled units -- adiabatic, isentropic compression
complete cooling during compression -- isothermal compression
large compressors or incomplete cooling -- polytropic compression
Before calculating a compressor cycle, gas properties (heat capacity ratio, compressibility, molecular weight, etc.) must be determined for the fluid to be compressed. For mixtures, use an appropriate weighted mean value for the specific heats and molecular weight.

Adiabatic, Isentropic Compression

If there is no heat transfer to or from the gas being compressed, the porocess is adiabatic and isentropic. From thermodynamics and the study of compressible flow, you are supposed to recall that an ideal gas compression path depends on:

Adiabatic Path
This can be rearranged to solve for density in terms of one known pressure and substituted into the work equation, which then can be integrated.
Adiabatic Work
The ratio of the isentropic work to the actual work is called the adiabatic efficiency (or isentropic efficiency). The outlet temperature may be calculated from
Adiabatic Temperature Change
Power is found by multiplying the work by the mass flow rate and adjusting for the units and efficiency.
Isothermal Compression

If heat is removed from the gas during compression, an isothermal compression cycle may be achieved. In this case, the work may be calculated from:

http://facstaff.cbu.edu/rprice/lectures/compress.html

A 2.0-in-thick slab is 10.0 in wide and 12.0 ft long. Thickness is to be reduced in three steps in a hot rolling operation. Each step will reduce the slab to 75% of its previous thickness. It is expected that for this metal and reduction, the slab will widen by 3% in each step. If the entry speed of the slab in the first step is 40 ft/min, and roll speed is the same for the three steps, determine: (a) length and (b) exit velocity of the slab after the final reduction

Answers

Answer:

26.02 ft

86.7690 ft/min

Explanation:

After 3 steps

0.75³(2.0 thickness)

T = 0.84375

W = (1+0.03)³10

= 10.92727 inches

A To get length

2.0 x 10 x 12 x12 = 0.84375 x 10.92727x lf

= 2880 = 9.21988Lf

Lf = 2880/9.21988

= 312.368 inches

Convert to feet

322.368 x 0.0833

= 26.02 ft

B.

= 2 x 10 x 40 = 0.84375 x 10.92727 x vf

800 = 9.21988vf

Vf = 800/9.21988

Vf = 86.7690 ft/min

are there engineering students here?​

Answers

Uh, I’d assume so because Brainly has a whole section of questions for them.
Probably...i mean....your asking in the engineering section

Things to be done before isolation

Answers

As in a lockdown? Or quarantine? Gets lots of canned food, toilet paper, water and clothing for the future.

A turbine of a fossil fuel burning installation delivers 1,500 hp of mechanical energy to a generator. The generator then converts 80.0% of the mechanical energy into electrical energy. If the terminal potential difference of the generator is 1790 V, what current does it deliver (in A)

Answers

Answer:

The generator delivers current of 500.11 A

Explanation:

Given the data in the question;

mechanical energy delivered to the generator = 1500 hp

efficiency η = 80.0 %

terminal potential difference of the generator = 1790 V

we know that;

1 hp = 746 W

so

the mechanical energy delivered to the generator will be

Generator Input = ( 1500 × 746 )W = 1119000 W

So the generator output will be;

Generator Output = Generator Input × η

we substitute

Generator Output = 1119000 W × 80.0 %

Generator Output = 1119000 W × 0.8

Generator Output = 895200 W

So the Current will be;

[tex]I[/tex] = Generator Output / terminal potential difference of the generator

we substitute

[tex]I[/tex] =  895200 W / 1790 V

[tex]I[/tex] =  500.11 A

Therefore, The generator delivers current of 500.11 A

to check for ripple voltage from the alternator, connect a digital multimeter and select

Answers

Answer:

isn't it summer? sjsushsiansudndd

Hi, can anyone draw me an isometric image of this shape?​

Answers

I can give me like 10 minutes I gotta find my book
I have 2 accounts here’s the answer

Can some people answer these questions so i can get to know the age group i an making my target market for DT-GSCE thankyou if you do my deadline is tomorrow :D

Answers

Answer:

I think you might have forgotten to post the problems

The purpose of the international residential code is to

Answers

Answer:

The International Building Code (IBC) is a model code that provides minimum requirements to safeguard the public health, safety and general welfare of the occupants of new and existing buildings and structures.

Explanation:

An intelligence signal is amplified by a 65% efficient amplifier before being combined with a 250W carrier to generate an AM signal. If it is desired to operate at 50% modulation, what must be the dc input power to the final intelligence signal amplifier

Answers

Answer:

"192.3 watt" is the right answer.

