Answer:
Answer: ±0.02 units or 20±0.02 units or 19.98-20.02 units depending on how they prefer its written (typically the first or second one)
Explanation:
says on the sheet. Unless otherwise stated 0.XX = ±0.02 tolerance
(based on image sent in other post)
If the diameter of the bar is 0.490 at this load, determine I. the engineering stress and strain, and [2] II. the true stress and strain is 1561. 84 MPa.
<h3>What is strain?</h3>
Strain is a unitless degree of ways a great deal an item receives larger or smaller from an implemented load. Normal stress happens while the elongation of an item is in reaction to an everyday pressure (i.e. perpendicular to a surface), and is denoted via way of means of the Greek letter epsilon.
- L = 20 cm d x 1 = 0.21 cm
- dx 2 = 0.25 cmF=5500 a) σ= F/A1= 5000/(π/4x(0.0025)^2)= 1018.5916 MPa lateral stress= Ad/d1= (0.0021-0.0025)/0.0025 = - 0.1 longitudinal stress (ɛ_l)= -lateral stress/v = -(-0.16)/0.3
- (assuming a poisson's ration of 0.3) ε_l=0.16/0.3 = 0.5333
- b) σ_true= σ(1+ ɛ_I)= 1018.5916(1+0.5333
- = 1561.84 MPa.
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Answer:
The inductance of the inductor is 0.051H
Explanation:
From Ohm's law;
V = IR .................. 1
The inductor has its internal resistance referred to as the inductive reactance, X
, which is the resistance to the flow of current through the inductor.
From equation 1;
V = IX
X
=
................ 2
Given that; V = 240V, f = 50Hz,
=
, I = 15A, so that;
From equation 2,
X
= 
= 16Ω
To determine the inductance of the inductor,
X
= 2
fL
L = 
= 
= 0.05091
The inductance of the inductor is 0.051H.
The initial void ratio is the <em>parameter </em>which is used to show the structural foundations for each <em>specimen of sand </em>so that the method and speed of compression would be <em>measured</em>.
Relative density is the mass per unit volume of each specimen of sand which is <em>measured </em>and it has to do with the<em> relative ratio</em> of the density of the sand.
Unit weight is the the exact weight per cubic foot of the sand which is measured.
Please note that your question is incomplete so I gave you a general overview to help you better understand the concept
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Answer:28.21 kJ/kg
Explanation:
Given
Process
Work done for Polytropic process
where n=Polytropic index
![W=\dfrac{0.296(405-300)}{1.3-1}\quad [R_{N_2}=\frac{8.314}{28}]](https://tex.z-dn.net/?f=W%3D%5Cdfrac%7B0.296%28405-300%29%7D%7B1.3-1%7D%5Cquad%20%5BR_%7BN_2%7D%3D%5Cfrac%7B8.314%7D%7B28%7D%5D)
Now Calculating change in Internal energy
Now applying First law concept