Answer:
Disc brake system use a slim rotor and small caliper to halt wheel movement but a drum brake system allow heat to build up inside the drum during heavy braking .
Answer:
There are three common methods of charging a battery; constant voltage, constant current and a combination of constant voltage/constant current with or without a smart charging circuit.
Constant voltage allows the full current of the charger to flow into the battery until the power supply reaches its pre-set voltage. The current will then taper down to a minimum value once that voltage level is reached. The battery can be left connected to the charger until ready for use and will remain at that “float voltage”, trickle charging to compensate for normal battery self-discharge.
Constant current is a simple form of charging batteries, with the current level set at approximately 10% of the maximum battery rating. Charge times are relatively long with the disadvantage that the battery may overheat if it is over-charged, leading to premature battery replacement. This method is suitable for Ni-MH type of batteries. The battery must be disconnected, or a timer function used once charged.
Constant voltage / constant current (CVCC) is a combination of the above two methods. The charger limits the amount of current to a pre-set level until the battery reaches a pre-set voltage level. The current then reduces as the battery becomes fully charged. The lead acid battery uses the constant current constant voltage (CC/CV) charge method. A regulated current raises the terminal voltage until the upper charge voltage limit is reached, at which point the current drops due to saturation.
Answer:
The elevation at the high point of the road is 12186.5 in ft.
Explanation:
The automobile weight is 2500 lbf.
The automobile increases its gravitational potential energy in 
. It means the mobile has increased its elevation.
The initial elevation is of 5183 ft.
The first step is to convert Btu of potential energy to adequate units to work with data previously presented.
British Thermal Unit - 
Now we have the gravitational potential energy in lbf*ft. Weight of the mobile is in lbf and the elevation is in ft. We can evaluate the expression for gravitational potential energy as follows:
Where m is the mass of the automobile, g is the gravity, W is the weight of the automobile showed in the problem.
is the final elevation and 
is the initial elevation.
Replacing W in the Ep equation
Finally, the next step is to replace the variables of the problem.
The elevation at the high point of the road is 12186.5 in ft.
Answer:
Superficial design improvements are typically only trivial changes to a design, while functional design improvements can change the way a product or process is used to significantly enhance performance.
Explanation:
As a PC board designer, I would sometimes spend a certain amount of time making traces have shorter routes, or fewer layer changes or bends. (I wanted to make the layout "pretty.") In some cases, these changes are superficial, affecting the appearance only. In some cases, they are functional, reducing crosstalk or emissions or susceptibility to interference.
I deal with a web site that seems to be changing all the time (Brainly). In many cases, the same information is rearranged on the page—a superficial change. In other cases, the information being displayed changes, or the way that certain information is accessed changes. These are functional changes. (Sometimes, they "enhance performance," and sometimes they don't, IMO.)
In short ...
<em>Superficial design improvements are typically only trivial changes to a design, while functional design improvements can change the way a product or process is used to significantly enhance performance.</em>
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
