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Katen [24]
3 years ago
10

If the specific surface energy for soda–lime glass is 0.30 J/m2 and its modulus of elasticity is (69 GPa), compute the critical

stress required for the propagation of an internal crack of length 0.8 mm.
Engineering
1 answer:
aliya0001 [1]3 years ago
6 0

Answer: the critical stress for the sodalime glass = 5.7MPa

Explanation:

Ist Step

we Calculate  half length   of internal crack as

2a =0.8 mm

a = 0.8/2 = 0.4 mm

Changing  to meters becomes = 0.4 / 1000 =0.0004m

2nd Step

Now  the critical stress required for the propagation of the internal crack can be calculated using the formulae

Critical Stress (σc) =      (2 E γs/ πa) 1/2

where E= modulus of elasticity

γs= specific surface energy for soda–lime glass

a= Length

= (2 x 69 x 10 ^9 x 0.30/ π x 0.0004)1/2

=\sqrt{ 32,940,802,036,919}

= 5,739,407.8= 5.7 x 10^6 N/m^2

= 5.7MPa

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Consider the base plate of an 800-W household iron with a thickness of L 5 0.6 cm, base area of A 5 160 cm2, and thermal conduct
N76 [4]

Answer:

a. \frac{-kdT(0)}{dx} =q_{0}=5000W/m^2

b.833.3(0.006-x)+112

c. 117 deg C

Explanation:

Consider the base plate of an 800-W household iron with a thickness of L 5 0.6 cm, base area of A 5 160 cm2, and thermal conductivity of k 5 60 W/m·K. The inner surface of the base plate is subjected to uniform heat flux generated by the resistance heaters inside. When steady operating conditions are reached, the outer surface temperature of the plate is measured to be 112°C. Disregarding any heat loss through the upper part of the iron,

<u>Assumption</u>

Heat conduction is steady state and unidimensional  2. thermal conductivity is constant. Heat supplied is not in the plate

4. we disregard heat loss

Heat flux=heat/area

\alpha/A=800W/160*10^-4

with direction to the surface been in the x direction,

the mathematical expression will be

\frac{d^2T}{dx^2}=0..............1

and \frac{-kdT(0)}{dx} =q_{0}=5000W/m^2

from fourier law, for conductivity

T(L)=T2=112C

b. integrating equation 1 twice we have\dT/dx=c1

T(x)=C1x+C2

C1 and C2 are arbitrary constant

at x=0 the boundary conditions become

-kC1=qo

C1=-(qo/k)

at x=L          

=T(L)=C1L+C2=T2

C2=T2-cL1

C2=T2+qoL/k

Juxtaposing C1 and C2 into the general equation , we have

T(x)=-qo/k+T2+qoL/k=qo(L-k)/k+T2

50000*(0.006-x)/60+112

833.3(0.006-x)+112

c. inner surface plate temperature is

T(0)=833.33(0.006-0)+112 ( using the derivation in answer b)

117 deg C

6 0
3 years ago
Select the properties and typical applications for the high carbon steels.
yanalaym [24]

Answer:

<u>Option-(A)</u>

Explanation:

<u>Typical applications for the high carbon steels includes the following;</u>

It is heat treatable, relatively large combinations of mechanical characteristics. Typical applications: railway wheels and tracks, gears, crankshafts, and machine parts.

3 0
3 years ago
Determine the time required for a car to travel 1 km along a road if the car starts from rest, reaches a maximum speed at some i
lbvjy [14]

Total time taken by car to travel 1 km distance is 48.2 seconds

Explanation:

Given data-

Total distance- 1 km= 1000m

Car starts from rest- Hence the initial velocity (u)= 0 m/s

Then, car accelerates at 1.5 m/s ²

Let us suppose with these acceleration car reaches the max speed of V

Car then decelerates at rate of 2 m/s ²

Finally, car comes to rest

We need to consider the question in two parts

Part 1= when car starts from rest and reaches the value V in the time tₐ at the distance s₁.

