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3 years ago

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2. BCD uses 6 bits to represent a symbol. a) True b) False

1 answer:

Goshia [24]3 years ago

4 0

Answer:

true because BCD used 6 bits to represent a symbol .

Explanation:

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A slab-milling operation is performed on a 0.7 m long, 30 mm-wide cast-iron block with a feed of 0.25 mm/tooth and depth of cut

denis23 [38]

Answer:

a) $T_m=1.787min$

b) $MRR=35259.7mm^3/min$

Explanation:

From the question we are told that:

Cast-iron block Dimension:

Length $l=0.7m=>700mm$

Width $w=30mm$

Feed $F=0.25mm/tooth$

Depth $dp=3mm$

Diameter $d=75mm$

Number of cutting teeth $n=8$

Rotation speed $N=200rpm$

Generally the equation for Approach is mathematically given by

$x=\sqrt{Dd-d^2}$

$X=\sqrt{75*3-3^2}$

$X=14.69mm$

Therefore

Effective length is given as

$L_e=Approach +object Length$

$L_e=700+14.69$

$L_e=714.69mm$

a)

Generally the equation for Machine Time is mathematically given by

$T_m=\frac{L_e}{F_m}$

Where

$F_m=F*n*N$

$F_m=0.25*8*200$

$F_m=400$

Therefore

$T_m=\frac{714.69}{400}$

$T_m=1.787min$

b)

Generally the equation for Material Removal Rate. is mathematically given by

$MRR=\frac{L*B*d}{t_m}$

$MRR=\frac{700*30*3}{1.787}$

$MRR=35259.7mm^3/min$

3 0

3 years ago

What is Not considered as metric system ? a. Length b. Mass c. Time d. Volume e. Temperature

vampirchik [111]

Answer:

D.) Volume

Explanation:

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A 150 MVA, 24 kV, 123% three-phase synchronous generator supplies a large network. The network voltage is 27 kV. The phase angle

Aleks04 [339]

Answer:

the generator induced voltage is 60.59 kV

Explanation:

Given:

S = 150 MVA

Vline = 24 kV = 24000 V

$X_{s} =1.23(\frac{V_{line}^{2} }{s} )=1.23\frac{24000^{2} }{1500} =4723.2 ohms$

the network voltage phase is

$V_{phase} =\frac{V_{nline} }{\sqrt{3} } =\frac{27}{\sqrt{3} } =15.58kV$

the power transmitted is equal to:

$|E|=\frac{P*X_{s} }{3*|V_{phase}|sinO } ;if-O=60\\|E|=\frac{300*4.723}{3*15.58*sin60} =34.98kV$

the line induced voltage is

$|E_{line} |=\sqrt{3} *|E|=\sqrt{3} *34.98=60.59kV$

7 0

3 years ago

An equal-tangent sag vertical curve (with a negative initial and a positive final grade) is designed for 55 mi/h. The PVI is at

Varvara68 [4.7K]

Answer:

The lowest point of the curve is at 239+42.5 ft where elevation is 124.16 ft.

Explanation:

Length of curve is given as

$L=2(PVT-PVI)\\L=2(242+30-240+00)\\L=2(230)\\L=460 ft$

$G_2$ is given as

$G_2=\frac{E_{PVT}-E_{PVI}}{0.5L}\\G_2=\frac{127.5-122}{0.5*460}\\G_2=0.025=2.5 \%$

The K value is given from the table 3.3 for 55 mi/hr is 115. So the value of A is given as

$A=\frac{L}{K}\\A=\frac{460}{115}\\A=4$

A is given as

$-G_1=A-G_2\\-G_1=4.0-2.5\\-G_1=1.5\\G_1=-1.5\%$

With initial grade, the elevation of PVC is

$E_{PVC}=E_{PVI}+G_1(L/2)\\E_{PVC}=122+1.5%(460/2)\\E_{PVC}=125.45 ft\\$

The station is given as

$St_{PVC}=St_{PVI}-(L/2)\\St_{PVC}=24000-(230)\\St_{PVC}=237+70\\$

Low point is given as

$x=K \times |G_1|\\x=115 \times 1.5\\x=172.5 ft$

The station of low point is given as

$St_{low}=St_{PVC}-(x)\\St_{low}=23770+(172.5)\\St_{low}=239+42.5 ft\\$

The elevation is given as

$E_{low}=\frac{G_2-G_1}{2L} x^2+G_1x+E_{PVC}\\E_{low}=\frac{2.5-(-1.5)}{2*460} (1.72)^2+(-1.5)*(1.72)+125.45\\E_{low}=124.16 ft$

So the lowest point of the curve is at 239+42.5 ft where elevation is 124.16 ft.

3 0

3 years ago