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IrinaVladis [17]
4 years ago
6

g Define the term Instruction Set Architecture, and answer each of the following questions about ISA styles: A. Tell what the ac

ronyms RISC and CISC stand for. B. Describe two characteristics that differentiate between these two ISA styles. C. Name one commercial processor (or processor family) that is a member of each style. D. Which style does the ARC processor used in the book use?
Engineering
1 answer:
allochka39001 [22]4 years ago
5 0

Answer:

Explanation:

a

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6. Question
valkas [14]

Answer:

Check  the 2nd, 3rd and 4th statements.

Explanation:

4 0
3 years ago
A model of a submarine, 1:15 scale, is to be tested at 180 ft/s in a wind tunnel with standard sea-level air, while theprototype
ella [17]

Answer:

Explanation:

Given

scale i.e. L_r=1:15

Using Reynolds number similarity

(Re)_m=(Re)_p

(\frac{Vl}{\nu })_m=(\frac{Vl}{\nu })_p

Properties of air

\nu _{air}=1.57\times 10^{-4} ft/s

Properties of sea water

\nu _{sea}=1.26\times 10^{-5} ft/s

\left ( \frac{V_ml_M}{\nu _m}\right )=\left ( \frac{V_pl_p}{\nu _p}\right )

V_p=V_m\left ( \frac{l_m}{l_p}\right )\left ( \frac{\nu _p}{\nu _m}\right )

V_p=180\times \frac{1}{15}\times \frac{1.26\times 10^{-5}}{1.57\times 10^{-4}}

V_p=0.963\ ft/s

8 0
4 years ago
Please follow meee ​
vitfil [10]

Answer:

wheeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeee

3 0
3 years ago
While walking across campus one windy day, an engineering student speculates about using an umbrella as a "sail" to propel a bic
makvit [3.9K]

Answer:

Given data:\\While walking across campus one windy day\\Frontal area, \(A=0.3 m ^{2}\)\\Wind speed \(V=24 Km / hr\)\\The drag coefficient \(C_{D, b}=1.2\)\\The combined mass \(m=75 kg\)\\Umbrella diameter, \(D=1.22 m\)\\Velocity of wind \(V=24 \frac{ km }{ hr }\)\\The rolling resistance \(C_{R}=0.75 \%\)

Solution:

Note: Refer the diagram

Basic equation:\\'s law of motion: \(\sum F_{x}=m a_{x}\)\\Lift coefficient, \(C_{L}=\frac{F_{L}}{\frac{1}{2} \rho V^{2} A_{p}}\)\\Drag coefficient, \(C_{D}=\frac{F_{D}}{\frac{1}{2} \rho V^{2} A_{p}}\)

From force balance equation:\\\(\sum F_{x}=F_{D}-F_{R}=0\)\\But \(F_{D}=\left(C_{D, \alpha} A_{u}+C_{D, B} A_{b}\right) \frac{1}{2} \rho\left(V_{\nu}-V_{b}\right)^{2}\)\(F_{R}=C_{R} m g\)\\Area of the Umbrella \(A_{u}=\frac{\pi D_{u}^{2}}{4}\)\(A_{x}=\frac{\pi \times 1.22^{2}}{4} m ^{2}\)\(A_{v}=1.17 m ^{2}\)

Drag coefficient data for selected objects table at

Hemisphere (open end facing flow), C_{D, x}=1.42

Substituting all parameters,

\begin{aligned}&F_{R}=0.0075 \times 75 \times 9.81\\&F_{R}=5.52 N\end{aligned}

Then,

\begin{aligned}&V_{b}=V_{w}-\left[\frac{2 F_{R}}{\rho\left(C_{D, w} A_{w}+C_{D, B} A_{b}\right)}\right]^{\frac{1}{2}} \dots\\&V_{w}=24 \times 1000 \times \frac{1}{3600}\\&V_{w}=6.67 \frac{ m }{ s }\end{aligned}

And the equation becomes,

\begin{aligned}&V_{b}=6.67-\left[\frac{2 \times 5.52}{1.23(1.42 \times 1.17+1.2 \times 0.3)}\right]^{\frac{1}{2}}\\&V_{b}=6.67-2.11\\&V_{b}=4.56 \frac{ m }{ s }\end{aligned}

Thus the floyds travels at 68.3^{\circ}wind speed.

7 0
4 years ago
Eigth people work in an office.they are paid hourly rates of £12 £15 £15 £14 £13 £14 £13 £13
jok3333 [9.3K]

Answer:

What about it?

Explanation:

8 0
2 years ago
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