Nar
- 1.
Da li se pojavio signal prekida CPU proverava:
- A.
Kada naidje na odredjenu komandu u programu
- B.
Na kraju svakog instr. seta
- C.
Kada dobije signal sa gen. takta
- D.
Option 4
Correct Answer
B. Na kraju svakog instr. setaExplanation
The correct answer is "Na kraju svakog instr. seta." This means that the CPU interrupt signal is checked at the end of each instruction set. This is important because the CPU needs to be able to handle interrupts from external devices or events that may require immediate attention and interrupt the normal execution of the program. By checking for interrupts at the end of each instruction set, the CPU can ensure that it can respond to these interrupts in a timely manner.Rate this question:
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- 2.
Kada je U/I bez tehnike prekida onda CPU:
- A.
Krace vreme ceka da se zavrsi U/I operacija
- B.
Ne prekida program ako je periferija brza
- C.
Ceka sve vreme dok traje U/I operacija
- D.
Option 4
Correct Answer
C. Ceka sve vreme dok traje U/I operacijaExplanation
When there is no interrupt technique, the CPU waits the entire time for the I/O operation to finish.Rate this question:
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- 3.
Arhitektura jednostruke magistrale je:
- A.
Smanjuje kapacitet i performanse i izbegava se
- B.
Najbolja, najefikasnija i najbrza
- C.
Najcesce u primeni u PC racunarima
- D.
Option 4
Correct Answer
A. Smanjuje kapacitet i performanse i izbegava seExplanation
The correct answer is "Smanjuje kapacitet i performanse i izbegava se" which means that the architecture of a single bus reduces capacity and performance and is generally avoided. This suggests that the single bus architecture is not the most efficient or fastest option, and it is commonly not preferred in PC computers.Rate this question:
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- 4.
Tradicionalna arhitektura ima:
- A.
Sistemsku magistralu velike brzine i magistralu za prosirenje
- B.
Lokalnu, sistemsku i magistralu za prosirenje
- C.
Lokalnu, sistemsku, magistralu velike brzine i mag. za prosirenje
- D.
Option 4
Correct Answer
B. Lokalnu, sistemsku i magistralu za prosirenje -
- 5.
T vreme prenosa ka/od diska =
- A.
Broj bajtova za prenos / (broj bajtova na stazi * brzina rotacije)
- B.
Broj bajtova za prenos * (broj bajtova na stazi / brzina rotacije)
- C.
(Broj bajtova za prenos / broj bajtova na stazi) * brzina rotacije
- D.
Option 4
Correct Answer
A. Broj bajtova za prenos / (broj bajtova na stazi * brzina rotacije)Explanation
The correct formula for calculating the transfer time from/to the disk is given by dividing the number of bytes to be transferred by the product of the number of bytes per track and the rotation speed. This formula takes into account the number of bytes to be transferred, the density of data on the track, and the speed at which the disk is rotating. By dividing the number of bytes to be transferred by the product of the number of bytes per track and the rotation speed, we get the time it takes to transfer the data.Rate this question:
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- 6.
Vincester diskovi imaju sledece osobine:
- A.
Univerzalni su, jeftini, najbrza spoljasnja memorija, sve veci kapacitet
- B.
Posebne namene, jeftini, najbrza spolj. memorija, sve manji po dimenzijama i kapacitetu
- C.
Univerzalni su, jeftini, najbrza spolj. memorija, sve veci po dim.
- D.
Option 4
Correct Answer
A. Univerzalni su, jeftini, najbrza spoljasnja memorija, sve veci kapacitetExplanation
The correct answer is option 4 because it mentions that Vincester disks are universal, cheap, the fastest external memory, and have increasing capacity. This aligns with the given characteristics of Vincester disks, which include being universal, cheap, the fastest external memory, and having increasing capacity.Rate this question:
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- 7.
Poboljsana uniformnost povrsine (povecana pouzdanost)m, smanjenje defekata povrsine (smanjen broj gresaka citanja/upisivanja), manje visine leta glave cvrsca struktura (smanjuje se dinamicnost diska) i veca otpornost na udare/ostecenja su vise izracene kod diskova sa:
- A.
Aluminijuskom podlogom
- B.
Staklenom podlogom
- C.
Plasticnih diskova
- D.
