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ella [17]
3 years ago
11

With 64 KB of memory and 8 bits in each memory location, how wide should the address bus be to access all 64 KB of memory? (k =

kilo = 1000, whereas K = 2^10 = 1024)
Engineering
1 answer:
marishachu [46]3 years ago
3 0

Answer:

16-bit wide

Explanation:

In order to find the width of the address bus, we need first to know how many memory cells it is needed to address.

If the size memory is 64 KB, this means that the memory size, in bytes, is equal to the following quantity:

64 KB = 2⁶ * 2¹⁰ bytes = 2¹⁶ bytes.

In order to address this quantity of cell positions, the address bus must be able to address 2¹⁶ bytes, so it must have 16-bit wide.

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4 0
1 year ago
An iron-carbon alloy initially containing 0.286 wt% C is exposed to an oxygen-rich and virtually carbon-free atmosphere at 1200°
Fantom [35]

Answer:

Explanation:

Given data:

initial construction co = 0.286 wt %

concentration at surface position cs = 0 wt %

carbon concentration cx = 0.215 wt%

time = 7 hr

D =  7.5 \times 10^{-11} m^2/s

for 0.225% carbon concentration following formula is used

\frac{cx -co}{cs -co} = 1 - erf(\frac{x}{2\sqrt{DT}})

where, erf stand for error function

\frac{cx -co}{cs -co} = \frac{0.215 -0.286}{0 -0.286} =0.248

0.248 = 1 - erf(\frac{x}{2\sqrt{DT}})

erf(\frac{x}{2\sqrt{DT}}) = 1 - 0.248

erf(\frac{x}{2\sqrt{DT}}) = 0.751

from the table erf(Z) value = 0.751 lie between (z) = 0.80 and z = 0.85 so by inteerpolation we have z = 0.815

from given table

\frac{x}{2\sqrt{DT}} = 0.815

x = 2\times 0.815 \times \sqrt{7.5 \times 10^{-11}\times (7\times 3600)

x = 2.39\times 10^{-3} m

x = 0.002395 mm

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3 years ago
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A communication systems

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