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
Not sure what programming language, but i'll use Java
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"creates a weekly backup from a remote web server"
This implies that he is the owner of the server since, one, he backs it up every week, and two, he has access to create a backup.
Answer:
Following are the answer to this question:
Explanation:
A)
The memory size is 1 Giga Bytes which is equal to 
B)
calculating the register Bits:
C)
Immediate value size while merging the additional benefit with the address field:
The range is accomplished by dividing the bits by 2 into the two sides of the o and the number is one short to 0.