Option 4
Correct Answer
B. Staklenom podlogomExplanation
Disks with a glass substrate have improved surface uniformity, reduced surface defects, lower head flying height, and higher resistance to impacts/damage compared to disks with aluminum or plastic substrates.Rate this question:
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- 8.
Podatak umesto kroz CPU:
- A.
Uvek prolazi kroz DMA
- B.
Zavisno od konf. moze a ne mora da prolazi kroz DMA
- C.
Nikada ne prolazi kroz DMA vec ga odrediste uzima sa magistrale
- D.
Option 4
Correct Answer
B. Zavisno od konf. moze a ne mora da prolazi kroz DMAExplanation
Depending on the configuration, the data may or may not pass through the DMA. The DMA (Direct Memory Access) is a feature that allows data to be transferred between devices without involving the CPU. In some configurations, the data may be directly accessed from the bus by the destination device, bypassing the DMA. However, in other configurations, the data may need to pass through the DMA for the transfer. Therefore, the statement "Zavisno od konf. moze a ne mora da prolazi kroz DMA" (Depending on the configuration, it may or may not pass through the DMA) is the correct answer.Rate this question:
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- 9.
U/I je komplikovan jer su periferije takve da su:
- A.
Sve sporije od CPU i RAM
- B.
Sve sporije od CPU a neke i od RAM
- C.
Sve sporije od CPU i SRAM a neke od RAM
- D.
Option 4
Correct Answer
A. Sve sporije od CPU i RAMExplanation
The correct answer is "Sve sporije od CPU i RAM." This means that all peripherals are slower than both the CPU and RAM. This implies that the peripherals, such as input/output devices, have slower data transfer rates compared to the CPU and RAM.Rate this question:
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- 10.
Procesor neposredno ucestvuje samo u:
- A.
U oba predhodno navedena nacina prenosa
- B.
Programiranom U/I prenosu
- C.
U/I prenosu tehnikom prekida
- D.
Option 4
Correct Answer
A. U oba predhodno navedena nacina prenosaExplanation
The correct answer is "U oba predhodno navedena nacina prenosa" which translates to "In both previously mentioned transmission methods." This suggests that the processor is involved in both programmed I/O transfer and interrupt-driven I/O transfer.Rate this question:
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- 11.
Glavni cilj OS sa raspodelom vremena je:
- A.
Minimizovati vreme odziva
- B.
Maksimalno iskoriscenje procesora
- C.
Dodela resursa
- D.
Option 4
Correct Answer
A. Minimizovati vreme odzivaExplanation
The main goal of a time-sharing operating system is to minimize response time. This means that the operating system aims to provide quick and efficient responses to user requests and actions. By minimizing response time, the operating system can enhance user experience and improve overall system performance. This can be achieved through various techniques such as efficient scheduling algorithms, prioritizing interactive tasks, and optimizing resource allocation.Rate this question:
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- 12.
OS je efikasan samo ako:
- A.
CPU ima odg. hardverske sklopove
- B.
Je CPU vrlo brz
- C.
Je vrlo mali
- D.
Option 4
Correct Answer
A. CPU ima odg. hardverske sklopoveExplanation
An operating system is efficient only if the CPU has appropriate hardware components. This means that the CPU should have the necessary circuits and components to handle the tasks and instructions efficiently. Without the proper hardware, the operating system may not be able to effectively utilize the CPU's capabilities, resulting in decreased performance and efficiency. Therefore, the presence of suitable hardware components in the CPU is crucial for the operating system to be efficient.Rate this question:
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- 13.
Vrste OS-a su:
- A.
Interaktivni, paketni, jednoprogramski i viseprogramski
- B.
Paketni, jednoprogramski, monitorski i viseprogramski
- C.
Interaktivni, paketni, monitorski i viseprogramski
- D.
Option 4
Correct Answer
A. Interaktivni, paketni, jednoprogramski i viseprogramskiExplanation
The correct answer is "Interaktivni, paketni, jednoprogramski i viseprogramski." This answer correctly identifies the different types of operating systems. An interactive operating system allows for user interaction and input. A batch operating system executes jobs in batches without user interaction. A single-program operating system allows for the execution of only one program at a time. A multi-program operating system allows for the execution of multiple programs simultaneously.Rate this question:
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- 14.
