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Tresset [83]
3 years ago
8

Why is it better for a CPU to have more than one cache?

Engineering
1 answer:
Tomtit [17]3 years ago
3 0

Answer:

In general a cache memory is useful because the speed of the processor is higher than the speed of the ram . so reducing the number of memory is desirable to increase performance .

Explanation:

.

.

#hope it helps you ..

(◕ᴗ◕)

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Need answers for these please ​
Step2247 [10]

Answer:

Following are given the answers one-by-one with explanation:

(a) 8 records

As records are the horizontal rows of data and we have 8 of them.

(b) 5 fields

As fields are vertical columns of data and we have 5 of them.

(c)E3000

As by sorting Makers_names in ascending order the last one would be Rany and by sorting Processor_type we have two option with Rany that are B1500 and E3000. We can see that E3000 will come at the end.

(d) Computer_type

As only two choices (Laptop and PC )  are being used under the column computer_type so it can be amended to contain Boolean data.

(e) format check

This will check that the field should contain one alphabet (capital letter) followed by 4 digits each.

i hope it will help you!

7 0
3 years ago
Which of the following circumstances call for a greater than normal following distance?
mariarad [96]

Explanation:

The three-second rule is recommended for passenger vehicles during ideal road and weather conditions. Slow down and increase your following distance even more during adverse weather conditions or when visibility is reduced. Also increase your following distance if you are driving a larger vehicle or towing a trailer.

7 0
3 years ago
He is going ___ in the hot air ballon​
Vladimir [108]

no artical shoul be used here

5 0
3 years ago
Consider a steam turbine, with inflow at 500oC and 7.9 MPa. The machine has a total-to-static efficiency ofηts=0.91, and the pre
sergiy2304 [10]

Answer: \dot m_{in} = 23.942 \frac{kg}{s}, \dot H_{out} = 39632.62 kW

Explanation:

Since there is no information related to volume flow to and from turbine, let is assume that volume flow at inlet equals to \dot V = 1 \frac{m^{3}}{s}. Turbine is a steady-flow system modelled by using Principle of Mass Conservation and First Law of Thermodynamics:

Principle of Mass Conservation

\dot m_{in} - \dot m_{out} = 0

First Law of Thermodynamics

- \dot W_{out} + \eta\cdot (\dot m_{in} \dot h_{in} - \dot m_{out} \dot h_{out}) = 0

This 2 x 2 System can be reduced into one equation as follows:

-\dot W_{out} + \eta \cdot \dot m \cdot ( h_{in}- h_{out})=0

The water goes to the turbine as Superheated steam and goes out as saturated vapor or a liquid-vapor mix. Specific volume and specific enthalpy at inflow are required to determine specific enthalpy at outflow and mass flow rate, respectively. Property tables are a practical form to get information:

Inflow (Superheated Steam)

\nu_{in} = 0.041767 \frac{m^{3}}{kg} \\h_{in} = 3399.5 \frac{kJ}{kg}

The mass flow rate can be calculated by using this expression:

\dot m_{in} =\frac{\dot V_{in}}{\nu_{in}}

\dot m_{in} = 23.942 \frac{kg}{s}

Afterwards, the specific enthalpy at outflow is determined by isolating it from energy balance:

h_{out} =h_{in}-\frac{\dot W_{out}}{\eta \cdot \dot m}

h_{out} = 1655.36 \frac{kJ}{kg}

The enthalpy rate at outflow is:

\dot H_{out} = \dot m \cdot h_{out}

\dot H_{out} = 39632.62 kW

3 0
3 years ago
A rocket is launched vertically from rest with a constant thrust until the rocket reaches an altitude of 25 m and the thrust end
hjlf

Answer:

a) v=19.6 m/s

b) H=19.58 m

c) v_{f}=29.57 m/s  

Explanation:

a) Let's calculate the work done by the rocket until the thrust ends.

W=F_{tot}h=(F_{thrust}-mg)h=(35-(2*9.81))*25=384.5 J

But we know the work is equal to change of kinetic energy, so:

W=\Delta K=\frac{1}{2}mv^{2}

v=\sqrt{\frac{2W}{m}}=19.6 m/s

b) Here we have a free fall motion, because there is not external forces acting, that is way we can use the free-fall equations.

v_{f}^{2}=v_{i}^{2}-2gh

At the maximum height the velocity is 0, so v(f) = 0.

0=v_{i}^{2}-2gH

H=\frac{19.6^{2}}{2*9.81}=19.58 m  

c) Here we can evaluate the motion equation between the rocket at 25 m from the ground and the instant before the rocket touch the ground.

Using the same equation of part b)

v_{f}^{2}=v_{i}^{2}-2gh

v_{f}=\sqrt{19.6^{2}-(2*9.81*(-25))}=29.57 m/s

The minus sign of 25 means the zero of the reference system is at the pint when the thrust ends.

I hope it helps you!

6 0
3 years ago
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