Explanation:

Given:

Efficient amplifier,

= 65%

or,

= 0.65

Power,

[tex]P_c=250 \ watt[/tex]

As we know,

⇒ [tex]P_t=P_c(1+\frac{\mu^2}{2} )[/tex]

By putting the values, we get

        [tex]=P_c(1+\frac{1}{2} )[/tex]

        [tex]=1.5 \ P_c[/tex]

Now,

⇒ [tex]P_i=(P_t-P_c)[/tex]

        [tex]=1.5 \ P_c-P_c[/tex]

        [tex]=\frac{P_c}{2}[/tex]

DC input (0.65) will be equal to "[tex](\frac{P_c}{2} )[/tex]".

hence,

The DC input power will be:

= [tex]\frac{250}{2}\times \frac{1}{0.65}[/tex]

= [tex]\frac{125}{0.65}[/tex]

= [tex]192.3 \ watt[/tex]

Other than applying a strain to the gauge, what is the primary external/environmental factor that will influence the readings of a strain gauge?

Answers

Answer:

Temperature of the environment

Explanation:

Aside the strain applied to the strain gauge there are several other factors that might affect the reading of the strain gauge, and they are either external or internal factors like ; resistivity, length, and material of the strain gauge.

But the primary external factor is the Temperature of the environment when reading is taken

Water flows through a converging pipe at a mass flow rate of 25 kg/s. If the inside diameter of the pipes sections are 7.0 cm and 5.0 cm, find the volume flow rate and the average velocity in each pipe section.

Answers

Answer:

volumetric flow rate = [tex]0.0251 m^3/s[/tex]

Velocity in pipe section 1 = [tex]6.513m/s[/tex]

velocity in pipe section 2 = 12.79 m/s

Explanation:

We can obtain the volume flow rate from the mass flow rate by utilizing the fact that the fluid has the same density when measuring the mass flow rate and the volumetric flow rates.

The density of water is = 997 kg/m³

density = mass/ volume

since we are given the mass, therefore, the  volume will be mass/density

25/997 = [tex]0.0251 m^3/s[/tex]

volumetric flow rate = [tex]0.0251 m^3/s[/tex]

Average velocity calculations:

Pipe section A:

cross-sectional area =

[tex]\pi \times d^2\\=\pi \times 0.07^2 = 3.85\times10^{-3}m^2[/tex]

mass flow rate = density X cross-sectional area X velocity

velocity = mass flow rate /(density X cross-sectional area)

[tex]velocity = 25/(997 \times 3.85\times10^{-3}) = 6.513m/s[/tex]

Pipe section B:

cross-sectional area =

[tex]\pi \times d^2\\=\pi \times 0.05^2= 1.96\times10^{-3}m^2[/tex]

mass flow rate = density X cross-sectional area X velocity

velocity = mass flow rate /(density X cross-sectional area)

[tex]velocity = 25/(997 \times 1.96\times10^{-3}) = 12.79m/s[/tex]

A square plate of titanium is 12cm along the top, 12cm on the right side, and 5mm thick. A normal tensile force of 15kN is applied to the top side of the plate. A normal tensile force of 20kN is applied to the right side of the plate. The elastic modulus, E, is 115 GPa for titanium.If the left and bottom edges of the plate are fixed, calculate the normal strain and elongation of both the TOP and RIGHT side of the plate.

Answers

Answer:

[tex]X_t=2.17391304*10^{-4}[/tex]

[tex]X_r=2.89855072*10^{-4}[/tex]

[tex]e_t=0.0026[/tex]

[tex]e_r=0.0035[/tex]

Explanation:

From the question we are told that:

Dimension [tex]12*12[/tex]

Thickness [tex]l_t=5mm=5*10^-3[/tex]

Normal tensile force on top side [tex]F_t= 15kN[/tex]

Normal tensile force on right side  [tex]F_r= 20kN[/tex]

Elastic modulus, [tex]E=115Gpap=>115*10^9[/tex]

Generally the equation for Normal Strain X is mathematically given by

 [tex]X=\frac{Force}{Area*E}[/tex]

Therefore

For Top

 [tex]X_t=\frac{Force_t}{Area*E}[/tex]

Where

 [tex]Area=L*B*T[/tex]

 [tex]Area=12*10^{-2}*5*10^{-3}[/tex]

 [tex]Area=6*10^{-4}[/tex]  

 [tex]X_t=\frac{15*10^3}{6*10^{-4}*115*10^9}[/tex]

 [tex]X_t=2.17391304*10^{-4}[/tex]

For Right side[tex]X_r=\frac{Force_r}{Area*E}[/tex]

Where

Area=L*B*T

 [tex]Area=12*10^{-2}*5*10^{-3}[/tex]

 [tex]Area=6*10^{-4}[/tex]  

 [tex]X_r=2.89855072*10^{-4}[/tex]

 [tex]X_r=2.89855072*10^{-4}[/tex]

Generally the equation for elongation is mathematically given by

 [tex]e=strain *12[/tex]

For top

 [tex]e_t=2.17391304*10^{-4}*12[/tex]

 [tex]e_t=0.0026[/tex]

For Right

 [tex]e_r=2.89855072*10^{-4} *12[/tex]

 [tex]e_r=0.0035[/tex]

Safety Issues for Operators of Oil and Gas Exploiting Equipment when working off rigs

Answers

Answer:

Safety First, Safety Always. Safety stands out as a core value for the oil and natural gas industry, embedded in every process and decision for operations. The oil and natural gas industry and the federal government are working together to continuously improve the safety of offshore operations. ...