Stanja jednog procesa su:
- A.
Nov, spreman, izvrsavanje, ceka, zaustavljen
- B.
Spreman, izvrsavanje, ceka
- C.
Spreman, izvrsavanje, ceka, zaustavljen
- D.
Option 4
Correct Answer
A. Nov, spreman, izvrsavanje, ceka, zaustavljenExplanation
The correct answer is "Nov, spreman, izvrsavanje, ceka, zaustavljen" because it includes all the possible states of a process: "Nov" (new), "spreman" (ready), "izvrsavanje" (running), "ceka" (waiting), and "zaustavljen" (stopped). The other options either miss one or more states or include additional states that are not mentioned in the given list.Rate this question:
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- 15.
Dispecer vrsi:
- A.
Srednjerocno rasporedjivanje procesora
- B.
Kratkorocno rasporedjivanje procesora
- C.
Dugorocno rasporedjivanje procesora
- D.
Option 4
Correct Answer
B. Kratkorocno rasporedjivanje procesoraExplanation
The given options refer to different types of processor scheduling. "Srednjerocno rasporedjivanje procesora" translates to "medium-term processor scheduling," "Kratkorocno rasporedjivanje procesora" translates to "short-term processor scheduling," and "Dugorocno rasporedjivanje procesora" translates to "long-term processor scheduling." The correct answer, "Kratkorocno rasporedjivanje procesora," refers to the short-term scheduling of processes on the processor.Rate this question:
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- 16.
Kada je proces u stanju cekanja on ceka na:
- A.
Procesor ili U/i
- B.
U/I
- C.
U/I ili dogadjaj
- D.
Option 4
Correct Answer
C. U/I ili dogadjajExplanation
When a process is in a waiting state, it is waiting for either an input/output operation (U/I) or an event to occur. This means that the process is not actively using the processor and is temporarily suspended until it receives the required input/output or until the specified event takes place.Rate this question:
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- 17.
Sistemi sa multiprogramiranjem mogu biti:
- A.
Jednoprogramski-jednoprocesni, viseprogramski-jednoprocesni, jednoprogramski-viseprocesni
- B.
Jednoprogramski-jednoprocesni, viseprogramski-viseprocesni
- C.
Jednoprogramski-jednoprocesni, viseprogramski-jednoprocesni, jednoprogramski-viseprocesni, vise programski-viseprocesni
- D.
Option 4
Correct Answer
C. Jednoprogramski-jednoprocesni, viseprogramski-jednoprocesni, jednoprogramski-viseprocesni, vise programski-viseprocesni -
- 18.
Prelazak sa jednog procesa na drugi obavlja se:
- A.
Uvek dugo traje jer je procesor vrlo brz
- B.
Uvek brzo, jer je dispecer mali
- C.
Zavisi od procesa koji se smenjuju
- D.
Option 4
Correct Answer
C. Zavisi od procesa koji se smenjujuExplanation
The transition from one process to another depends on the processes being switched. The time it takes to switch processes is not always the same because different processes may have different requirements and priorities. Therefore, the statement "Zavisi od procesa koji se smenjuju" (It depends on the processes being switched) is the correct answer.Rate this question:
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- 19.
Dodela memorije u particijama je:
- A.
Sa fragmenacijom i kod din i fiksne
- B.
Bez fragmentacije ako je din
- C.
Sa fragmentacijom ako su particije fiksne
- D.
Option 4
Correct Answer
A. Sa fragmenacijom i kod din i fiksneExplanation
The correct answer is "Sa fragmenacijom i kod din i fiksne". This means that memory allocation in partitions can lead to fragmentation in both dynamic and fixed partitions. This suggests that regardless of the type of partition used, there is a possibility of fragmentation occurring during the memory allocation process.Rate this question:
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- 20.
Tehnika preklapanja omogucava da programi:
- A.
Budu veci od virtuelne memorije
- B.
Budu brzi
- C.
Budu veci od fiz memorije
- D.
Option 4
Correct Answer
C. Budu veci od fiz memorijeExplanation
The correct answer is "Budu veci od fiz memorije." The technique of overlapping allows programs to be larger than the physical memory available. This technique involves dividing the program into smaller segments and loading them into memory as needed, allowing the program to exceed the physical memory size. This is achieved by swapping segments in and out of memory, utilizing secondary storage such as a hard disk. By utilizing overlapping, programs can effectively utilize more memory than what is physically available.Rate this question:
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- 21.
