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3 years ago

How can you use the IPDE Process to help protect motorcyclists while driving?

1 answer:

3 0

Answer: you can use the IPDE process to protect motorcyclists while driving by identifying when a motorcycle is present, predicting what it will do next, deciding how to react to the other drivers decision, and exacting that decision.

I hope this helped

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Write definitions for the following two functions:______

lawyer [7]

Answer:

Written in Python

def SumN(n):

total = 0

for i in range(1,n+1):

total = total + i

print("Total: ",total)

def SumNCubes(n):

total = 0

for i in range(1,n+1):

total = total + i**3

print("Cube: ",total)

n = int(input("User Input: "))

if n > 0:

SumN(n)

SumNCubes(n)

Explanation:

The SumN function is defined here

def SumN(n):

This line initializes the total to 0

total = 0

The following iteration compute the required sum

for i in range(1,n+1):

total = total + i

This line outputs the calculated sum

print("Total: ",total)

The SumNCubes function is defined here

def SumNCubes(n):

This line initializes the total to 0

total = 0

The following iteration compute the required sum of cubes

for i in range(1,n+1):

total = total + i**3

This line outputs the calculated sum of cubes

print("Cube: ",total)

The main starts here; The first line prompts user for input

n = int(input("User Input: "))

The next line checks if input is greater than 0; If yes, the two defined functions are called

if n > 0:

SumN(n)

SumNCubes(n)

6 0

3 years ago

A heated long cylindrical rod is placed in a cross flow of air at 20°C (1 atm) with velocity of 10 m/s. The rod has a diameter o

postnew [5]

Answer:

Ts = 413.66 K

Explanation:

given data

temperature = 20°C

velocity = 10 m/s

diameter = 5 mm

surface emissivity = 0.95

surrounding temperature = 20°C

heat flux dissipated = 17000 W/m²

to find out

surface temperature

solution

we know that here properties of air at 70°C

k = 0.02881 W/m.K

v = 1.995 × $10^{-5}$ m²/s

Pr = 0.7177

we find here reynolds no for air flow that is

Re = $\frac{\rho V D }{\mu } = \frac{VD}{v}$

Re = $\frac{10*0.005}{1.99*10^{-5}}$

Re = 2506

now we use churchill and bernstein relation for nusselt no

Nu = $\frac{hD}{k}$ = 0.3 + $\frac{0.62 Re6{0.5}Pr^{0.33}}{[1+(0.4/Pr)^{2/3}]^{1/4}} [1+ (\frac{2506}{282000})^{5/8}]^{4/5}$

h = $\frac{0.02881}{0.005}$ 0.3 + $\frac{0.62*2506{0.5}0.7177^{0.33}}{[1+(0.4/0.7177)^{2/3}]^{1/4}} [1+ (\frac{2506}{282000})^{5/8}]^{4/5}$

h = 148.3 W/m².K

q conv = h∈(Ts- T∞ )

17000 = 148.3 ( 0.95) ( Ts - (20 + 273 ))

Ts = 413.66 K

5 0

4 years ago

Who wants fight with me

vladimir1956 [14]

Yea, ‘Who wants to fight with me’

3 0

3 years ago

A 55-μF capacitor has energy ω (t) = 10 cos2 377t J and consider a positive v(t). Determine the current through the capacitor.

mart [117]

Given :

Capacitor , C = 55 μF .

Energy is given by :

$\omega(t)=10cos^2 (377t)\ J$ .

To Find :

The current through the capacitor.

Solution :

Energy in capacitor is given by :

$\omega=\dfrac{Cv^2}{2}\\\\v=\sqrt{\dfrac{2\omega}{C}}\\\\v=\sqrt{\dfrac{2\times 10cos^2 (377t)}{55\times 10^{-6}}}\\\\v=cos(337t)\sqrt{\dfrac{2\times 10}{55\times 10^{-6}}}\\\\v=603.02\ cos( 337t)$

Now , current i is given by :

$i=C\dfrac{dv}{dt}\\\\i=C\dfrac{d[603.02cos(337t)]}{dt}\\\\i=-55\times 10^{-6}\times 603.03\times 337\times sin(337t)\\\\i=-11.18\ sin(337t)$

( differentiation of cos x is - sin x )

Therefore , the current through the capacitor is -11.18 sin ( 377t).

Hence , this is the required solution .

6 0

3 years ago

Five kg of nitrogen gas (N2) in a rigid, insulated container fitted with a paddle wheel is initially at 300 K, 150 kPa. The N2 g

andrew-mc [135]

Answer:

A) attached below

B) 743 KJ

C) 1.8983 KJ/K

Explanation:

A) Diagram of system schematic and set up states

attached below

B) Calculate the amount of work received from the paddle wheel

assuming ideal gas situation

v1 = v2 ( for a constant volume process )

work generated by paddle wheel = system internal energy

dw = mCv dT . where ; Cv = 0.743 KJ/kgk

= 5 * 0.743 * ( 500 - 300 )

= 3.715 * 200 = 743 KJ

C) calculate the amount of entropy generated ( KJ/K )

S2 - S1 = 1.8983 KJ/K

attached below is the detailed solution

4 0

3 years ago