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2 years ago

What is 7-?=4 i need help

Engineering

2 answers:

aleksandr82 [10.1K]2 years ago

6 0

Answer:

Explanation:

7-3=4

SashulF [63]2 years ago

5 0

Answer:

Explanation:

Because when you do 7 - 3 it gives you 4. You check this by adding 4 to 3 and you will get 7. Also you can figure it out the number by subtracting 7 and 4 to get your answer.

Your answer will be 3.

I hope this helped. I am sorry if you get this wrong. Have a wonderful day.

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A long corridor has a single light bulb and two doors with light switch at each door. design logic circuit for the light; assume

sattari [20]

Answer and Explanation:

Let A denote its switch first after that we will assume B which denotes the next switch and then we will assume C stand for both the bulb. we assume 0 mean turn off while 1 mean turn on, too. The light is off, as both switches are in the same place. This may be illustrated with the below table of truth:

A B C (output)

0 0 0

0 1 1

1 0 1

1 1 0

The logic circuit is shown below

C = A'B + AB'

If the switches are in multiple places the bulb outcome will be on on the other hand if another switches are all in the same place, the result of the bulb will be off. This gate is XOR. The gate is shown in the diagram adjoining below.

3 0

2 years ago

Can I get a hand? Thanks y’all much luv

Lilit [14]

Answer:

Explanation:

3 0

2 years ago

Calculate the radius of gyration for a bar of rectangular cross section with thickness t and width w for bending in the directio

SVEN [57.7K]

Answer:

a)R= sqrt( wt³/12wt)

b)R=sqrt(tw³/12wt)

c)R= sqrt ( wt³/12xcos45xwt)

Explanation:

Thickness = t

Width = w

Length od diagonal =sqrt (t² +w²)

Area of raectangle = A= tW

Radius of gyration= r= sqrt( I/A)

Moment of inertia in the direction of thickness I = w t³/12

R= sqrt( wt³/12wt)

Moment of inertia in the direction of width I = t w³/12

R=sqrt(tw³/12wt)

Moment of inertia in the direction of diagonal I= (w t³/12)cos 45=( wt³/12)x 1/sqrt (2)

R= sqrt ( wt³/12xcos45xwt)

4 0

2 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

2 years ago

Three identical fatigue specimens (denoted A, B, and C) are fabricated from a nonferrous alloy. Each is subjected to one of the

Law Incorporation [45]

Answer:

B A and C

Explanation:

Given:

Specimen σ $_{max}$ σ $_{min}$

A +450 -150

B +300 -300

C +500 -200

Solution:

Compute the mean stress

σ $_{m}$ = (σ $_{max}$ + σ $_{min}$ )/2

σ $_{mA}$ = (450 + (-150)) / 2

= (450 - 150) / 2

= 300/2

σ $_{mA}$ = 150 MPa

σ $_{mB}$ = (300 + (-300))/2

= (300 - 300) / 2

= 0/2

σ $_{mB}$ = 0 MPa

σ $_{mC}$ = (500 + (-200))/2

= (500 - 200) / 2

= 300/2

σ $_{mC}$ = 150 MPa

Compute stress amplitude:

σ $_{a}$ = (σ $_{max}$ - σ $_{min}$ )/2

σ $_{aA}$ = (450 - (-150)) / 2

= (450 + 150) / 2

= 600/2

σ $_{aA}$ = 300 MPa

σ $_{aB}$ = (300- (-300)) / 2

= (300 + 300) / 2

= 600/2

σ $_{aB}$ = 300 MPa

σ $_{aC}$ = (500 - (-200))/2

= (500 + 200) / 2

= 700 / 2

σ $_{aC}$ = 350 MPa

From the above results it is concluded that the longest fatigue lifetime is of specimen B because it has the minimum mean stress.

Next, the specimen A has the fatigue lifetime which is shorter than B but longer than specimen C.

In the last comes specimen C which has the shortest fatigue lifetime because it has the higher mean stress and highest stress amplitude.

7 0

2 years ago