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Rudiy27
3 years ago
11

Two infinite extent current sheets exist at z = −3.0 m and at z = +3.0 m. The top sheet has a uniform current

Engineering
1 answer:
Pachacha [2.7K]3 years ago
5 0

[ find the attachment ]

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What are the four categories of engineering materials used in manufacturing?
alexgriva [62]

Answer:

metals, composite, ceramics and polymers.

Explanation:

The four categories of engineering materials used in manufacturing are metals, composite, ceramics and polymers.

i) Metals: Metals are solids made up of atoms held by matrix of electrons. They are good conductors of heat and electricity, ductile and strong.

ii) Composite: This is a combination of two or more materials. They have high strength to weight ratio, stiff, low conductivity. E.g are wood, concrete.

iii) Ceramics: They are inorganic, non-metallic crystalline compounds with high hardness and strength as well as poor conductors of electricity and heat.

iv) Polymers: They  have low weight and are poor conductors of electricity and heat

8 0
3 years ago
Assume a program requires the execution of 50 x 106 FP instructions, 110 x 106 INT instructions, 80 x 106 L/S instructions, and
Pavlova-9 [17]

Answer:

Part A:

1.3568*10^{-5}=\frac{5300* New\  CPI_1+11660*1+8480*4+1696*2}{2*10^9\ Hz} \\ New\ CPI_1=-4.12

CPI cannot be negative so it is not possible to for program to run two times faster.

Part B:

1.3568*10^{-5}=\frac{5300*1+11660*1+8480*New\ CPI_3+1696*2}{2*10^9\ Hz} \\ New\ CPI_3=0.8

CPI reduced by 1-\frac{0.8}{4} = 0.80=80%

Part C:

New Execution Time=\frac{5300*0.6+11660*0.6+8480*2.8+1696*1.4}{2*10^9\ Hz}=1.81472*10^{-5}\ s

Increase in speed=1-\frac{1.81472*10^{-5}}{2.7136*10^{-5}} =0.33125= 33.125\%

Explanation:

FP Instructions=50*106=5300

INT  Instructions=110*106=11660

L/S  Instructions=80*106=8480

Branch  Instructions=16*106=1696

Calculating Execution Time:

Execution Time=\frac{\sum^4_{i=1} Number\ of\ Instruction*\ CPI_{i}}{Clock\ Rate}

Execution Time=\frac{5300*1+11660*1+8480*4+1696*2}{2*10^9\ Hz}

Execution Time=2.7136*10^{-5}\ s

Part A:

For Program to run two times faster,Execution Time (Calculated above) is reduced to half.

New Execution Time=\frac{2.7136*10^{-5}}{2}=1.3568*10^{-5}\ s

1.3568*10^{-5}=\frac{5300* New\  CPI_1+11660*1+8480*4+1696*2}{2*10^9\ Hz} \\ New\ CPI_1=-4.12

CPI cannot be negative so it is not possible to for program to run two times faster.

Part B:

For Program to run two times faster,Execution Time (Calculated above) is reduced to half.

New Execution Time=\frac{2.7136*10^{-5}}{2}=1.3568*10^{-5}\ s

1.3568*10^{-5}=\frac{5300*1+11660*1+8480*New\ CPI_3+1696*2}{2*10^9\ Hz} \\ New\ CPI_3=0.8

CPI reduced by 1-\frac{0.8}{4} = 0.80=80%

Part C:

New\ CPI_1=0.6*Old\ CPI_1=0.6*1=0.6\\New\ CPI_2=0.6*Old\ CPI_2=0.6*1=0.6\\New\ CPI_3=0.7*Old\ CPI_3=0.7*4=2.8\\New\ CPI_4=0.7*Old\ CPI_4=0.7*2=1.4

New Execution Time=\frac{\sum^4_{i=1} Number\ of\ Instruction*\ CPI_{i}}{Clock\ Rate}

New Execution Time=\frac{5300*0.6+11660*0.6+8480*2.8+1696*1.4}{2*10^9\ Hz}=1.81472*10^{-5}\ s

Increase in speed=1-\frac{1.81472*10^{-5}}{2.7136*10^{-5}} =0.33125= 33.125\%

8 0
3 years ago
River models are used to study many different types of flow situations. A certain small river has an average width and depth of
slavikrds [6]

Answer: 7ft x21 I’d be right but yes I am

Explanation: because it is Welty

4 0
3 years ago
The oil system is:
kirill [66]

Answer:

From the main bearings, the oil passes through feed-holes into drilled passages in the crankshaft and on to the big-end bearings of the connecting rod.

3 0
2 years ago
The Acme threading tool forms an inc luded angle of how many degrees? A. 30 B. 55 C. 29 D. 60
Schach [20]

Answer:

(C) 29°

Explanation:

ACME THREAD ANGLE : it is the angle measured between the thread axis and thread the thread flanks, with the help of thread angle we calculate the shape of screw thread it is the mean of v thread and square thread  the acme thread angle is denoted by β every thread has its own charactersistics which depends on the pitch and diameter of the thread

4 0
3 years ago
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