Answer:
D. Axial stress divided by axial strain
Explanation:
Lets take rod is pulled by force P
Stress σ = P/A
We know that
σ = ε E
E= Lets take rod is pulled by force P
Stress σ = P/A
We know that
σ = ε E
σ=Axial stress
ε =Axial Strain = ΔL/L
E= σ/ε
E= Axial stress/Axial Strain
So the modulus of elasticity is the ratio of axial stress to axial strain.
σ=Axial stress
ε =Axial Strain = ΔL/L
E= σ/ε
E= Axial stress/Axial Strain
So the modulus of elasticity is the ratio of axial stress to axial strain.
The option D is correct.
Answer:
as the ball is thrown, the enegry rises up, when it falls, the ball relseases all energy
Explanation:
Oh jk haha LOL bye hahaha
Since the system itself is giving off heat, this is a
reduction in the internal energy.
heat = - 25,000 J
Since work is being done on the system, therefore it is
an additional energy to the system. Work is given as:
work = - P dV
work = - 1.50 atm (6 L – 12 L)
work = 9 L atm
Since it is given that 1 L atm is equivalent to 101.3 J,
therefore the total energy added is:
energy due to work = 9 L atm (101.3 J / 1 L atm)
energy due to work = 911.7 J
Therefore the total change in internal energy is the sum
of heat and energy due to work:
Change in internal energy = - 25,000 J + 911.7 J
Change in internal energy = - 24,088.3 J
<span>Therefore, approximately 24.1 kJ of energy is lost by the
system in the total process.</span>
<span>
</span>
<span>Answer:</span>
<span>-24.1 kJ</span>