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2 years ago

What are the elements of parallel computing

2 answers:

5 0

Answer:
Large problems can often be divided into smaller ones which can then be solved at the several different forms of parallels computing bit-level, instruction-level, data and task parellelism.

liraira [26]2 years ago

4 0

Elements of Parallel Computing:-

Computer systems organization. Dependable and fault-tolerant systems and networks.

Computing methodologies. Parallel computing methodologies. ...

General and reference. Cross-computing tools and techniques. ...

Networks. Network performance evaluation.

Software and its engineering. ...

Theory of computation.

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A four-lane divided multilane highway (two lanes in each direction) in rolling terrain has five access points per mile and 11-ft

grandymaker [24]

Answer:

peak-hour volume = 1890 veh/h

Explanation:

<u>Determine the peak-hour Volume </u>

Applying the equation below

Vp = v / ( PHF * N * Fg * Fdp ) -------------- ( 1 )

where :

Vp = 1250

v ( peak - hour volume ) = ?

PHF ( peak hour factor ) = 0.84

N = 2 lanes per direction

Fg ( grade adjustment for rolling terrain ) = 0.99 ≈ 1

Fdp = 0.90

<u>Back to equation 1 </u>

v = Vp ( PHF * N * Fg * Fdp )

= 1250 ( 0.84 * 2 * 1 * 0.90 )

= 1890 veh/h

5 0

3 years ago

Jack has been concerned about the rapidly changing green regulations in his state and his ability as a mechanical engineer to ke

uysha [10]

Answer:

Option A, B and D

Explanation:

Jack can easily convince boss if he focus around two major aspects of the company

a) Revenue enhancement - Jack must outline the benefits of his research that can be used to improvise customer offerings and hence can be further used to devise more energy-efficient options to customer

b) Reduction in mistakes - Issues such as poor implementation can be avoided with better approach and understanding.

Hence, option A, B and D are correct

3 0

3 years ago

In a semiconductor manufacturing process, three wafers from a lot are tested. Each wafer is classified as pass or fail. Assume t

gladu [14]

Answer:

F(x) = 0 ; x < 0

0.064 ; 0 ≤ x < 1

0.352 ; 1 ≤ x < 2

0.784 ; 2 ≤ x < 3

1 ; x ≥ 3

Explanation:

Each wafer is classified as pass or fail.

The wafers are independent.

Then, we can modelate X : ''Number of wafers that pass the test'' as a Binomial random variable.

X ~ Bi(n,p)

Where n = 3 and p = 0.6 is the success probability

The probatility function is given by :

$P(X=x)=f(x)=nCx.p^{x}.(1-p)^{n-x}$

Where $nCx$ is the combinatorial number

$nCx=\frac{n!}{x!(n-x)!}$

Let's calculate f(x) :

$f(0)=3C0.(0.6^{0}).(0.4^{3})=0.4^{3}=0.064$

$f(1)=3C1.(0.6^{1}).(0.4^{2})=0.288$

$f(2)=3C2.(0.6^{2}).(0.4^{1})=0.432$

$f(3)=3C3.(0.6^{3}).(0.4^{0})=0.6^{3}=0.216$

For the cumulative distribution function that we are looking for :

$P(X\leq x)=F(x)$

$F(0)=f(0)\\F(1)=f(0)+f(1)\\F(2)=f(0)+f(1)+f(2)\\F(3)=f(0)+f(1)+f(2)+f(3)=1$

$F(0)=0.064\\F(1)=0.064+0.288=0.352\\F(2)=0.064+0.288+0.432=0.784\\F(3)=0.064+0.288+0.432+0.216=1$

The cumulative distribution function for X is :

F(x) = 0 ; x < 0

0.064 ; 0 ≤ x < 1

0.352 ; 1 ≤ x < 2

0.784 ; 2 ≤ x < 3

1 ; x ≥ 3

5 0

3 years ago

Design roller chain drive. Specify the chain size, the sizes and number of teeth in the sprockets, the number of chain pitches,

Rom4ik [11]

Answer:

The total design can be summarized as follows:

AC motor 55 hp, 750 rpm as a driver

Service factor 1.0

Design power 5 hp

No. 40 chain, 0.50 in pitch, 1 strands

17 teeth, 2.72 in pitch dia, 1 strand small sprocket

39 teeth, 7.46 in pitch dia, 1 strand large sprocket

1 strand of length 54 in

Center distance of 19.9 in

Actual Output speed of 326.9 rpm

Type B lubrication

Explanation:

4 0

3 years ago

A long, circular aluminum rod is attached at one end to a heated wall and transfers heat by convection to a cold fluid.

Trava [24]

Answer:

a. Heat removal rate will increase

b. Heat removal rate will decrease

Explanation:

Given that

One end of rod is connected to the furnace and rod is long.So this rod can be treated as infinite long fin.

We know that heat transfer in fin given as follows

$Q_{fin}=\sqrt{hPKA}\ \Delta T$

We know that area

$A=\dfrac{\pi}{4}d^2$

Now when diameter will triples then :

$A_f=\dfrac{\pi}{4}{\left (3d \right )}^2$

$A_f=9A$

$Q'_{fin}=\sqrt{9hPKA}\ \Delta T$

$Q'_{fin}=3\sqrt{hPKA}\ \Delta T$

$Q'_{fin}=3Q$

So the new heat transfer will increase by 3 times.

Now when copper rod will replace by aluminium rod :

As we know that thermal conductivity(K) of Aluminium is low as compare to Copper .It means that heat transfer will decreases.

3 0

3 years ago