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3 years ago

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14. A digital computer has a memory unit with 32 bits per word. The instruction set consists of 110 different operations. All in

structions have an operation code part (opcode) and two address fields: one for a memory address and one for a register address. This particular system includes eight general-purpose, user-addressable registers. Registers may be loaded directly from memory, and memory may be updated directly from the registers. Direct memory-to-memory data movement operations are not supported. Each instruction is stored in one word of memory.
How many bits are needed for the opcode?

1 answer:

Fofino [41]3 years ago

5 0

Answer:

7 bits

Explanation:

Given

Instruction Set = 110 operation

Memory unit = 32 bits per word.

We get the required bits by using the following formula

2^n = 110

But 110 is not a factor of 2.

So, we pick the nearest decimal number greater than 110 that is a power of 2.

The number is 128

2^n = 110 becomes

2^n = 128

2^n = 2^7 ---- 2 cancels out

So,

n = 7

Hence, the required number of bits needed for the opcode is 7 bits

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A system has a level 1 cache and a level 2 cache. The hit rate of the level 1 cache is 90% and the hit rate of the level 2 cache

Hatshy [7]

Answer:

B) 2.22

Explanation:

In a first step, the system will look in cache number 1. If it is not found in cache number 1, then the system will look in cache number 2. Finally (if not in cache 2 also) the system will look in main memory.

The average access time will take into consideration success in cache 1, failure in cache number 1 but success in cache number 2, failure in cache number 1 and 2 but success in main memory.

Mathematically we can write it into the following equation :

$AAT=(H1.T1)+(1-H1).H2.T2+(1-H1).(1-H2).Hm.Tm$

Where AAT is the average access time

H1,H2 and Hm are the hit rate of cache 1,cache 2 and main memory respectively.

T1,T2 and Tm are the access time of cache 1, cache 2 and main memory respectively

Hm = 1

$AAT=(0.9).(1)+(1-0.9).(0.8).(4)+(1-0.9).(1-0.8).(1).(50)=0.9+0.32+1=2.22$

$AAT=2.22$

Therefore, option b) is the correct.

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The wall of drying oven is constructed by sandwiching insulation material of thermal conductivity k = 0.05 W/m°K between thin me

masha68 [24]

Answer:

86 mm

Explanation:

From the attached thermal circuit diagram, equation for i-nodes will be

$\frac {T_ \infty, i-T_{i}}{ R^{"}_{cv, i}} + \frac {T_{o}-T_{i}}{ R^{"}_{cd}} + q_{rad} = 0$ Equation 1

Similarly, the equation for outer node “o” will be

$\frac {T_{ i}-T_{o}}{ R^{"}_{cd}} + \frac {T_{\infty, o} -T_{o}}{ R^{"}_{cv, o}} = 0$ Equation 2

The conventive thermal resistance in i-node will be

$R^{"}_{cv, i}= \frac {1}{h_{i}}= \frac {1}{30}= 0.033 m^{2}K/w$ Equation 3

The conventive hermal resistance per unit area is

$R^{"}_{cv, o}= \frac {1}{h_{o}}= \frac {1}{10}= 0.100 m^{2}K/w$ Equation 4

The conductive thermal resistance per unit area is

$R^{"}_{cd}= \frac {L}{K}= \frac {L}{0.05} m^{2}K/w$ Equation 5

Since $q_{rad}$ is given as 100, $T_{o}$ is 40 $T_ \infty$ is 300 $T_{\infty, o}$ is 25

Substituting the values in equations 3,4 and 5 into equations 1 and 2 we obtain

$\frac {300-T_{i}}{0.033} +\frac {40-T_{i}}{L/0.05} +100=0$ Equation 6

$\frac {T_{ i}-40}{L/0.05}+ \frac {25-40}{0.100}=0$

$T_{i}-40= \frac {L}{0.05}*150$

$T_{i}-40=3000L$

$T_{i}=3000L+40$ Equation 7

From equation 6 we can substitute wherever there’s $T_{i}$ with 3000L+40 as seen in equation 7 hence we obtain

$\frac {300- (3000L+40)}{0.033} + \frac {40- (3000L+40)}{L/0.05}+100=0$

The above can be simplified to be

$\frac {260-3000L}{0.033}+ \frac {(-3000L)}{L/0.05}+100=0$

$\frac {260-3000L}{0.033}=50$

-3000L=1.665-260

$L= \frac {-258.33}{-3000}=0.086*10^{-3}m= 86mm$

Therefore, insulation thickness is 86mm

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An automobile has a mass of 1200 kg. What is its kinetic energy, in ki, relative to the road when traveling at a velocity of 50

netineya [11]

Answer:

a)Ek=115759.26J

b)23.15kW

Explanation:

kinetic energy(Ek) is understood as that energy that has a body with mass when it travels with a certain speed, it is calculated by the following equation

$Ek=0.5mv^{2}$

for the first part

m=1200Kg

V=50km/h=13.89m/s

solving for Ek

$Ek=0.5(1200)(13.89)^2$

Ek=115759.26J

For the second part of the problem we calculate the kinetic energy, using the same formula, then in order to find the energy change we find the difference between the two, finally divide over time (15s) to find the power.

m=1200kg

V=100km/h=27.78m/s

$Ek=0.5(1200)(27.78)^2=462963J\\$

taking into account all of the above the following equation is inferred

ΔE= $\frac{Ek2-EK1}{T}=\frac{462963-115759.26}{15} =23146.916W=23.15kW$

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