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vova2212 [387]
3 years ago
15

Q10. Select the correct option for the following questions – (10 points, 2 each) a. After an edge dislocation has passed through

some region of a crystal, the atomic arrangement of that region is disordered, Justify your answer - • True • False b. The process by which plastic deformation is produced by dislocation motion is called ____. • Interstitial Dislocation • Slip • Additional plain • Screw c. How does the theoretical strength of a solid material compare with its experimental strength? • Strength theoretical < Strength experimental • Strength theoretical = Strength experimental • Strength theoretical > Strength experimental d. The atoms surrounding a screw dislocation experience what kind(s) of strain(s)? • Shear strain • Compressive strain • Tensile strain e. For a particular crystal structure, the slip direction is that direction in the slip plane having the • highest linear density. • lowest linear density.
Engineering
1 answer:
Gnom [1K]3 years ago
4 0

Answer:

a. True - Because the atomic arrangements of that region is disorderer because of the extra half plane atoms in between the line

b.Slip

C. Strength theoretical is greater than strength experimental

d. Shear stress

e. Highest linear density

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What is the air change rate (ACH) for a 100 ft^2 (9.3 m^2) space with a 10 ft (3.0 m) ceiling and an airflow rate of 200 cfm (95
kakasveta [241]

Answer:

The ACH is 12/h

Solution:

As per the question:

Area of the space, A_{s} = 100 ft^{2}

Height of the given space, h = 10 ft

Air flow rate, Q_{a} = 200 cfm

Now, to find the Air Change Rate (ACH):

We calculate the Volume of the given space:

V_{s} = A_{s}\times h = 100\times 10 = 1000 ft^{3}

Now, the ACH per min:

= \frac{V_{s}}{Q_{a}} = \frac{1000}{200} = 5/min

Now, ACH per hour:

= \frac{60}{5} = 12/h

3 0
4 years ago
list out main types of material used in design and Manufacture of product give one example for each in engineering application ?
cluponka [151]
One of the types of product design is interface design. Another type of product design is called visual product design.
7 0
3 years ago
Atmospheric pressure is measured to be 14.769 psia. a. What would be the equivalent reading of a water barometer (inches of H20)
Fofino [41]

Answer:

(a) water height =408.66 in.

(b) mercury height=30.04 in.

Explanation:

Given: P=14.769 psi     ( 1 psi= 6894.76 \frac{N}{m^2} )

we know that   P=\rho\times g\timesh

where \rho =Density,g=9.81\frac{m}{s^2}

     h=height.

Given that P=14.769 psi ⇒P= 101828.6 7\dfrac{N}{m^2}

(a) P=\rho_{w}\times g\times h_{w}  

     \rho_{w}=1000\frac{Kg}{m^3}

⇒101828.67=1000\times 9.81\times h_{w}

h_{w}=10.38 m

So water barometer will read 408.66 in.            (1 m=39.37 in)

(b)  P=\rho_{hg}\times g\times h_{hg}

     \rho_{hg}=13600

So 101828.67=13600\times 9.81\times h_{hg}

h_{hg}=0.763 m

So mercury barometer will read 30.04 in.

6 0
3 years ago
the voltage across a 5mH inductor is 5[1-exp(-0.5t)]V. Calculate the current through the inductor and the energy stored in the i
Serggg [28]

Given Information:

Inductance = L = 5 mH = 0.005 H

Time = t = 2 seconds

Required Information:

Current at t = 2 seconds = i(t) = ?

Energy at t = 2 seconds = W = ?

Answer:

Current at t = 2 seconds = i(t) = 735.75 A

Energy at t = 2 seconds = W = 1353.32 J

Explanation:

The voltage across an inductor is given as

V(t) = 5(1-e^{-0.5t})

The current flowing through the inductor is given by

i(t) = \frac{1}{L} \int_0^t \mathrm{V(t)}\,\mathrm{d}t \,+ i(0)

Where L is the inductance and i(0) is the initial current in the inductor which we will assume to be zero since it is not given.

i(t) = \frac{1}{0.005} \int_0^t \mathrm{5(1-e^{-0.5t}}) \,\mathrm{d}t \,+ 0\\\\i(t) = 200 \int_0^t \mathrm{5(1-e^{-0.5t}}) \,\mathrm{d}t \\\\i(t) = 200 \: [ {5\: (t + \frac{e^{-0.5t}}{0.5})]_0^t \\i(t) = 200\times5\: \: [ { (t + 2e^{-0.5t} + 2 )] \\

i(t) = 1000t +2000e^{-0.5t} -2000\\

So the current at t = 2 seconds is

i(t) = 1000(2) +2000e^{-0.5(2)} -2000\\\\i(t) = 735.75 \: A

The energy stored in the inductor at t = 2 seconds is

W = \frac{1}{2}Li(t)^{2}\\\\W = \frac{1}{2}0.005(735.75)^{2}\\\\W = 1353.32 \:J

4 0
3 years ago
A slab-milling operation is performed on a 0.7 m long, 30 mm-wide cast-iron block with a feed of 0.25 mm/tooth and depth of cut
denis23 [38]

Answer:

a)  T_m=1.787min

b)  MRR=35259.7mm^3/min

Explanation:

From the question we are told that:

Cast-iron block Dimension:

Lengthl=0.7m=>700mm

Width w=30mm

FeedF=0.25mm/tooth

Depth dp=3mm

Diameter d=75mm

Number of cutting teeth n=8

Rotation speed N=200rpm

Generally the equation for Approach is mathematically given by

x=\sqrt{Dd-d^2}

X=\sqrt{75*3-3^2}

X=14.69mm

Therefore

Effective length is given as

L_e=Approach +object Length

L_e=700+14.69

L_e=714.69mm

a)

Generally the equation for Machine Time is mathematically given by

T_m=\frac{L_e}{F_m}

Where

F_m=F*n*N

F_m=0.25*8*200

F_m=400

Therefore

T_m=\frac{714.69}{400}

T_m=1.787min

b)

Generally the equation for Material Removal Rate. is mathematically given by

MRR=\frac{L*B*d}{t_m}

MRR=\frac{700*30*3}{1.787}

MRR=35259.7mm^3/min

3 0
3 years ago
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