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2 years ago

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Question 29: Returns a string based on input string The function below takes a single string parameter: input_string. If the inp

ut contains the lowercase letter z, return the string 'has the letter z'. Otherwise, return the string 'not worthwhile'. contain.py 1. def string contains(input_string): Que Best s Availat Awarde Restore original file Save & Grade Save only Atta No attach Attacha Attache

1 answer:

kirill [66]2 years ago

8 0

Answer:

Two Python codes are explained for the problem. Modify as appropriate

Explanation:

<u>CODE 1:</u>

def string_contains(input_string): # called function

if(input_string.__contains__('z')): # Check input_string contains 'z'

print('has the letter z.') # print input_string contains 'z'

else:

print('not worthwhile.') # print if input_string not contains 'z'

input_string = input('Please enter the string: ') # ACeept string from user

string_contains(input_string) # calling function where we pass input_string as actual parameter

<u>CODE 2:</u>

def string_contains(input_string):

for x in input_string:

if x=='z':

return 'has the letter z'

return 'not worthwhile'

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Michelle is the general manager of a power plant. This morning, she will meet with city officials to discuss environmental issue

Irina-Kira [14]

Answer:

interpersonal.

Explanation:

Out of all the activities performed by Michelle, three activities involves the interpersonal skills.

1. Meeting with city officials

2. Meeting with section managers

3. Handling the complaint filed by an employee

All these activities involves interpersonal skills. Hence, we can say that she had spent her most of the day by using the interpersonal skills.

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2 years ago

The air contained in a room loses heat to the surroundings at a rate of 50 kJ/min while work is supplied to the room by computer

Artemon [7]

Answer:

net amount of energy change of the air in the room during a 30-min period = 660KJ

Explanation:

The detailed calculation is as shown in the attached file.

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2 years ago

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A six- lane freeway ( three lanes in each direction) in a scenic area has a measured free- flow speed of 55 mi/ h. The peak- hou

Novosadov [1.4K]

Answer:

0.867

Explanation:

The driver population factor ( $f_{p}$ )can be estimated using the equation below:

$f_{p} = \frac{V}{PHF*N*f_{HV}*v_{p}}$

The value of the heavy vehicle factor ( $f_{HV}$ ) is determined below:

The values of the $E_{T}$ = 2 and $E_{R}$ = 3 are gotten from the tables for the RVs, trucks and buses upgrades for passenger-car equivalents. Therefore:

$f_{HV}$ = 1/[1+0.08(2-1)+0.06(3-1)] = 1/[1+0.08+0.12] = 1/1.2 = 0.833

Furthermore, the vp is taken as 2250 pc/(h*In) from the table of LOS criteria for lane freeway using the 15 minutes flow rate. Therefore:

$f_{p}$ = 3900/[0.8*3*0.833*2250] = 3900/4498.2 = 0.867

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2 years ago

A thermal energy storage unit consists of a large rectangular channel, which is well insulated on its outer surface and encloses

yaroslaw [1]

Answer:

the temperature of the aluminum at this time is 456.25° C

Explanation:

Given that:

width w of the aluminium slab = 0.05 m

the initial temperature $T_1$ = 25° C

$T{\infty} =600^0C$

h = 100 W/m²

The properties of Aluminium at temperature of 600° C by considering the conditions for which the storage unit is charged; we have ;

density ρ = 2702 kg/m³

thermal conductivity k = 231 W/m.K

Specific heat c = 1033 J/Kg.K

Let's first find the Biot Number Bi which can be expressed by the equation:

$Bi = \dfrac{hL_c}{k} \\ \\ Bi = \dfrac{h \dfrac{w}{2}}{k}$

$Bi = \dfrac{hL_c}{k} \\ \\ Bi = \dfrac{100 \times \dfrac{0.05}{2}}{231}$

$Bi = \dfrac{2.5}{231}$

Bi = 0.0108

The time constant value $\tau_t$ is :

$\tau_t = \dfrac{pL_cc}{h} \\ \\ \tau_t = \dfrac{p \dfrac{w}{2}c}{h}$

$\tau_t = \dfrac{2702* \dfrac{0.05}{2}*1033}{100}$

$\tau_t = \dfrac{2702* 0.025*1033}{100}$

$\tau_t = 697.79$

Considering Lumped capacitance analysis since value for Bi is less than 1

Then;

$Q= (pVc)\theta_1 [1-e^{\dfrac {-t}{ \tau_1}}]$

where;

$Q = -\Delta E _{st}$ which correlates with the change in the internal energy of the solid.

So;

$Q= (pVc)\theta_1 [1-e^{\dfrac {-t}{ \tau_1}}]= -\Delta E _{st}$

The maximum value for the change in the internal energy of the solid is :

$(pVc)\theta_1 = -\Delta E _{st}max$

By equating the two previous equation together ; we have:

$\dfrac{-\Delta E _{st}}{\Delta E _{st}{max}}= \dfrac{ (pVc)\theta_1 [1-e^{\dfrac {-t}{ \tau_1}}]} { (pVc)\theta_1}$

Similarly; we need to understand that the ratio of the energy storage to the maximum possible energy storage = 0.75

Thus;

$0.75= [1-e^{\dfrac {-t}{ \tau_1}}]}$

So;

$0.75= [1-e^{\dfrac {-t}{ 697.79}}]}$

$1-0.75= [e^{\dfrac {-t}{ 697.79}}]}$

$0.25 = e^{\dfrac {-t}{ 697.79}}$

$In(0.25) = {\dfrac {-t}{ 697.79}}$

$-1.386294361= \dfrac{-t}{697.79}$

t = 1.386294361 × 697.79

t = 967.34 s

Finally; the temperature of Aluminium is determined as follows;

$\dfrac{T - T _{\infty}}{T_1-T_{\infty}}= e ^ {\dfrac{-t}{\tau_t}}$

$\dfrac{T - 600}{25-600}= e ^ {\dfrac{-967.34}{697.79}$

$\dfrac{T - 600}{25-600}= 0.25$

$\dfrac{T - 600}{-575}= 0.25$

T - 600 = -575 × 0.25

T - 600 = -143.75

T = -143.75 + 600

T = 456.25° C

Hence; the temperature of the aluminum at this time is 456.25° C

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2 years ago

A plane wall of thickness 2L = 60 mm and thermal conductivity k= 5W/m.K experiences uniform volumetric heat generation at a rate

aniked [119]

Answer:

Explanation:

A plane wall of thickness 2L=40 mm and thermal conductivity k=5W/m⋅Kk=5W/m⋅K experiences uniform volumetric heat generation at a rateq

˙

q

q

˙

, while convection heat transfer occurs at both of its surfaces (x=-L, +L), each of which is exposed to a fluid of temperature T∞=20∘CT

∞

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2

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∘

C,b=−210

∘

C/m,c=−2×10

4

C/m

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′′

x

(−L)q

x

′′

(−L) and q

′′

x

(+L)q

x

′′

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′′

x

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x

′′

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2

) to reach this state? The density and specific heat of the wall material are 2600kg/m32600kg/m

3

and 800J/kg⋅K800J/kg⋅K, respectively.

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2 years ago