Problemi sa fragmentacijom particija mogu:
- A.
Eliminisati tehnikom stranicinje
- B.
Eliminisati tehnikom segmentacije
- C.
Minimizirati primenom stranicenja i segmentacije
- D.
Option 4
Correct Answer
C. Minimizirati primenom stranicenja i segmentacijeExplanation
The correct answer is "Minimizirati primenom stranicenja i segmentacije." This means that the problems with partition fragmentation can be minimized by implementing both paging and segmentation techniques. This suggests that using a combination of these two memory management techniques can help reduce fragmentation and improve overall system performance.Rate this question:
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- 22.
Tabela stranica se raspodeljuje po mem. hijerarhiji. Njeni delovi se nalaze:
- A.
U TLB baferu, u glavnoj memoriji, na disku
- B.
U TLB baferu, u kes memoriji, u glavnoj memoriji, na disku
- C.
U TLB baferu, u kes memorijii, u glavnoj memoriji
- D.
Option 4
Correct Answer
A. U TLB baferu, u glavnoj memoriji, na diskuExplanation
The correct answer is "U TLB baferu, u glavnoj memoriji, na disku." This answer states that the parts of the page table are located in the TLB buffer, in the main memory, and on the disk. This suggests that the page table is distributed across different levels of the memory hierarchy, with some parts residing in the TLB buffer for faster access, some parts in the main memory for medium-speed access, and some parts on the disk for slower access when needed.Rate this question:
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- 23.
Tipicna velicina jedne stranice je:
- A.
2 do 16 sektora
- B.
2 do 16 KB
- C.
2 do 18 KB
- D.
Option 4
Correct Answer
B. 2 do 16 KBExplanation
The typical size of one page is 2 to 16 KB.Rate this question:
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- 24.
Ako se virtuelna adresa poziva na deo programa ili podatak koji je trenutno u fizickoj memoriji (kes, glavna memorija), onda se:
- A.
Taj deo prog. / podataka se prvo prenosi sa diska
- B.
Onda se odg. lokaciji pristupa odmah, koristeci repektivnu fiz. adresu
- C.
Onda se odg. lokaciji pristupa odmah, jer je to ujedno i fiz. adresa
- D.
Option 4
Correct Answer
B. Onda se odg. lokaciji pristupa odmah, koristeci repektivnu fiz. adresuExplanation
When a virtual address refers to a part of the program or data that is currently in physical memory (cache, main memory), it is accessed immediately using the respective physical address. This means that there is no need to transfer the part of the program or data from the disk, and the corresponding location is accessed directly using the corresponding physical address.Rate this question:
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- 25.
Kod adresiranja steka koristimo:
- A.
Indirektno adresiranje
- B.
Implicitno adresiranje
- C.
Neposredno adresiranje
- D.
Option 4
Correct Answer
B. Implicitno adresiranjeExplanation
Implicitno adresiranje se koristi kada se operand za instrukciju automatski određuje na osnovu konteksta ili prethodno definisanih pravila. Ovo znači da se adresa operand-a ne mora eksplicitno navesti u instrukciji. Umesto toga, adresa se dobija implicitno iz registra ili nekog drugog dela memorije. U ovom slučaju, implicitno adresiranje se koristi za adresiranje steka.Rate this question:
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- 26.
Indirektno adresiranje adresira:
- A.
Konstante u memoriji
- B.
Promenljive u memoriji i/ili registrima CPU
- C.
Promenljive u memoriji
- D.
Option 4
Correct Answer
C. Promenljive u memorijiExplanation
Indirektno adresiranje adresira promenljive u memoriji. This means that indirect addressing is used to access variables stored in memory. It allows the program to access the value of a variable indirectly, by using a memory address instead of directly referencing the variable itself. This can be useful in situations where the exact memory location of a variable may not be known or may change dynamically during program execution. By using indirect addressing, the program can access and manipulate variables in memory more flexibly.Rate this question:
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- 27.
Implicitno adresiranje adresira:
- A.
Promenljive u memoriji
- B.
Konstante u memoriji
- C.
Promenljive u registrima CPU
- D.