Determine the horsepower required to compress 1 lbm/min of ethylene oxide from 70 oF and 1 atm to 250 psia. The compressor has an efficiency of 75%. The molar heat capacity of ethylene oxide is given by Cp

Answers

Complete Question:

Problem 8 Determine the horsepower required to compress 1 lbm/min of ethylene oxide from 70 °F and 1 atm to 250 psia. The compressor has an efficiency of 75%. The molar heat capacity of ethylene oxide is given by

C_p=10.03+0.0184T  C_p[=]Btu/lbmole- "F ; T[=] °F C,

Answer:

[tex]P'=0.377hp[/tex]

Explanation:

From the question we are told that:

Initial Temperature T_1=70 F

Final Temperature [tex]T_2=250pisa =114.94F[/tex]

Efficiency [tex]E=75\%=0.75[/tex]

Generally the equation for Work-done  is mathematically given by

 [tex]W=\int C_pT[/tex]

 [tex]W=10.03(114.94-70 )+0.0184((114.94)^2-70^2 )[/tex]

 [tex]W=527.21btu/ibmole[/tex]

 [tex]W=11.982btu/ibm[/tex]

Generally the equation for Efficiency  is mathematically given by

 [tex]E=\frac{isotropic Power}{Actual P'}[/tex]

 [tex]E=\frac{P}{P'}[/tex]

Since

Isotropic Power

  [tex]P=0.0167*11.982btu/ibm[/tex]

 [tex]P=0.2btu/s[/tex]

Therefore

 [tex]P'=\frac{0.2}{0.75}[/tex]

 [tex]P'=0266btu/s[/tex]

Since

 [tex]1btu/s=1.4148hp[/tex]

Therefore

 [tex]P'=0.377hp[/tex]

A cylindrical block of wood 1 m in diameter and 1 m long has a specific weight of 7500 N/m^3. Will it float in water with its axis vertical?

Answers

Answer:

The block will float with its axis vertical.

Explanation:

For it to float, the upward force on the cylindrical block must be equal to the weight of an equal volume of water. Also, this upward force must be greater than or equal to the weight of the cylindrical block for it to float.

So, weight of cylindrical block, W = specific weight × volume

specific weight = 7500 N/m³

volume = πd²h/4 where d = diameter of block and h = height of block

volume = π(1 m)² × 1 m/4 = π/4 m³ = 0.7854 m³

W = 7500 N/m³ × 0.7854 m³ = 5890.5 N

Since the density of water = 1000 kg/m³, its specific weight W' = 1000 kg/m³ × 9.8 m/s² = 9800 N/m³

Since the volume of the cylinder = volume of water displaced, the weight of water displaced W' = upward force = specific weight of water × volume of water displaced =  9800 N/m³ × 0.7854 m³ = 7696.92 N

Since W' = 7696.92 N > W = 5890.5 N, the block will float with its axis vertical since the upward force is greater than the weight of the cylindrical block.

If you deposit $ 1000 per month into an investment account that pays interest at a rate of 9% per year compounded quarterly.how much will be in your account at the end of 5 years ?assume no interpèriod compounding

Answers

Answer:

5,465.4165939453

Explanation:

formula

A=P(1+r/n)^n(t)

p=1000

r=0.09

n=4

t=5

An engineer is applying dimensional analysis to study the flow of air through this horizontal sudden contraction for the purpose of characterizing the pressure drop. The flow is being modeled as constant density and steady. What is the functional relationship of the variables that characterize this situation

Answers

Answer:

The answer is " [tex]\Delta p = f(V1, p, V2, d, D, L)[/tex]"

Explanation:

Please find the complete question in the attached file.

Its change in temperature in pipes depends on rate heads and loss in pipes owing to pipe flow, contractual loss, etc.

The temperature change thus relies on V1 v2 p d D L.

How to Cancel prescription

Answers

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A brittle material is subjected to a tensile stress of 1.65 MPa. If the specific surface energy and modulus of elasticity for this material are 0.60 J/m2 and 2.0 GPa, respectively. What is the maximum length of a surface flaw that is possible without fracture

Answers

Answer:

The maximum length of a surface flaw that is possible without fracture is

[tex]2.806 \times 10^{-4} m[/tex]

Explanation:

The given values are,

σ=1.65 MPa

γs=0.60 J/m2

E= 2.0 GPa

The maximum possible length is calculated as:

[tex]\begin{gathered}a=\frac{2 E \gamma_{s}}{\pi \sigma^{2}}=\frac{(2)\left(2 \times 10^{9} \mathrm{~N} / \mathrm{m}^{2}\right)(0.60 \mathrm{~N} / \mathrm{m})}{\pi\left(1.65\times 10^{6} \mathrm{~N} / \mathrm{m}^{2}\right)^{2}} \\=2.806 \times 10^{-4} \mathrm{~m}\end{gathered}[/tex]

The maximum length of a surface flaw that is possible without fracture is

[tex]2.806 \times 10^{-4} m[/tex]

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