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Alexxandr [17]
1 year ago
11

Five programs are currently being run in a computer. Program 1 is using 10 GiB of RAM, program 2 is using 5 GiB of RAM, program

3 is using 12 GiB of RAM and program 4 is using 4 GiB of RAM. The programs are at the stage where program 5 now needs to access RAM, but RAM is presently full (RAM has a 32 GiB maximum capacity). Explain how virtual memory could be used to allow program 5 to access RAM without any of the data from the other four programs being lost.
Computers and Technology
1 answer:
muminat1 year ago
6 0

Virtual memory could be used to allow program 5 to access RAM without any of the data from the other four programs being lost because it is one that tend to allows the system to give all of the process its own memory space that is said to be  isolated from the other processes.

<h3>How is virtual memory used instead of RAM?</h3>

A system is known to make use of a virtual memory and this is one that tend to make use of a section of the hard drive to act like the RAM.

With the use of virtual memory, a system can be able to load bigger or a lot of programs running at the same time, and this is one that tends to hep one to work as if it has more space, without having to buy more RAM.

Therefore, Virtual memory could be used to allow program 5 to access RAM without any of the data from the other four programs being lost because it is one that tend to allows the system to give all of the process its own memory space that is said to be  isolated from the other processes.

Learn more about virtual memory from

brainly.com/question/13088640

#SPJ1

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Se tiene un pilar de hormigón de 30 cm de lado y 5 m de altura. Presenta una densidad de 2400 kg/m3 y una resistencia a la compr
kakasveta [241]

Answer:

El pilar no podrá soportar una masa de 10 toneladas

Explanation:

La dimensión del pilar de hormigón se da de la siguiente manera;

La longitud del lado, s = 30 cm = 0,3 m

La altura del pilar, h = 5 m

La densidad del pilar, ρ = 2,400 kg / m³

La resistencia a la compresión del pilar, σ = 500 kg / m²

El área de la sección transversal del pilar, A = s² = 0.3 m × 0.3 m = 0.09 m²

La masa del pilar, m = ρ × A × h = 2,400 × 0.09 × 5 = 1,080

La masa del pilar, m = 1.080 kg

Tenemos;

\sigma = \dfrac{F}{A}

Dónde;

F = La carga aplicada

A = El área de la sección transversal

∴ F = A × σ

F = 0,09 m² × 500 kg / m² = 45 kg

Por tanto, la carga que el pilar puede soportar sin compresión = 45 kg <10 toneladas = 9.071,847 kg

El pilar no podrá soportar una masa de 10 toneladas.

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3 years ago
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The current calendar, called the Gregorian calendar, was introduced in 1582. Every year divisible by four was declared to be a l
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Solution :

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End Function

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