Explanation:
1) Wind energy is generated through a wind turbine. When wind passes through the blades of wind mill, the blades of the wind mill tend to rotate. Due to the spinning of the rotor across the turbine, the kinetic energy from the wind is converted to electrical energy
2) Incase of wind energy, the consumption gets higher when there is more wind and would be zero incase of no movement of blades.
Incase of hydroelectric power, the generation is rather stable and consumption depends on the usage of power from the consumers
Incase of gasoline generator, the generation is also stable subject to availability of gasoline and consumption again depends on the usage of power from the consumers
3) Pros of Hydroelectric power
Cost of electricity generation is less
Can produce green energy
Produce mass volume of electricity
Cons of Hydroelectric power
Requires massive initial investement
Can be installed on certain demographical area
Answer:
(a) Surface energy is greater than grain boundary energy due to the fact that the bonds of the atoms on the surface are lower than those of the atoms at the grain boundary. The energy is also directly proportional to the number of bonds created.
(b) The energy of a high-angle grain boundary is higher than that of a small-angle grain boundary because the high-angle grain boundary has a higher misalignment and smaller number of bonds than a small-angle grain boundary.
Explanation:
(a) Surface energy is greater than grain boundary energy due to the fact that the bonds of the atoms on the surface are lower than those of the atoms at the grain boundary. The energy is also directly proportional to the number of bonds created.
(b) The energy of a high-angle grain boundary is higher than that of a small-angle grain boundary because the high-angle grain boundary has a higher misalignment and smaller number of bonds than a small-angle grain boundary.
Answer:
radius = 9.1 × 
m
Explanation:
given data
applied load = 5560 N
flexural strength = 105 MPa
separation between the support = 45 mm
solution
we apply here minimum radius formula that is
radius = ![\sqrt[3]{\frac{FL}{\sigma \pi}}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B%5Cfrac%7BFL%7D%7B%5Csigma%20%5Cpi%7D%7D)
.................1
here F is applied load and is length
put here value and we get
radius = ![\sqrt[3]{\frac{5560\times 45\times 10^{-3}}{105 \times 10^6 \pi}}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B%5Cfrac%7B5560%5Ctimes%2045%5Ctimes%2010%5E%7B-3%7D%7D%7B105%20%5Ctimes%2010%5E6%20%5Cpi%7D%7D)
solve it we get
radius = 9.1 × m
