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GREYUIT [131]
3 years ago
15

NIST 800-14's Principles for Securing Information Technology Systems, can be used to make sure the needed key elements of a succ

essful effort are factored into the design of an information security program and to produce a blueprint for an effective security architecture.a) trueb) false
Physics
1 answer:
mariarad [96]3 years ago
5 0

Answer:

True

Explanation:

-NIST 800-14's are generally accepted principles for securing information technology systems.

- The defined principles, if adhered to and continously improved, will will ensure sytem security over it's lifetime as desired.

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Write a hypothesis about how the number of half-lives affects the number of radioactive atoms. Use the "if . . . then . . . beca
Roman55 [17]

Answer:

Explanation:

Answer:

Explanation:

The half life is the time taken for half of a radioactive substance to disintegrate.

The shorter the half life, the larger the decay constant and the faster the decay process.

For a very large half life, it would take a very long time for the radioactive nuclide to decay to half.

With each half life reached, a new set of daughter cell is formed. Atoms that have short half life would decay rapidly. Every radionuclide has its own characteristic half-life.

If the number of half-lives increases, then the number of radioactive atoms decreases, because approximately half of the atoms' nuclei decay with each half-life. With this observation, we can hypothesise and conduct experiment to support the assertion that as the number of half-lives increases then the number of radioactive atoms decreases.

8 0
4 years ago
Read 2 more answers
Why is the monolith el capitan in yosemite national park more resistant to erosion than other igneous plutons in the park?
Harman [31]
Because it is intact and unfractured
4 0
3 years ago
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evablogger [386]

Answer:

b

Explanation:

3 0
3 years ago
Just wanna make sure im right
NikAS [45]
It is right have a good day

explanation step by step
6 0
3 years ago
What is the change in length of a 1400. m steel, (12x10^-6)/(C0) , pipe for a temperature change of 250.0 degrees Celsius? Remem
8090 [49]

Answer:

\Delta L = 4.2 m

Explanation:

As per the formula of thermal expansion we know that

L = L_o(1 + \alpha\Delta T)

so here we will have

L_o = 1400 m

\alpha = 12 \times 10^{-6} per ^oC

\Delta T = 250 degree C

so here change in the length of the rod is given as

\Delta L = L - L_o

\Delta L = L_o \alpha \Delta T

\Delta L = 1400 (12 \times 10^{-6})(250)

\Delta L = 4.2 m

8 0
4 years ago
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