Option 4
Correct Answer
C. Promenljive u registrima CPUExplanation
Implicitno adresiranje adresira promenljive u registrima CPU. Implicitno adresiranje je način adresiranja u kojem se adresa operanda ne navodi eksplicitno, već se podrazumeva da se operand nalazi u registru CPU. Ovo znači da se vrednost promenljive direktno koristi iz registra, bez potrebe za eksplicitnim navođenjem adrese u memoriji. U ovom slučaju, implicitno adresiranje se koristi za adresiranje promenljivih koje su smeštene u registrima CPU.Rate this question:
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- 28.
Operacija OR:
- A.
Zeljene bitove invertuje a ostale ne menja
- B.
Zeljene bitove postavlja na 1 a ostale ne menja
- C.
Zeljene bitove postavlja na 0 a ostale ne menja
- D.
Option 4
Correct Answer
B. Zeljene bitove postavlja na 1 a ostale ne menjaExplanation
The correct answer is option 4 because the operation OR sets the desired bits to 1 and does not change the rest of the bits.Rate this question:
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- 29.
Operacije za kontrolu toka programa:
- A.
Menjaju sadrzaj programskog brojaca
- B.
Menjaju sadrzaj registra stanja
- C.
Menjaju sadrzaj pokazivaca steka
- D.
Option 4
Correct Answer
A. Menjaju sadrzaj programskog brojacaExplanation
The correct answer is "Menjaju sadržaj programskog brojača." This answer is correct because operations for controlling the flow of a program often involve changing the value of the program counter. The program counter is a register that keeps track of the memory address of the next instruction to be executed. By modifying the program counter, the control flow of the program can be altered, allowing for branching, looping, and other control structures.Rate this question:
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- 30.
Manje adresa, prednosti i mane:
- A.
Jednostavnije i krace instrukcije, jednostavniji procesor, brze donosenje/izvrsavanje instrukcija, Vise instr. po progamu, Duzi programi, slozeniji programi
- B.
Jednostavnije i krace instrukcije, jednostavniji procesor, brze donosenje/izvrsavanje instrukcija, Manje instrukcija po progamu, Kraci programi, jednostavniji programi
- C.
Jednostavnije i krace instrukcije, jednostavniji procesor, Sporije donosenje/izvrsavanje instr. , Vise instr. po programu, Duzi programi, slozeniji programi
- D.
Option 4
Correct Answer
A. Jednostavnije i krace instrukcije, jednostavniji procesor, brze donosenje/izvrsavanje instrukcija, Vise instr. po progamu, Duzi programi, slozeniji programiExplanation
The correct answer is option 4 because it states that shorter and simpler instructions, along with a simpler processor, result in faster instruction execution and decision-making. Additionally, having more instructions per program, longer programs, and more complex programs are mentioned as disadvantages.Rate this question:
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- 31.
Tacan iskaz je:
- A.
Izmedju kesa i RAM se prenose okviri
- B.
Izmedju kesa i RAM se prenose reci
- C.
Izmedju kesa i RAM se prenose blokovi
- D.
Option 4
Correct Answer
C. Izmedju kesa i RAM se prenose blokoviExplanation
The correct answer is "Izmedju kesa i RAM se prenose blokovi." This answer states that blocks are transferred between cache and RAM. This is a common practice in computer architecture, where the cache is used to store frequently accessed blocks of data from RAM, allowing for faster access times. By transferring blocks instead of individual words or frames, the cache can effectively store and retrieve larger chunks of data, improving overall performance.Rate this question:
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- 32.
Vise kesa je uvek:
- A.
Jeftino i do neke velicine brze
- B.
Skupo i do neke vel. brze
- C.
Skupo i uvek brze
- D.
Option 4
Correct Answer
B. Skupo i do neke vel. brzeExplanation
The correct answer is "Skupo i do neke vel. brze". This answer suggests that the quality or value of something is expensive but it is faster up to a certain size. This implies that although it may be costly, it provides a faster outcome within a specific range or limit.Rate this question:
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- 33.
Najbolja dva algoritma za zamenu su:
- A.
Najduze nije koriscen, LRU
- B.
Slucajna zamena, LRU
- C.
Najredje koriscen, LFU
- D.
