Finally, the time has come to say good bye. I had so much experiences in ComSci1. I've gained lots of knowledge and cool ideas. I never imagined that I would survive here in Pisay. The whole year was amazing! Yes, it's hard for me to think that this entry would be my last but in every ending comes a new begining, right? So let's just meet in my next adventure in Computer Science 2. Hooray! Kumsay1 will always be really unforgettable.
To the coolest teacher (hahahaha!), Sir Tom, thank you so much po and more challenges to come!
This is Kate Therese P. Noel of 7-Diamond signing off. 'Til next time!
#ClearanceNaThis
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#G8!HereWeCome
Pisay, ComSci and I
Magandang Buhay! This Blog is all about my Computer Science 1 experiences as a Grade 7 student in Pisay.
Linggo, Marso 15, 2015
The Final Chapter: Installing an OS
So, here it is:( Our last lesson for this school year - installing an Operating System. We are going to have a practical exam on how we can do it. It is indeed another challenge for us.
Here are the links of Youtube tutorials on how you can install:
Here are the links of Youtube tutorials on how you can install:
The Final Chapter: More about Computers
Sir Tom introduced the basic computer hardware troubleshooting. He also taught us what causes them and how we can fix them. Here they are:
Computer System Unit Components was also discussed.
I really had fun during our practical exam wherein we connected some parts of the CPU. It was thrilling but my group mates and I enjoyed it.
- No internet connection
- Blue Screen of Death
- Input/Output devices aren't working
- No screen display
- Computer is not working and produces beeping sound
Computer System Unit Components was also discussed.
- Motherboard
- Power box
- RAM (Main Memory)
- Processor (CPU)
- Hard disk (Secondary Memory)
- DVD Drive
- SATA Cables
- Fan
- Video card
- Network card
- Audio Card
I really had fun during our practical exam wherein we connected some parts of the CPU. It was thrilling but my group mates and I enjoyed it.
The Final Chapter: It's the New Generation
We really are in the 21st century with the 5 generations of computer. Join me as I travel from the past to the present and take a look from where all the computers started up to the modern ones. We can also have an exciting sneak peek of the future generation of computers.
- 1ST GENERATION: VACUUM TUBES
This was used from 1940 to 1956 and could solve one problem at a time. The computation time was on milliseconds.
- 2ND GENERATION: TRANSISTORS
- 3RD GENERATION: INTEGRATED CIRCUITS
- 4TH GENERATION: MICROPROCESSORS
- 5TH GENERATION: ARTIFICIAL INTELLIGENCE
"Each generation of computer is designed based on a new technological development, resulting in better, cheaper and smaller computers that are more powerful, faster and efficient than their predecessors."
- Goel, Ashish
Sabado, Marso 14, 2015
The Final Chapter: Machine Cycle + Networking
Machine Cycle, also known as Instruction Cycle or Processor Cycle, is the operation made by a central processing unit which is the main logic unit of a computer. There are three steps:
- Fetch the instruction from the memory
- Decode or translate the retrieved instruction into a series of computer commands in the control unit
- Execute the computer commands in the arithmethic logic unit
Also, the data are being stored in the memory after being executed by ALU. The process goes on and on after the storage.
The Final Chapter: Confusing Conversions Part 2
Continuation...
C51=___
C = 11 , 8+0+2+1=11 , 1011 = C
5 = 5 , 0+4+0+1=5 , 0101 = 5
1 =1 , 0+0+0+1=1 , 0001 = 1
Thus, it is equal to 1011 0101 0001.
- Hexadecimal - Binary
C51=___
C = 11 , 8+0+2+1=11 , 1011 = C
5 = 5 , 0+4+0+1=5 , 0101 = 5
1 =1 , 0+0+0+1=1 , 0001 = 1
Thus, it is equal to 1011 0101 0001.
- Decimal - Hexadecimal
- Hexadecimal - Decimal
Linggo, Marso 1, 2015
The Final Quarter: Confusing Conversions
Really, converting is really confusing. It's easy once you are used to it. But the only way to master it is to practice and be careful. Always check the base. The most common case of mistakes is converting it wrongly. For example, you must convert a binary into a hexadecimal but you didn't notice that you turned it into a decimal. Here are the ways on how you can convert:
128 64 32 16 8 4 2 1
1 1 1 0 0 1 1 1
The numbers above the binary are the powers of two. The numbers that point to 1 must be added together to get the decimal form.
128+64+32+4+2+1=231
Therefore, 11100111 is equal to 231.
Subtract the highest power of two that is less than the binary from the binary. Continue on subtracting the powers of two from the binary, making sure that the difference is not negative. Whenever the power of two can be subtracted, put 1 under it, but if it can't be, put 0.
128 64 32 16 8 4 2 1
1 1 1 0 1 1 1 1
In this case, 16 got a 0 under it because the difference of 32 and 47 (which you get from subtracting 239 to 64 and 128) is 15. And if you subtract 15 to 16, you'll get a negative one.
So, 239=11101111
Divide the binary every 4 bits starting from the right. For every set of bits, put the four lowest power of two from right to left. Add the ones which are pointed to 1.
1110 0111 1101
8421 8421 8421
= 14 7 13
Because the sum could only be up to 15, here are the corresponding values in hexadecimal:
Therefore, 111001111101 is equal to E7D in hexadecimal.
0010 be continued...
- Binary - Decimal
128 64 32 16 8 4 2 1
1 1 1 0 0 1 1 1
The numbers above the binary are the powers of two. The numbers that point to 1 must be added together to get the decimal form.
128+64+32+4+2+1=231
Therefore, 11100111 is equal to 231.
- Decimal - Binary
Subtract the highest power of two that is less than the binary from the binary. Continue on subtracting the powers of two from the binary, making sure that the difference is not negative. Whenever the power of two can be subtracted, put 1 under it, but if it can't be, put 0.
128 64 32 16 8 4 2 1
1 1 1 0 1 1 1 1
In this case, 16 got a 0 under it because the difference of 32 and 47 (which you get from subtracting 239 to 64 and 128) is 15. And if you subtract 15 to 16, you'll get a negative one.
So, 239=11101111
- Binary - Hexadecimal
Divide the binary every 4 bits starting from the right. For every set of bits, put the four lowest power of two from right to left. Add the ones which are pointed to 1.
1110 0111 1101
8421 8421 8421
= 14 7 13
Because the sum could only be up to 15, here are the corresponding values in hexadecimal:
0010 be continued...
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