All categories

3 years ago

CO! Oher...9

Computers and Technology

1 answer:

kondor19780726 [428]3 years ago

6 0

Answer:

Altitude of the town(h1) = 1,196.8 m

Explanation:

Given:

Height of barometer(h) = 65 cm Hg = 0.65

Standard atmospheric pressure = 76 cm Hg = 0.76

Density of mercury(Pm) = 13,600 kg/m³

Density of air (Pa) = 1.25 kg/m³

Find:

Altitude of the town(h1)

Computation:

Pressure due column = Δp mercury column

(Pa)(h1) = (Pm)(h)

(1.25)(h1) = (13,600)(0.76-0.65)

(1.25)(h1) = 1,496

Altitude of the town(h1) = 1,196.8 m

You might be interested in

What is output? x = 10 if (x > 10): print(1) elif (x = 10): print(3) else: print(4)

nikdorinn [45]

Answer:

There are three logical operators: and, or, and not. The semantics (meaning) of these operators is similar to their meaning in English. For example, x > 0 and x < 10 is true only if x is greater than 0 and less than 10.

n % 2 == 0 or n % 3 == 0 is true if either of the conditions is true, that is, if the number is divisible by 2 or 3.

Finally, the not operator negates a boolean expression, so not(x > y) is true if (x > y) is false, that is, if x is less than or equal to y.

Explanation:

3 0

3 years ago

Filmmakers must often establish information quickly in a film in order to give the audience context for the plot. In this film,

Dimas [21]

One thing evident about the two individuals is that Don and Lina are famous. The filmmaker, in the beginning uses the announcer to introduce them as "known all over the world." Another thing to note is that Don and Lina personified themselves a certain way to their fans. The announcer believed sincerely that they were courting. While the announcer played a huge part in giving the audience information that Don and Lina were courting, the filmmaker made sure that audience (crowd in the scene) played a vital part too. Once Don and Lina posed for a picture together, the crowd went wild. This implies that the audience believed that they were together.

8 0

3 years ago

It takes 2 seconds to read or write one block from/to disk and it also takes 1 second of CPU time to merge one block of records.

Alexxx [7]

Answer:

Part a: For optimal 4-way merging, initiate with one dummy run of size 0 and merge this with the 3 smallest runs. Than merge the result to the remaining 3 runs to get a merged run of length 6000 records.

Part b: The optimal 4-way merging takes about 249 seconds.

Explanation:

The complete question is missing while searching for the question online, a similar question is found which is solved as below:

Part a

For optimal 4-way merging, we need one dummy run with size 0.

Merge 4 runs with size 0, 500, 800, and 1000 to produce a run with a run length of 2300. The new run length is calculated as follows $L_{mrg}=L_0+L_1+L_2+L_3=0+500+800+1000=2300$
Merge the run as made in step 1 with the remaining 3 runs bearing length 1000, 1200, 1500. The merged run length is 6000 and is calculated as follows

$L_{merged}=L_{mrg}+L_4+L_5+L_6=2300+1000+1200+1500=6000$

The resulting run has length 6000 records.

Part b

For step 1

Input Output Time

Input Output Time is given as

$T_{I.O}=\frac{L_{run}}{Size_{block}} \times Time_{I/O \, per\, block}$

Here

L_run is 2300 for step 01
Size_block is 100 as given
Time_{I/O per block} is 2 sec

$T_{I.O}=\frac{L_{run}}{Size_{block}} \times Time_{I/O \, per\, block}\\T_{I.O}=\frac{2300}{100} \times 2 sec\\T_{I.O}=46 sec$

So the input/output time is 46 seconds for step 01.

CPU Time

CPU Time is given as

$T_{CPU}=\frac{L_{run}}{Size_{block}} \times Time_{CPU \, per\, block}$

Here

L_run is 2300 for step 01
Size_block is 100 as given
Time_{CPU per block} is 1 sec

$T_{CPU}=\frac{L_{run}}{Size_{block}} \times Time_{CPU \, per\, block}\\T_{CPU}=\frac{2300}{100} \times 1 sec\\T_{CPU}=23 sec$

So the CPU time is 23 seconds for step 01.

Total time in step 01

$T_{step-01}=T_{I.O}+T_{CPU}\\T_{step-01}=46+23\\T_{step-01}=69 sec\\$

Total time in step 01 is 69 seconds.

For step 2

Input Output Time

Input Output Time is given as

$T_{I.O}=\frac{L_{run}}{Size_{block}} \times Time_{I/O \, per\, block}$

Here

L_run is 6000 for step 02
Size_block is 100 as given
Time_{I/O per block} is 2 sec

$T_{I.O}=\frac{L_{run}}{Size_{block}} \times Time_{I/O \, per\, block}\\T_{I.O}=\frac{6000}{100} \times 2 sec\\T_{I.O}=120 sec$

So the input/output time is 120 seconds for step 02.

CPU Time

CPU Time is given as

$T_{CPU}=\frac{L_{run}}{Size_{block}} \times Time_{CPU \, per\, block}$

Here

L_run is 6000 for step 02
Size_block is 100 as given
Time_{CPU per block} is 1 sec

$T_{CPU}=\frac{L_{run}}{Size_{block}} \times Time_{CPU \, per\, block}\\T_{CPU}=\frac{6000}{100} \times 1 sec\\T_{CPU}=60 sec$

So the CPU time is 60 seconds for step 02.

Total time in step 02

$T_{step-02}=T_{I.O}+T_{CPU}\\T_{step-02}=120+60\\T_{step-02}=180 sec\\$

Total time in step 02 is 180 seconds

Merging Time (Total)

Now the total time for merging is given as

$T_{merge}=T_{step-01}+T_{step-02}\\T_{merge}=69+180\\T_{merge}=249 sec\\$

Total time in merging is 249 seconds seconds

5 0

4 years ago

Quick Search lets you refine or narrow your search results using links on the right side of the screen. Do a search on nano-mate

Fofino [41]

Answer:

C. by topic.

D. by collection.

4 0

4 years ago

One of the files you should now have in your fileAsst directory is TweedleDee/hatter.txt. Suppose that you wished to copy that f

Paha777 [63]

Answer:

cp /path/to/source.txt .

is the general format of copying a file to the current directory

cp ~/UnixCourse/fileAsst/TweedleDee/hatter.txt .

You can refer to the current directory with a dot (.)

Explanation:

First, we would like to delete all files in our current directory (commandsAsst directory). To delete the files inside the directory, we would run the following command:

rm -r commandsAsst/*

This command delete the files recursively. Usually, the command for deleting is rm -r. The operator * run rm -r on every file or directory within commandsAsst.

5 0

3 years ago