Option 4
Correct Answer
B. Slucajna zamena, LRUExplanation
The correct answer is "Slucajna zamena, LRU." This means that the two best algorithms for replacement are "Slucajna zamena" and "LRU." The LRU (Least Recently Used) algorithm replaces the least recently used page, while the "Slucajna zamena" algorithm randomly selects a page to replace. Both of these algorithms are considered to be effective strategies for page replacement in computer systems.Rate this question:
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- 34.
Direktno preslikavanje je:
- A.
Jednostavno i jeftino, polje za tag je, pristup je veoma brz a zamena bloka fleksibilna
- B.
Jednostavno i jeftino, polje za tag je kratko, pristup veoma brz, dobro iskoriscenje kesa
- C.
Jednostavno i jeftino, polje za tag je kratko, pristup je veoma brz, lose iskoriscenje kesa
- D.
Option 4
Correct Answer
C. Jednostavno i jeftino, polje za tag je kratko, pristup je veoma brz, lose iskoriscenje kesaExplanation
The correct answer states that "Direktno preslikavanje" (Direct mapping) is simple and inexpensive, the tag field is short, access is very fast, but the cache utilization is poor. This means that in direct mapping, each block of main memory can be mapped to only one specific cache line, which leads to a high risk of cache conflicts and lower cache utilization.Rate this question:
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- 35.
Princip lokalnosti:
- A.
Verovatno je da ce vecina buducih pristupanja procesora glavnoj memoriji biti ka onim lokacijama kojima se nije davno pristupalo
- B.
Verovatno je da ce vecina buducih pristupanja procesora glavnoj memoriji biti ka onim lokacijama kojima se nedavno pristupalo
- C.
Verovatno je da nece vecina buducih pristupanja procesora glavnoj memoriji biti ka onim lokacijama kojima se nije davno pristupalo
- D.
Option 4
Correct Answer
B. Verovatno je da ce vecina buducih pristupanja procesora glavnoj memoriji biti ka onim lokacijama kojima se nedavno pristupaloExplanation
The correct answer suggests that most future accesses by the processor to the main memory will be to locations that have been recently accessed. This aligns with the principle of locality, which states that memory accesses tend to cluster in time and space. The processor is likely to access locations that have been recently accessed because of temporal locality, meaning that recently accessed locations are more likely to be accessed again in the near future.Rate this question:
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- 36.
Lokalnost reference je:
- A.
Prostorna za instrukcije i za podatke
- B.
Vremenska za instrukcije, prostorna za podatke
- C.
Vremenska i prostorna za instukcije i podatke
- D.
Option 4
Correct Answer
C. Vremenska i prostorna za instukcije i podatkeExplanation
The correct answer is "Vremenska i prostorna za instrukcije i podatke." This means that the locality of reference refers to both the temporal (time-related) and spatial (location-related) aspects of instructions and data in a computer system. In other words, it encompasses both the time and location of where instructions and data are accessed and used within the system.Rate this question:
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- 37.
Podaci koji se drze u registrima su pod direktnim upravljanjem:
- A.
Operativnog sistema uz hardversku podrsku
- B.
Hardvera
- C.
Programera i kompajlera ili asemblera
- D.
Option 4
Correct Answer
C. Programera i kompajlera ili asembleraExplanation
The data stored in registers are directly managed by programmers and compilers or assemblers. Registers are small, high-speed memory locations within the CPU that hold data that is frequently accessed by the processor. Programmers and compilers or assemblers are responsible for writing and optimizing the code that uses these registers to store and manipulate data. They determine how data is stored in registers and how it is accessed and manipulated by the CPU. Therefore, the correct answer is "Programmers and compilers or assemblers".Rate this question:
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- 38.
Karakteristike memorije su:
- A.
Lokacija, kapacitet, jedinica prenosa, broj reci performansa, fizicka vrsta, fizicke karakteristike, organizacija
- B.
Lokacija, kapacitet, jedinica prenosa, metod pristupa, performansa, fizicka vrsta, fizicke karakteristike, organizacija
- C.
Lokacija, kapacitet, broj bajtova, jedinica prenosa, metod pristupa, fizicka vrsta, fiz. karakteristike, organizacija
- D.
