It seems like <span>xK6D(#dsswuJD would be best because it has several "special characters" as well as capitals, lower cases, and numbers.</span><span /><span>
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Metaphors are considered in a presentation as a statement that considers one idea in terms of another. This literary use has a deeper function as it entertains and motivates our minds. Therefore, metaphors are a great tool in presentations. They help with simplifying ideas, and make concepts easier to understand. Metaphors bring a creative touch to the content of the presentation, as well. Especially in multimedia, where you have visual metaphors to stir up the audiences senses t the content being presented.
The answer is <span> tunnel mode encryption. This</span><span> describes encryption that protects the entire original ip packet's header and payload. A tunnel mode encryption </span><span>protects the internal routing information by encrypting the IP header of the original packet. The original packet is encapsulated by a another set of IP headers. .<span>Additional headers are added to the packet; so the payload MSS is less.</span></span>
Answer:
2.342m
Explanation:
Given
Time = 0.5 s
Height of Window = 2m
Because the pot was in view for a total of 0.5 seconds, we can assume that it took the cat 0.25 seconds to go from the bottom of the window to the top
Using this equation of motion
S = ut - ½gt²
Where s = 2
u = initial velocity = ?
t = 0.25
g = 9.8
So, we have.
2 = u * 0.25 - ½ * 9.8 * 0.25²
2 = 0.25u - 0.30625
2 + 0.30625 = 0.25u
2.30625 = 0.25u
u = 2.30625/0.25
u = 9.225 m/s ------------ the speed at the bottom of the pot
Using
v² = u² + 2gs to calculate the height above the window
Where v = final velocity = 0
u = 9.225
g = 9.8
S = height above the window
So, we have
0² = 9.225² - 2 * 9.8 * s
0 = 85.100625 - 19.6s
-85.100625 = -19.6s
S = -85.100625/19.6
S = 4.342
If 4.342m is the height above the window and the window is 2m high
Then 4.342 - 2 is the distance above the window
4.342 - 2 = 2.342m
Answer:
Given address = 94EA6
tag = 0 * 94 ( 10010100 )
line = 0 * 1 D 4 ( 111010100 )
word position = 0*6 ( 110 )
Explanation:
using the direct mapping method
Number of lines = 512
block size = 8 words
word offset = 
= 3 bit
index bit = 
= 9 bit
Tag = 20 - ( index bit + word offset ) = 20 - ( 3+9) = 8 bit
Given address = 94EA6
tag = 0 * 94 ( 10010100 )
line = 0 * 1 D 4 ( 111010100 )
word position = 0*6 ( 110 )
