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stealth61 [152]
3 years ago
11

When cleaning a firearm, what end of the firearm should you generally clean from?

Computers and Technology
2 answers:
weqwewe [10]3 years ago
8 0

Answer:

Explanation:

You should always clean from the breech (chamber) end. On firearms that you don't have access to the breech, you should put the cleaning rod through the muzzle to the breech. Then place the cleaning patch or brush on the end of the rod. Then PULL the rod through the barrel and out of the muzzle.

tester [92]3 years ago
8 0
The äss end of the gun
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Since the rules cannot address all circumstances, the Code includes a conceptual framework approach for members to use to evalua
Veseljchak [2.6K]

Answer:

more than one safeguard may be necessary.

Explanation:

The conceptual framework can be used to developed as well as construct through a process of the qualitative analysis. The approach includes the in the frameworks for identifying and evaluating the threats to compliance with the rules.

But since the rules formed cannot always address all the circumstances, the Code includes to evaluate the threats to the compliance of more than one safeguards that are necessary.

8 0
3 years ago
Computer system with a 32-bit logical address and 4k-byte page size. assume that each entry of a page table consists of 4bytes.
Annette [7]

The number of bits in the physical address is 26 bits. The number of entries in a page table is \mathbf{2^{20}} entries. The size of the page table in a one-level paging scheme is 4MB.

<h3>
What is paging in Operating System?</h3>

Paging is a storage method used in operating systems to recover activities as pages from secondary storage and place them in primary memory. The basic purpose of pagination is to separate each procedure into pages.

We are given the following parameters:

  • 32-bit logical address
  • Page size = 4 KB = \mathbf{2^{12} \ bytes }
  • Size of each Page table entry = 4 bytes

Suppose the system supports physical memory size = 64 MB = \mathbf{2^{26} \ bytes}

Thus, the number of bits in the physical address is computed as:

= \mathbf{log_2 \{Physical-memory-size\}}

=\mathbf{log_2(2^{26})}}

= 26 bits

The number of entries in a page table = logical address space size/page size

The number of entries in a page table \mathbf{= \dfrac{2^{32}}{2^{12}}}

\mathbf{=2^{20}} entries

In a one-level paging scheme, the size of the table is:

= entire no. of page entries × page table size

= \mathbf{2^{20}\times 4 \ bytes}

= 4 MB

b.

suppose that this system supports up to 2^30 bytes of physical memory.

  • The size of the page table will be the same as 4 MB due to the fact that the number of entries, as well as, the page table entry size is the same.

Since the size of the page table surpasses that of a single page. A page cannot include a whole page table. Therefore, the page table must be broken into parts to fit onto numerous pages, and an additional level of the page table is required to access this page table.

  • This is called the Multi-Level Paging system.

Therefore, we can conclude that the number of bits in the physical address is 26 bits, the number of entries in a page table is \mathbf{2^{20}} entries, and the size of the page table in a one-level paging scheme is 4MB.

Learn more about Paging in Operating System here:

brainly.com/question/17004314

#SPJ1

4 0
2 years ago
Point giveaway and brainliest
melamori03 [73]

Thank you, pal!

You are invited to my clubhouse!

5 0
3 years ago
Read 2 more answers
In this lab, 172.30.0.0 represents the __________ network and 10.20.1.0 represents the _________ network.
umka21 [38]
172.30.0.0: private network
10.20.1.0: public network
6 0
3 years ago
Write a program that accepts a time as an hour and minute. Add 15 minutes to the time
Anastaziya [24]

Answer:

The code is given below

hours = int(input("Enter time in hour: "))

minutes = int(input("Enter time in minute: "))

total time = (hours * 60) + (minutes + 15 )

total hours = int(total minutes  / 60)

minutes  = total hours/ 60

print("Hours: " + str(hours))

print("Minutes: " + str(minutes))

5 0
3 years ago
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