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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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A wood pole with a diameter of 10 in. has a moisture content of 5%. The fiber saturation point (FSP) for this wood is 30%. The w
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a) Δd(change in wood diameter) = 5%

b) The wood would swell since the moisture content is increasing which will also led to increase in the wood's diameter

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A) The percentage change in the wood's diameter

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