Answer:
Yield strength, tensile strength decreases with increasing temperature and modulus of elasticity decreases with increasing in temperature.
Explanation:
The modulus of elasticity of a material is theoretically a function of the shape of curve plotted between the potential energy stored in the material as it is loaded versus the inter atomic distance in the material. The temperature distrots the molecular structure of the metal and hence it has an effect on the modulus of elasticity of a material.
Mathematically we can write,
![E(t)=E_o[1-a\frac{T}{T_m}]](https://tex.z-dn.net/?f=E%28t%29%3DE_o%5B1-a%5Cfrac%7BT%7D%7BT_m%7D%5D)
where,
E(t) is the modulus of elasticity at any temperature 'T'
is the modulus of elasticity at absolute zero.
is the mean melting point of the material
Hence we can see that with increasing temperature modulus of elasticity decreases.
In the case of yield strength and the tensile strength as we know that heating causes softening of a material thus we can physically conclude that in general the strength of the material decreases at elevated temperatures.
Answer:
The correct options are;
C. Alternating current and direct current
E. Electrolyte
Explanation:
Electricity is generated in either AC or DC depending on the source of the energy which aids the electricity generation;
The most common method through which electricity is generated and through which the principle of electric generation was discovered by Michael Faraday, is the rotation of a wire loop or coil between the North and South poles of a magnet to produce Alternating Current
Electricity is also generated in Direct Current by solar cells which convert sunlight into electric energy by photovoltaic effect.
20. Electrolysis is a chemical reaction that is brought about by the application or passing of a direct electric current through an electrolyte, which is an ion containing liquid.
Answer:
47.91 sec
Explanation:
it is given that ![\alpha =\frac{1}{4v^{2}}](https://tex.z-dn.net/?f=%5Calpha%20%3D%5Cfrac%7B1%7D%7B4v%5E%7B2%7D%7D)
at t=0 velocity =0 ( as it is given that it is starting from rest )
we have to find time at which velocity will be 3.3 ![\frac{m}{sec^{2}}](https://tex.z-dn.net/?f=%5Cfrac%7Bm%7D%7Bsec%5E%7B2%7D%7D)
we know that ![\alpha =\frac{dv}{dt}=\frac{1}{4v^{2}}](https://tex.z-dn.net/?f=%5Calpha%20%3D%5Cfrac%7Bdv%7D%7Bdt%7D%3D%5Cfrac%7B1%7D%7B4v%5E%7B2%7D%7D)
![4v^{2}dv=dt](https://tex.z-dn.net/?f=4v%5E%7B2%7Ddv%3Ddt)
integrating both side
---------------eqn 1
at t=o it is given that v=0 putting these value in eqn 1 c=0
so ![\frac{4v^{3}}{3}=t](https://tex.z-dn.net/?f=%5Cfrac%7B4v%5E%7B3%7D%7D%7B3%7D%3Dt)
when v= 3.3 ![\frac{m}{sec^{2}}](https://tex.z-dn.net/?f=%5Cfrac%7Bm%7D%7Bsec%5E%7B2%7D%7D)
t=![\frac{4}{3}\times 3.3^{3}](https://tex.z-dn.net/?f=%5Cfrac%7B4%7D%7B3%7D%5Ctimes%203.3%5E%7B3%7D)
=47.91 sec
Answer:
winter viscosity grades
Explanation:
The “W”/winter viscosity grades describe the oil's viscosity under cold temperature engine starting conditions. There's a Low Temperature Cranking Viscosity which sets a viscosity requirement at various low temperatures to ensure that the oil isn't too thick so that the starter motor can't crank the engine over.