Answer:
Shearing stresses are the stresses generated in any material when a force acts in such a way that it tends to tear off the material.
Generally the above definition is valid at an armature level, in more technical terms shearing stresses are the component of the stresses that act parallel to any plane in a material that is under stress. Shearing stresses are present in a body even if normal forces act on it along the centroidal axis.
Mathematically in a plane AB the shearing stresses are given by
Yes the shearing force which generates the shearing stresses is similar to frictional force that acts between the 2 surfaces in contact with each other.
External depreciation may be defined as a loss in value caused by an undesirable or hazardous influence offsite.
<h3>What is depreciation?</h3>
Depreciation may be defined as a situation when the financial value of an acquisition declines over time due to exploitation, fray, and incision, or obsolescence.
External depreciation may also be referred to as "economic obsolescence". It causes a negative influence on the financial value gradually.
Therefore, it is well described above.
To learn more about Depreciation, refer to the link:
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Answer:
a) m=336.18N
b) Vn=16.67m/kmol
Vm=0.1459m^3/kg
Explanation:
To calculate the mass of the octane(m):
Number of mole of octane (n) =0.3kmol(given)
Molarmass of octane (M) =114.23kg/kmol
m=n*M
m=(0.3kmol)*(114.23kg/kmol)
m=34.269kg
To calculate for the weight of octane(W):
W=g*m
W=(9.81m/s^2)*(34.269kg)
W=336.18N
b) For specific volumes of Vn and Vm:
Given volume of octane (V) =5m^3
Vm=V/m
Vm=5m^3/34.269kg
Vm=0.1459m^3/kg
And Vn will be :
Vn=V/m=5m^3/0.3kmol
Vn=16.67m/Kmol
Therefore, the answers are:
a) m=336.18N
b) Vn=16.67m/kmol
Vm=0.1459m^3/kg
Answer:
work is 50 kj
Explanation:
Given data
heat (Q) = 50 kj
To find out
work input for the compression stroke per kilogram of air
Solution
we will apply here "first law of thermodynamics" i.e.
The First Law of Thermodynamics states that heat is a form of energy, subject to the principle of conservation of energy, that heat energy cannot be created or destroyed. It can be transferred from one location to another location. i.e.
ΔU = Q – W ................1
here ΔU is change in internal energy, Q is heat and W is work done
here U = 0 because air compressor the compression takes place at a constant internal energy in question
so that by equation 1
Q = W
and Q = 50
so work will be 50 kj
