Answer:
Steel can rusts easily in the presence of moisture and oxygen, and tarnishes as the rust progresses.
Explanation:
Steel is an alloy of carbon and iron. it is a very useful alloy that is found in almost any engineered piece. The problem with steel is that it is very susceptible to rust, and the cost of maintaining it in order to prevent rust is very high. Steel rusts in the presence of moisture and oxygen (rust is an oxidation-reduction process). Using it as a water slide exposes it constantly to moisture, and to prevent it from rusting in this case will involve a lot of maintenance cost, which is why steel is not advisable to be used in this case.
Answer:
Skeletal muscle
Explanation:
These classifications describe three distinct muscle types: skeletal, cardiac and smooth. Skeletal muscle is voluntary and striated, cardiac muscle is involuntary and striated, and smooth muscle is involuntary and non-striated.
Answer:
See explaination
Explanation:
please kindly see attachment for the step by step solution of the given problem.
A effective ground-fault current path is an intentionally constructed, low-impedance electrically conductive path designed and intended to carry current during ground-fault conditions from the point of grounding on a wiring system to the electrical supply source.
<h3>Is earth an effective ground fault current path?</h3>
- Sticking the wire in the ground is not sufficient since the earth is not thought to be a reliable ground-fault current channel.
- The electrical system of a building or other structure is based on grounding.
- To give a fault current a secure path to travel, grounding is used.
- When installing switches, light fixtures, appliances, and receptacles, a complete ground route must be kept.
- The undesired current flow trips circuit breakers or blows fuses in a system that is correctly grounded.
- Through the use of a grounding bank, effective grounding maintains voltages within predetermined limits during a line-to-ground fault (short-circuit condition).
To learn more about ground-fault current channel refer,
brainly.com/question/28498355
#SPJ4
Answer:
The attached system shows that there’s an integrator between the point where disturbance enters the system and error measuring element. A any time when R(s)=0 then
and considering that 
then
For ramp disturbance d(t)=at
therefore, the steady state error is given by
![e(\infty)= \lim_{s \to 0} s [\frac {D(s)}{1+G_c(s)G(s)}]](https://tex.z-dn.net/?f=e%28%5Cinfty%29%3D%20%5Clim_%7Bs%20%5Cto%200%7D%20s%20%5B%5Cfrac%20%7BD%28s%29%7D%7B1%2BG_c%28s%29G%28s%29%7D%5D)
![e(\infty)= \lim_{s \to 0} s [\frac {a}{s^{2}+s^{2}G_c(s)G(s)}]](https://tex.z-dn.net/?f=e%28%5Cinfty%29%3D%20%5Clim_%7Bs%20%5Cto%200%7D%20s%20%5B%5Cfrac%20%7Ba%7D%7Bs%5E%7B2%7D%2Bs%5E%7B2%7DG_c%28s%29G%28s%29%7D%5D)
![e(\infty)= \lim_{s \to 0} s [\frac {a}{s+sG_c(s)G(s)}]](https://tex.z-dn.net/?f=e%28%5Cinfty%29%3D%20%5Clim_%7Bs%20%5Cto%200%7D%20s%20%5B%5Cfrac%20%7Ba%7D%7Bs%2BsG_c%28s%29G%28s%29%7D%5D)
![e(\infty)= \lim_{s \to 0} s [\frac {a}{sG_c(s)G(s)}]](https://tex.z-dn.net/?f=e%28%5Cinfty%29%3D%20%5Clim_%7Bs%20%5Cto%200%7D%20s%20%5B%5Cfrac%20%7Ba%7D%7BsG_c%28s%29G%28s%29%7D%5D)
Whenever 
has a double intergrator, the error 
becomes zero