Option 4
Correct Answer
B. Lokacija, kapacitet, jedinica prenosa, metod pristupa, performansa, fizicka vrsta, fizicke karakteristike, organizacijaExplanation
The correct answer is option 2. This is because it includes all the characteristics of memory mentioned in the question, such as location, capacity, data transfer unit, access method, performance, physical type, physical characteristics, and organization. Option 1 does not include the access method and performance, option 3 does not include the data transfer unit, and option 4 is not a valid option.Rate this question:
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- 39.
Mala po bitu, jeftina, potrebna elektronska kola za osvezavanje, spora, koristi se za glavnu memoriju, u sustini analogna je:
- A.
Obe pomenute vrste RAM memorija
- B.
DRAM
- C.
SRAM
- D.
Option 4
Correct Answer
B. DRAMExplanation
DRAM stands for Dynamic Random Access Memory, which is a type of volatile memory used in computers for the main memory. It is cheaper compared to SRAM (Static Random Access Memory) and is slower in terms of access speed. DRAM stores each bit of data in a separate capacitor within an integrated circuit, which needs to be constantly refreshed to maintain the data. This makes DRAM essentially analog in nature. Therefore, the given statement suggests that DRAM is analog in nature.Rate this question:
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- 40.
Ako sindrom sadrzi sve 0 onda:
- A.
Sindrom neispravan ali nije potrebna ispravka podataka
- B.
Su sindrom i podatak ispravni
- C.
Sindrom ispravan ali je potrebna ispravka podataka
- D.
Option 4
Correct Answer
B. Su sindrom i podatak ispravniExplanation
The correct answer is "sindrom neispravan ali nije potrebna ispravka podataka." This means that the syndrome is incorrect, but there is no need to correct the data.Rate this question:
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- 41.
Vrste poluprovodnickih memorija su:
- A.
RAM, ROM, EPROM, EPROM, EEPROM
- B.
RAM, ROM, PROM, EAROM, EEPROM
- C.
RAM, ROM, PROM, EPROM, EEPROM
- D.
Option 4
Correct Answer
C. RAM, ROM, PROM, EPROM, EEPROMExplanation
The correct answer is RAM, ROM, PROM, EPROM, EEPROM. This is because these are all types of semiconductor memories commonly used in computer systems. RAM (Random Access Memory) is a volatile memory that allows data to be read from and written to. ROM (Read-Only Memory) is a non-volatile memory that contains permanent instructions or data. PROM (Programmable Read-Only Memory) is a type of ROM that can be programmed once. EPROM (Erasable Programmable Read-Only Memory) is a type of ROM that can be erased and reprogrammed using ultraviolet light. EEPROM (Electrically Erasable Programmable Read-Only Memory) is a type of ROM that can be erased and reprogrammed electronically.Rate this question:
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- 42.
Velika po bitu, skupa, nisu potrebna elektronska kola za osvezavanje, brza za kes, digitalna je:
- A.
Obe pomenute vrste RAM memorija
- B.
DRAM
- C.
SRAM
- D.
Option 4
Correct Answer
C. SRAMExplanation
SRAM (Static Random Access Memory) is a type of RAM that is faster and more expensive than DRAM (Dynamic Random Access Memory). Unlike DRAM, SRAM does not require electronic circuits for refreshing the data, making it more efficient and faster for cache memory. Therefore, the given statement implies that SRAM is a type of digital RAM that is both mentioned and meets the mentioned criteria.Rate this question:
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- 43.
Programski brojac (PC) sadrzi:
- A.
Instrukciju koja treba da pribavi
- B.
Adresu instukcije koja treba da se pribavi
- C.
Instrukciju koja treba da izvrsi
- D.
Option 4
Correct Answer
B. Adresu instukcije koja treba da se pribaviExplanation
The correct answer is "adresu instukcije koja treba da se pribavi". A program counter (PC) is a register in a computer processor that holds the memory address of the next instruction to be executed. It keeps track of the current position in the program's sequence of instructions. Therefore, it contains the address of the instruction that needs to be fetched next.Rate this question:
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- 44.
Arhitektura racunara zavisi od:
- A.
Sako od instr. i nacina adresiranja
- B.
Skupa instrukcija, broja bitova koji se koriste za predst. podataka, mehanizama U/I, tehnika adresiranja
- C.
Od hardvera i tehnologija
- D.
