Complete Question
For some metal alloy, a true stress of 345 MPa (50040 psi) produces a plastic true strain of 0.02. How much will a specimen of this material elongate when a true stress of 411 MPa (59610 psi) is applied if the original length is 470 mm (18.50 in.)?Assume a value of 0.22 for the strain-hardening exponent, n.
Answer:
The elongation is 
Explanation:
In order to gain a good understanding of this solution let define some terms
True Stress
A true stress can be defined as the quotient obtained when instantaneous applied load is divided by instantaneous cross-sectional area of a material it can be denoted as 
.
True Strain
A true strain can be defined as the value obtained when the natural logarithm quotient of instantaneous gauge length divided by original gauge length of a material is being bend out of shape by a uni-axial force. it can be denoted as 
.
The mathematical relation between stress to strain on the plastic region of deformation is
Where K is a constant
n is known as the strain hardening exponent
This constant K can be obtained as follows
No substituting 
from the question we have
Making the subject from the equation above
From the definition we mentioned instantaneous length and this can be obtained mathematically as follows
Where
is the instantaneous length
is the original length
We can also obtain the elongated length mathematically as follows
Answer:
a. The very first liquid process, when heated from 1250 degree Celsius, is expected to form at the temperature by which the vertical line crosses the phase boundary (a -(a + L)) which is about <em>1310 degree Celsius. </em>
b. The structure of that first liquid is identified by the intersection with ((a+ L)-L) phase boundary; <em>47wt %of Ni</em> is of a tie line formed across the (a+ L) phase area <em>at 1310 degrees.</em>
c. To find the alloy's full melting, it is determined that the intersection of the same vertical line at 60 wt percent Ni with (a -(a+L)) phase boundary is around <em>1350 degrees.</em>
c. The structure of the last remaining solid before full melting correlates to the intersection with the phase boundary (a -(a + L), of the tie line built at 1350 degrees across the (a + L) phase area, <em>being 72wt % of Ni.</em>
Friction losses in pipes can be reduced by decreasing the length of the pipes, reducing the surface roughness of the pipes, and increasing the pipe diameter. Thus, options (c),(e), and (f) hold correct answers.
Friction loss is a measure of the amount of energy a piping system loses because flowing fluids meet resistance. As fluids flow through the pipes, they carry energy with them. Unfortunately, whenever there is resistance to the flow rate, it diverts fluids, and energy escapes. These opposing forces result in friction loss in pipes.
Friction loss in pipes can decrease the efficiency of the functions of pipes. These are a few ways by which friction loss in pipes can be reduced and the efficiency of the piping system can be boosted:
- <u><em>Decrease the length of the pipes</em></u>: By decreasing pipe lengths and avoiding the use of sharp turns, fittings, and tees, whenever possible result in a more natural path for fluids to flow.
- <u><em>Reduce the surface roughness of the pipes</em></u>: By reducing the interior surface roughness of pipes, a smooth and clearer path is provided for liquids to flow.
- <u><em>Increase the pipe diameter: </em></u>By widening the diameters of pipes, it is ensured that fluids squeeze through pipes easily.
You can learn more about friction losses at
brainly.com/question/13348561
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Answer:
Maximum number that can be represented by 13 bits is 8192 Instructions
Explanation:
number of instructions = 1000
number of bits = log(1000) x number of register
= 6 bits
Since the complete instruction must have 32 bits, then
remaining number of bits = 32 - 6 = 236
number of registers in instruction = 2
number of bits per register = 26/2 = 13
Maximum number that can be represented by 13 bits = 
= 2¹³ = 8192
