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4 years ago

15

In the text box below, enter the original text you've reconstructed from the compressed version above. Make sure you use _ (unde

rscore) instead of spaces in your answer.

1 answer:

Ierofanga [76]4 years ago

8 0

Answer:

the_big_bug_bit_the_bull_but_the_bull_bit_the_big_bug_back

Explanation:

The image used as sample is attached. The image has the representation for each symbol.

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An electric circuit has a total resistance of 5ohms.If a current of 3A flows through the circuit,calculate the potential differe

Ratling [72]

Answer:

I=V/R

V=Ir

V=5×3

V=15

therefore answer is 15V

3 0

2 years ago

Read 2 more answers

Consider a magnetic disk consisting of 16 heads and 400 cylinders. This disk has four 100-cylinder zones with the cylinders in d

e-lub [12.9K]

Answer:

Disk Capacity = 45056000 bytes

Optimal skew = 26.455 ≈ 27

Maximum Data transfer rate = 262.5 GB per second

Explanation:

given data

heads = 16

cylinders = 400

cylinder zones = 100

each sector contains = 512 bytes

average seek time = 1 msec

disk rotates = 7200 RPM

to find out

disk capacity, optimal track skew, and maximum data transfer rate

solution

first we get total number of sectors that is

total number of sector = number of zones × (number of sectors in different zones

total number of sector = 100 × (160+200+240+280)

total number of sector = 88000

so

Disk Capacity = total number of sectors × size of each sector

Disk Capacity = 88000 × 512

Disk Capacity = 45056000 bytes

and

Rotation time = $\frac{60}{7200}$

Rotation time = 8.33 milli seconds

so Optimal number of sectors in a track = average of ( 160,200,240,280 )

Optimal number of sectors in a track = 220

now New sector is read every $\frac{8.33}{220}$ i.e = 0.0378 ms

here Optimal skew = seek time ÷ new sector read time

Optimal skew = $\frac{1}{0.0378}$

Optimal skew = 26.455 ≈ 27

and

here we know that for maximum transfer rate we will select cylinder with maximum number of sectors i.e here 280 sectors

so

capacity of one track with maximum = 280 × 512

capacity of one track with maximum = = 143360 bytes

and Number of rotations in 1 second is = $\frac{7200}{60}$

Number of rotations in 1 second is = 120

so Data transfer rate = Number of heads × Capacity of one track × Number of rotations in one second

Data transfer rate = 16 × 143360 × 120

Data transfer rate = 275251200 bytes per second

Data transfer rate = 262.5 GB per second

3 0

3 years ago

if you want to withdraw $10000 at the end of two years and $35000 at the end of four years, how much should you deposit now into

Katen [24]

Answer:

490000 dollars

Explanation:

3 0

2 years ago

Rod of steel, 200 mm length reduces its diameter (50 mm) by turning by 2 mm with feed speed 25 mm/min. You are required to calcu

diamong [38]

Answer:

125 cm³/min

Explanation:

The material rate of removal is usually given by the formula

Material Rate of Removal = Radial Depth of Cut * Axial Depth of Cut * Feed Rate, where

Radial Depth of Cut = 25 mm

Axial depth of cut = 200 mm

Feed rate = 25 mm/min

On multiplying all together, we will then have

MRR = 25 mm * 200 mm * 25 mm/min

MRR = 125000 mm³/min

Or we convert it to cm³/min and have

MRR = 125000 mm³/min ÷ 1000

MRR = 125 cm³/min

4 0

3 years ago

Write a function called largest3 which takes 3 numbers as parameters and returns the largest of the 3. Write a program which tak

tamaranim1 [39]

Answer:

The solution code is written in Python.

def largest3(num1, num2, num3):
largest = num1
if(largest < num2):
largest = num2
if(largest < num3):
largest = num3
return largest
first_num = int(input("Enter first number: "))
second_num = int(input("Enter second number: "))
third_num = int(input("Enter third number: "))
largest_number = largest3(first_num, second_num, third_num)
print("The largest number is " + str(largest_number))

Explanation:

<u>Create function largest3</u>

Firstly, we can create a function <em>largest3 </em>which take 3 numbers (<em>num1, num2, num3</em>) as input. (Line 1).
Please note Python uses keyword <em>def </em>to denote a function. The code from Line 2 - 10 are function body of <em>largest3</em>.
Within the function body, create a variable,<em> largest</em>, to store the largest number. In the first beginning, just tentatively assign<em> num1 </em>to<em> largest</em>. (Line 2)
Next, proceed to check if the current "<em>largest</em>" value smaller than the<em> num2 </em>(Line 4). If so, replace the original value of largest variable with <em>num2</em> (Line 5).
Repeat the similar comparison procedure to<em> </em><em>num3</em> (Line 7-8)
At the end, return the final value of "<em>largest</em>" as output

<u>Get User Input</u>

Prompt use input for three numbers (Line 13 -15) using Python built-in <em>input</em> function.
Please note the input parts of codes is done outside of the function <em>largest3</em>.

<u>Call function to get largest number and display</u>

We can simply call the function<em> largest </em>by writing the function name <em>largest</em> and passing the three user input into the parenthesis as arguments. (Line 17)
The function <em>largest </em>will operate on the three arguments and return the output to the variable <em>largest_number</em>.
Lastly, print the output using Python built-in <em>print</em> function. (Line 18)
Please note the output parts of codes is also done outside of the function<em> largest3</em>.

6 0

3 years ago