Option 4
Correct Answer
B. Skupa instrukcija, broja bitova koji se koriste za predst. podataka, mehanizama U/I, tehnika adresiranjaExplanation
The architecture of a computer depends on various factors, including the set of instructions and addressing modes used, the number of bits used to represent data, input/output mechanisms, and addressing techniques. These factors determine how the computer processes and stores data, communicates with other devices, and performs operations. Therefore, the correct answer is that the architecture of a computer depends on the set of instructions, the number of bits used to represent data, input/output mechanisms, and addressing techniques.Rate this question:
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- 45.
Strukturni hazard se eliminise
- A.
Dupliranjem resursa
- B.
Prosledivanjem rezultata na ulaz ALU
- C.
Ne moze se eliminisati ni smanjiti kazna
- D.
Option 4
Correct Answer
A. Dupliranjem resursaExplanation
The correct answer is "dupliranjem resursa" because duplicating resources is a way to eliminate structural hazards. By duplicating resources, such as registers or functional units, multiple instructions can be executed simultaneously without causing structural hazards. This helps to ensure that the pipeline runs smoothly and efficiently, avoiding any delays or conflicts in resource usage.Rate this question:
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- 46.
Hazard podataka se ublazava:
- A.
Ne moze se eliminisati ni smanjiti kazna
- B.
Dupliranjem resursa
- C.
Prosledjivanjem rezultata na ulaz ALU
- D.
Option 4
Correct Answer
C. Prosledjivanjem rezultata na ulaz ALUExplanation
The correct answer is "prosledjivanjem rezultata na ulaz ALU." This means that the data hazard is mitigated by forwarding the results to the input of the Arithmetic Logic Unit (ALU). By doing this, the ALU can directly use the results without having to wait for them to be stored and retrieved from memory, reducing the delay and potential hazards in the process.Rate this question:
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- 47.
DMA sadrzi brojac podataka, adresni registar, registar podataka i upravljacku logiku:
- A.
Vise kanala i svaki ima sve posebno
- B.
Vise kanala sa zajednickim brojacem podataka
- C.
Jedan kanal
- D.
Option 4
Correct Answer
A. Vise kanala i svaki ima sve posebno -
- 48.
Gospodari magistrala su:
- A.
Samo CPU
- B.
CPU i DMA
- C.
CPU, DMA i arbitar
- D.
Option 4
Correct Answer
B. CPU i DMAExplanation
The correct answer is "CPU i DMA" because "Gospodari magistrala" translates to "masters of the bus" in English. The CPU (Central Processing Unit) and DMA (Direct Memory Access) are the two components responsible for controlling and managing the bus in a computer system. They work together to transfer data between the CPU and memory without involving the CPU directly in the process, thus improving overall system performance. The inclusion of the CPU and DMA in the answer is supported by the context provided in the question.Rate this question:
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- 49.
Jednoprogramski sistemi treba da obezbede:
- A.
Da ceo program i podaci stanu u memoriju
- B.
Se programi brzo izvrsavaju
- C.
Zastitu dela memorije sa operativnim sistemom
- D.
Option 4
Correct Answer
C. Zastitu dela memorije sa operativnim sistemomExplanation
Jednoprogramski sistemi trebaju obezbediti zaštitu dela memorije sa operativnim sistemom. Ova zaštita je važna kako bi se sprečilo da programi ili korisnici nenamerno ili zlonamerno pristupe ili menjaju delove memorije koji nisu namenjeni njima. Ova mera obezbeđuje integritet podataka i programa, što je ključno za sigurno i pouzdano izvršavanje jednoprogramskih sistema.Rate this question:
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- 50.
Kontinualna dodela memorije u particijama je:
- A.
Jedini nacin da se ostvari multiprogramiranje
- B.
Je tipicna za stare operativne sisteme
- C.
Najefikasniji nacin dodele memorije
- D.
Option 4
Correct Answer
B. Je tipicna za stare operativne sistemeExplanation
The continuous allocation of memory in partitions is typical for old operating systems. This means that in these systems, memory is divided into fixed-sized partitions and each process is allocated a specific partition for its execution. This method was commonly used in early operating systems but has been replaced by more efficient memory allocation techniques in modern systems.Rate this question:
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Mar 21, 2023Quiz Edited by
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Feb 03, 2017Quiz Created by
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