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3 years ago

Example 12: Write an algorithm and draw a flowchart to calculate

Engineering

1 answer:

Ilya [14]3 years ago

7 0

Answer:

An algotherum is a finite set of sequential instructions to accomplish a task where instructions are written in a simple English language

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What is the command in the java script that receives data from the user

solmaris [256]

Answer:ummmmmmmmmmmm

Explanation:

Uuaysgggahanwnyecegvxg

6 0

3 years ago

You want to know your grade in Computer Science, so write a program that continuously takes grades between 0 and 100 to standard

Margarita [4]

Answer:

The code is given below in python

# Code Block 1

count = 0 # count variable

total = 0 # total variable

enter = '' # input variable

while enter != 'stop':

enter = input('Enter a grade:' )

if enter != 'stop' and enter.isdigit():

total += int(enter) # add to total value

count = count + 1 # then to the count

print float(total) / count

# Code Block 2

numbers = []

while enter != 'stop':

enter = input('Enter a grade:' )

if enter != 'stop':

numbers.append(int(enter))

print sum(numbers) / float(len(numbers))

5 0

3 years ago

g A smooth pipeline with a diameter of 5 cm carries glycerin at 20 degrees Celsius. The flow rate in the pipe is 0.01 m3/s. What

earnstyle [38]

Answer:

The friction factor is 0.303.

Explanation:

The flow velocity ( $v$ ), measured in meters per second, is determined by the following expression:

$v = \frac{4\cdot \dot V}{\pi \cdot D^{2}}$ (1)

Where:

$\dot V$ - Flow rate, measured in cubic meters per second.

$D$ - Diameter, measured in meters.

If we know that $\dot V = 0.01\,\frac{m^{3}}{s}$ and $D = 0.05\,m$ , then the flow velocity is:

$v = \frac{4\cdot \left(0.01\,\frac{m^{3}}{s} \right)}{\pi\cdot (0.05\,m)^{2}}$

$v \approx 5.093\,\frac{m}{s}$

The density and dinamic viscosity of the glycerin at 20 ºC are $\rho = 1260\,\frac{kg}{m^{3}}$ and $\mu = 1.5\,\frac{kg}{m\cdot s}$ , then the Reynolds number ( $Re$ ), dimensionless, which is used to define the flow regime of the fluid, is used:

$Re = \frac{\rho\cdot v \cdot D}{\mu}$ (2)

If we know that $\rho = 1260\,\frac{kg}{m^{3}}$ , $\mu = 1.519\,\frac{kg}{m\cdot s}$ , $v \approx 5.093\,\frac{m}{s}$ and $D = 0.05\,m$ , then the Reynolds number is:

$Re = \frac{\left(1260\,\frac{kg}{m^{3}} \right)\cdot \left(5.093\,\frac{m}{s} \right)\cdot (0.05\,m)}{1.519 \frac{kg}{m\cdot s} }$

$Re = 211.230$

A pipeline is in turbulent flow when $Re > 4000$ , otherwise it is in laminar flow. Given that flow has a laminar regime, the friction factor ( $f$ ), dimensionless, is determined by the following expression:

$f = \frac{64}{Re}$

If we get that $Re = 211.230$ , then the friction factor is:

$f = \frac{64}{211.230}$

$f = 0.303$

The friction factor is 0.303.

4 0

2 years ago

The base class Pet has attributes name and age. The derived class Dog inherits attributes from the base class Pet class and incl

Nonamiya [84]

Answer:

Explanation:

class Pet:

def __init__(self):

self.name = ''

self.age = 0

def print_info(self):

print('Pet Information:')

print(' Name:', self.name)

print(' Age:', self.age)

class Dog(Pet):

def __init__(self):

Pet.__init__(self)

self.breed = ''

def main():

my_pet = Pet()

my_dog = Dog()

pet_name = input()

pet_age = int(input())

dog_name = input()

dog_age = int(input())

dog_breed = input()

my_pet.name = pet_name

my_pet.age = pet_age

my_pet.print_info()

my_dog.name = dog_name

my_dog.age = dog_age

my_dog.breed = dog_breed

my_dog.print_info()

print(' Breed:', my_dog.breed)

main()

3 0

3 years ago

A cylindrical specimen of some metal alloy having an elastic modulus of 124 GPa and an original cross-sectional diameter of 4.2

IrinaVladis [17]

Answer:

the maximum length of the specimen before deformation is 0.4366 m

Explanation:

Given the data in the question;

Elastic modulus E = 124 GPa = 124 × 10⁹ Nm⁻²

cross-sectional diameter D = 4.2 mm = 4.2 × 10⁻³ m

tensile load F = 1810 N

maximum allowable elongation Δl = 0.46 mm = 0.46 × 10⁻³ m

Now to calculate the maximum length $l$ for the deformation, we use the following relation;

$l$ = [ Δl × E × π × D² ] / 4F

so we substitute our values into the formula

$l$ = [ (0.46 × 10⁻³) × (124 × 10⁹) × π × (4.2 × 10⁻³)² ] / ( 4 × 1810 )

$l$ = 3161.025289 / 7240

$l$ = 0.4366 m

Therefore, the maximum length of the specimen before deformation is 0.4366 m

5 0

3 years ago