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Katena32 [7]
3 years ago
12

a. Do you think the U.S. government should censor information on the Web, such as instructions for making weapons, to protect th

e general public? Why or why not? If you think there should be some censorship, do you think such a law would violate the First Amendment right to free speech? Explain your answer.
Computers and Technology
1 answer:
MrRissso [65]3 years ago
5 0

Answer:

The answer is YES.

Explanation:

<em>Things pertaining to weapons, destructive materials should be shared with precautions or much better if these things should not be put online. Since internet is widely available to almost all types of users and almost to all places, we cannot be sure of who will get hold of these information. Kids could easily see this and gain knowledge and out of curiosity will try these dangerous things. Terrorist groups will also have access to these information and who knows might happen to poor innocent souls around.  </em>

<em> </em>

<em>I don't think it would violate the right of First Amendment regarding freedom of speech. Freedom should be granted on matters that are most sensible. And this time, we are just protecting the rights of children and other innocent people who can be a victim including you. We are talking about weapons and destructive materials here which is non beneficial if one could get hold of this information, in my opinion. If there is really a need to construct these things, it is better to consult an expert or an authority to ensure that nobody gets harm.</em>  

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Interest

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Read 2 more answers
1. How many bits would you need to address a 2M × 32 memory if:
Dominik [7]

Answer:

  1. a) 23       b) 21
  2. a) 43        b) 42
  3. a) 0          b) 0

Explanation:

<u>1) How many bits is needed to address a 2M * 32 memory </u>

2M = 2^1*2^20, while item =32 bit long word

hence ; L = 2^21 ; w = 32

a) when the memory is byte addressable

w = 8;  L = ( 2M * 32 ) / 8 =  2M * 4

hence number of bits =  log2(2M * 4)= log2 ( 2 * 2^20 * 2^2 ) = 23 bits

b) when the memory is word addressable

W = 32 ; L = ( 2M * 32 )/ 32 = 2M

hence the number of bits = log2 ( 2M ) = Log2 (2 * 2^20 ) = 21 bits

<u>2) How many bits are required to address a 4M × 16 main memory</u>

4M = 4^1*4^20 while item = 16 bit long word

hence L ( length ) = 4^21 ; w = 16

a) when the memory is byte addressable

w = 8 ; L = ( 4M * 16 ) / 8 = 4M * 2

hence number of bits = log 2 ( 4M * 2 ) = log 2 ( 4^1*4^20*2^1 ) ≈ 43 bits

b) when the memory is word addressable

w = 16 ; L = ( 4M * 16 ) / 16 = 4M

hence number of bits = log 2 ( 4M ) = log2 ( 4^1*4^20 ) ≈ 42 bits

<u>3) How many bits are required to address a 1M * 8 main memory </u>

1M = 1^1 * 1^20 ,  item = 8

L = 1^21 ; w = 8

a) when the memory is byte addressable

w = 8 ; L = ( 1 M * 8 ) / 8 = 1M

hence number of bits = log 2 ( 1M ) = log2 ( 1^1 * 1^20 ) = 0 bit

b) when memory is word addressable

w = 8 ; L = ( 1 M * 8 ) / 8 = 1M

number of bits = 0

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