Answer:
Im guessing this is for CEA for PLTW, if so look up the exact assignment number and look at online examples of the exact same assignment.
Explanation:
They all share the way that they are fundamentally designed: if they are quite complex, they will share the same basic logic foundations, like the way that the programming languages work. They also all share the method of construction and common and fundamental electronic components, like resistors, capacitors and transistors. As we humans design them, they make logical sense to at least someone, and probably only discounting the internet, you can probably draw logic diagrams and whatever to represent how they work.
Because they are designed by Humans, in a way they all mimic how our brains and society work. Also, as yet there are no truly intelligent technological systems, and are only able to react to a situation how they have been programmed to do so.
Answer:
<h2>True Most Especially in the field of Automotive Engineering</h2>
Explanation:
Normally, before the introduction of vehicle diagnostics when a vehicle, mostly automobile/car break down, one could be the vehicle mechanic would only suspect one or two related faults based on the present working condition of the car, the mechanic would perform some trial and error before he could fix the car.
But in recent times, the introduction of vehicle diagnostics devices and software has changed the order as vehicles can be connected to a computer that will scan and tell what the problem is before a possible fix.
Answer:
Now find the temperature of each surface, we have that the the temperature on the left side of the wall is T∞₁ - Q/h₁A and the temperature on the right side of the wall is T∞₂ + Q/h₂A.
Note: kindly find an attached diagram to the complete question given below.
Sources: The diagram/image was researched and taken from Slader website.
Explanation:
Solution
Let us consider the rate of heat transfer through the plane wall which can be obtained from the relations given below:
Q = T∞₁ -T₁/1/h₁A = T₁ -T₂/L/kA =T₂ -T∞₂/1/h₂A
= T∞₁ - T∞₂/1/h₁A + L/kA + 1/h₂A
Here
The convective heat transfer coefficient on the left side of the wall is h₁, while the convective heat transfer coefficient on the right side of the wall is h₂. the thickness of the wall is L, the thermal conductivity of the wall material is k, and the heat transfer area on one side of the wall is A. Q is refereed to as heat transfer.
Thus
Let us consider the convection heat transfer on the left side of the wall which is given below:
Q = T∞₁ -T₁/1/h₁A
T₁ = T∞₁ - Q/h₁A
Therefore the temperature on the left side of the wall is T∞₁ - Q/h₁A
Now
Let us consider the convection heat transfer on the left side of the wall which is given below:
Q= T₂ -T∞₂/1/h₂A
T₂ = T∞₂ + Q/h₂A
Therefore the temperature on the right side of the wall is T∞₂ + Q/h₂A
