Answer:




Explanation:
From the question we are told that:
Dimension 
Thickness 
Normal tensile force on top side 
Normal tensile force on right side 
Elastic modulus, 
Generally the equation for Normal Strain X is mathematically given by

Therefore
For Top

Where



For Right side
Where
Area=L*B*T



Generally the equation for elongation is mathematically given by

For top


For Right


Answer:
the nation will suffer terrible consequences
Explanation:
I did that and got it right
Answer:
See explaination
Explanation:
LM358 is the useful IC which works as buffer. It enables circuit to remove overloading effect on each other. Image is in attachment.
We can define a light-emitting diode (LED) as a semiconductor light source that emits light when current flows through it. Electrons in the semiconductor recombine with electron holes, releasing energy in the form of photons
See attached file for detailed solution of the given problem.
Answer:
See explanation below
Explanation:
Hypo-eutectoid steel has less than 0,8% of C in its composition.
It is composed by pearlite and α-ferrite, whereas Hyper-eutectoid steel has between 0.8% and 2% of C, composed by pearlite and cementite.
Ferrite has a higher tensile strength than cementite but cementite is harder.
Considering that hypoeutectoid steel contains ferrite at grain boundaries and pearlite inside grains whereas hypereutectoid steel contains a higher amount of cementite, the following properties are obtainable:
Hypo-eutectoid steel has higher yield strength than Hyper-eutectoid steel
Hypo-eutectoid steel is more ductile than Hyper-eutectoid steel
Hyper-eutectoid steel is harder than Hyper-eutectoid steel
Hypo-eutectoid steel has more tensile strength than Hyper-eutectoid steel.
When making a knife or axe blade, I would choose Hyper-eutectoid steel alloy because
1. It is harder
2. It has low cost
3. It is lighter
When making a die to press powders or stamp a softer metals, I will choose hypo-eutectoid steel alloy because
1. It is ductile
2. It has high tensile strength
3. It is durable
Answer:
Hello your question is incomplete attached below is the complete question
answer:
Considering Laminar flow
Q ( heat ) will be independent of diameter
Considering Turbulent flow
The heat transfer will increase with decreasing "dia" for the turbulent
heat transfer = f(d^-0.8 )
Explanation:
attached below is the detailed solution
Considering Laminar flow
Q ( heat ) will be independent of diameter
Considering Turbulent flow
The heat transfer will increase with decreasing "dia" for the turbulent
heat transfer = f(d^-0.8 )