Part 2= When car starts from V and finally stops at 1 km mark in the time tₙ in distance 1000-s₁.

For the part 1

We know the formula  

v= u + a*tₐ

where v= final velocity

u- initial velocity

a= acceleration

tₐ= time period

At the starting u= 0  

Hence the equation reduces to V=0+1.5tₐ

Or V= 1.5tₐ                             Eq 1

We also know that s= u*tₐ+ ¹/₂*a*tₐ²

Where s₁= distance covered  (other symbols same meaning)

Since u=0   (u*tₐ=0)

s₁ = ¹/₂*1.5*tₐ²

s₁= ½* 1.5*tₐ²                    -------------Eq 2

Now considering Part 2

Here the case is deceleration hence the equation would change (symbols same)

v= V-a*tₙ     final velocity(v)=0 (car stops finally) & initial velocity (part 2)= V

V= a*tₙ

V=2*tₙ                                 -----------Eq 3

Similarly  

1000-s₁= V* tₙ+¹/₂*(-2) *(tₙ)²

1000-s₁= V* tₙ-tₙ²                      -------Eq 4

Comparing Eq 1 and Eq 3

V= 1.5tₐ  and V=2tₙ                                                              

1.5tₐ = 2 tₙ

tₐ=1.33 tₙ  

Using the above value of tₐ in Eq 1

V= 1.5 tₐ and tₐ= 1.33 tₙ

V= 2tₙ

Similarly from Eq 2 and putting the value of tₙ

s₁= ½*1.5*tₐ²      

s₁=  1.33*(tₙ)²

Substituting the above values in equation 4

1000-s₁= V* tₙ-tₙ²                      

1000- 1.33(tₙ)²=2*(tₙ)*(tₙ)- tₙ²

1000=2.33 (tₙ)²

tₙ=      1000/2.333    

tₙ= 20.7 sec

Similarly putting the value of tₙ in tₐ= 1.33 tₙ

tₙ= 27.5 sec

Hence total time is tₐ +tₙ  

T= 20.7+27.5= 48.2 sec                

8 0
3 years ago
A blizzard is a massive snowstorm. Definitions vary, but for our purposes, we will assume that a blizzard is characterized by bo
natali 33 [55]

Answer:

The weather conditions do not suggest a blizzard. It is explained below.

Explanation:

wind_speed=randi([12 56],24,1);

visibility=randi([1 10],24,1)/10;

storm_data=[wind_speed,visibility];

save stormtrack.dat storm_data -ascii

clear

load stormtrack.dat

fprintf('Below is the storm data from the file stormtrack.dat \n');

disp(stormtrack)

wind_speed=stormtrack(:,1);

visibility=stormtrack(:,2);

L=length(wind_speed);

count=0;

i=0;

while count<4 && i<L

i=i+1;

if wind_speed(i)>=30 && visibility(i)<=0.5

count=count+1;

else

count=0;

end

end

if count==4

fprintf('The weather conditions suggest a blizzard.\n')

else

fprintf('The weather conditions do not suggest a blizzard.\n')

end

Results:

Below is the storm data from the file stormtrack.dat  

49.0000 0.1000

56.0000 0.4000

44.0000 0.6000

27.0000 0.7000

38.0000 0.5000

16.0000 0.9000

52.0000 0.8000

1 51.0000 1.0000

48.0000 0.6000

23.0000 0.4000

38.0000 0.2000

13.0000 0.7000

31.0000 0.8000

26.0000 0.5000

19.0000 0.1000

20.0000 0.3000

31.0000 0.2000

16.0000 0.3000

38.0000 0.5000

33.0000 0.6000

43.0000 0.5000

43.0000 0.9000

40.0000 0.6000

13.0000 1.0000

The weather conditions do not suggest a blizzard.

6 0
3 years ago
Aqueous cleaners are ________ parts cleaning agents.
GREYUIT [131]

Answer:water based

Explanation:

4 0
2 years ago
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