U have to find the bill it into the internet in order for me to get the answer. Also u have to make it change into the call.
A good visual lead is 20-30 seconds from the front of the vehicle, focusing in the center of the path of travel. Searching 20 to 30 seconds ahead, gives you time to assess within the next ..12-15 seconds, actions you may need to take to control an approaching risk.
Discussion:
Keeping eye focus centered in a path of travel at an interval of 20 to 30 seconds away from the vehicle is critical to gaining enough info. as possible in the driving scene. Good targeting sets up good sight lines for referencing and good peripheral fields for observing changes.
- It is important to look ahead 12-15 seconds into your target area as one drives. Compromise. space by giving as much space to the greater of two hazards.
Read more on driving visual leads:
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Pseudocodes are used as a prototype of an actual program.
The error in the pseudocode is that, the while loop in the pseudocode will run endlessly.
From the pseudocode, the first line is:
<em>Declare Boolean finished = false</em>
The while loop is created to keep running as long as <em>finished = false.</em>
So, for the while loop to end, the finished variable must be updated to true.
This action is not implemented in the pseudocode.
Hence, the error in the pseudocode is that, the while loop is an endless loop
Read more about pseudocodes at:
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Answer:
<em>Object-oriented</em>
Explanation:
<em>Object Oriented programming</em> <em>(OOP)</em> is a specific way of programming, where the code is organized in units called classes, from which objects are created that are related to each other to achieve the objectives of the applications. Object-oriented programming took over as the dominant programming style in the mid-1980s, largely due to the influence of C ++. Its dominance was consolidated thanks to the rise of graphical user interfaces, for which object-oriented programming is particularly well suited. Its most important characteristics are the following:
Answer:
The change in kinetic energy per unit mass of water flowing through the valve is - ΔKE = 0.025 KJ/Kg
Explanation:
Knowing
-Fluid is air
-inlet 1: P1 = 300 kPa
-exit 2: P2 = 275 kPa
density - rho= 1000 kg/m3
Using the formula
Δh = cΔT + Δp/rho
as change in temperature is neglected then change in enthalpy becomes
Δh = Δp/rho
energy equation could be defined by
Q - W = m(out) [h(out) 
(out)/2 + g Z(out)] - m(in) [h(in) (in)/2 + g Z(in)]
Q - W = m2 [h2 2/2 + g Z2] - m1 [h1 1/2 + g Z1]
as for neglecting potential energy effects
Q - W = m2(h2) - m1(h1)
as the system is adiabatic and has no work done
0 = m2 [h2 2/2] - m1 [h1 1/2]
from mass balance m1 = m2
0 = [h2 2/2] - [h1 1/2]
Change in kinetic energy could be defined by
ΔKE = 2/2 - 1/2
Change in specific enthalpy could be defined by
Δh = h2 - h1
Then the change in kinetic energy per unit mass of water flowing through the valve could be calculated as following
ΔKE = -Δh = ΔP/rho
-(275 - 300)/1000 = 0.025 KJ/Kg
- ΔKE = 0.025 KJ/Kg
