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:
Check the explanation
Explanation:
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Answer:
1500Ω
Explanation:
Given data
voltage = 15 V
total Resistance = 4000Ω
potential drop V = 9.375 V
To find out
R2
Solution
we know R1 +R2 = 4000Ω
So we use here Ohm's law to find out current I
current = voltage / total resistance
I = 15 / 4000 = 3.75 ×
A
Now we apply Kirchhoffs Voltage Law for find out R2
R2 = ( 15 - V ) / current
R2 = ( 15 - 9.375 ) / 3.75 ×
R2 = 1500Ω
Answer:
A vehicle is considered to be legally parked if it is parked 20 feet (6 m) or more from a pedestrian crosswalk or a marked or unmarked intersection.
Explanation:
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