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Lelechka [254]
3 years ago
10

"A computer architect redesigns the pipeline above to enable branch prediction. When PCSrc is asserted (branch taken) IF/ID is f

lushed, and speculative instructions are flushed by deasserting some of their control signals. No other modifications are made. Which control signals of speculative instructions must be deasserted when PCSrc is asserted, to ensure correct operation

Engineering
1 answer:
Sindrei [870]3 years ago
3 0

Answer:

Explanation:

Find attach the solution

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One gram of Strontium-90 has an activity of 5.3 terabecquerels (TBq), what will be the activity of 1 microgram?
noname [10]

1 micro gram of Strontium-90 has an activity of

0.0000053 terabecquerels (TBq),

Explanation:

Given information denotes that .,one gram of Strontium-90 has an activity of 5.3 terabecquerels (TBq)

the activity of 1 micro gram is

1 gram = 1,000,000 micro gram has activities of 5.3 terabecquerels

therefore 1 micro gram has the activity of (5.3 ÷  1,000,000 = 0.0000053 )

= (5.3 ÷  1,000,000 = 0.0000053 )

Hence ., 1 micro gram of Strontium-90 has an activity of

0.0000053 terabecquerels (TBq),

8 0
3 years ago
Showing or hiding records in a database is called “filtering.”<br> True<br> False
agasfer [191]

Answer:

TRUE

Explanation:

4 0
3 years ago
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Milk has a density of as much as 64.6 lb/ft3. What is the gage pressure at the bottom of the straw 6.1 inches deep in the milk?
gregori [183]

Answer:

Explanation:

1 inch is 0.0833333feet

6.1 inches is 0.5083 feet

Density = mass/volume

64.6 = mass/0.50833

mass = 64.6 x 0.5083 =32.83618lb

3 0
3 years ago
If i eat myself will I get twice as big or disappear completely?
Butoxors [25]
Disappear completely
5 0
3 years ago
Water is the working fluid in an ideal Rankine cycle. Saturated vapor enters the turbine at 12 MPa, and the condenser pressure i
Brilliant_brown [7]

Answer:

\dot Q_{in} = 372.239\,MW

Explanation:

The water enters to the pump as saturated liquid and equation is modelled after the First Law of Thermodynamics:

w_{in} + h_{in}- h_{out} = 0

h_{out} = w_{in}+h_{in}

h_{out} = 12\,\frac{kJ}{kg} + 191.81\,\frac{kJ}{kg}

h_{out} = 203.81\,\frac{kJ}{kg}

The boiler heats the water to the state of saturated vapor, whose specific enthalpy is:

h_{out} = 2685.4\,\frac{kJ}{kg}

The rate of heat transfer in the boiler is:

\dot Q_{in} = \left(150\,\frac{kg}{s}\right)\cdot \left(2685.4\,\frac{kJ}{kg}-203.81\,\frac{kJ}{kg} \right)\cdot \left(\frac{1\,MW}{1000\,kW} \right)

\dot Q_{in} = 372.239\,MW

3 0
3 years ago
Read 2 more answers
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