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ZanzabumX [31]
3 years ago
13

A metal with a BCC structure, such as iron, usually exhibits which mechanical property?

Engineering
1 answer:
elena-s [515]3 years ago
6 0

Answer:

C can i have brainliest pleaseee

Explanation:

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The purpose of adjusting your mirrors is to _________.
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Its c . to reduce blind spots
7 0
3 years ago
4.7 If the maximum tensile force in any of the truss members must be limited to 22 kN, and the maximum compressive force must be
ki77a [65]

Answer:

588.55 kg

attached below is a sketch of the members of the truss

Explanation:

Given data:

Maximum tensile force = 22 KN

maximum compressive force = 20 KN

To determine the largest permissible mass m which may be supported by the truss

First we have have to find the summation of forces in the Y and X direction along Joint C

<u>Summation of forces in the Y direction</u>

∑Fy = 0

= BC sin30° - W = 0.   hence BC = 2WT

<u>summation of forces in the X direction </u>

∑Fx = 0

= -CD - BC cos30°

therefore CD = -2W cos30°

hence ; CD = - 1.732W

ignoring the compression on CD. therefore CD = 1.732W C

Next we have to analyze Joint B

summation of forces in the X direction

∑Fx = 0

= - AB + BC cos30° = 0

therefore AB = BC cos30°

                AB = 1.732W T

∑Fy = 0

- BD - BC sin30° = 0

therefore BD = -W = W C

Analyzing Joint D

∑Fy = 0

AD sin30° + BD = 0

hence ;AD = 2W T

∑Fx = 0

-DE - ADcos30° + CD = 0

hence DE = -3.464W  =  3.464W C

At Joint E

∑Fy = 0

i.e. AE = 0

from the above analysis The tensile force is greatest on members AD and BC

AD = BC = 2W

from the question the maximum tensile force in any of the truss members = 22KN

Hence ;

2W = 22KN

2mg = 22000N

m = 22000 / ( 2 * 9.81 )

   = 1121.30 kg

from the above analysis the greatest compressive force is found in DE which is the critical part of the Truss hence the maximum mass it can carry is the largest permissible mass which may be supported by the Truss

DE = 3.464 W

from the question the maximum compressive force = 20 KN

hence ;

3.464 W = 20KN

3.464 mg = 20000N

m = 20000 / ( 3.464 * 9.81 )

m = 588.55 kg

4 0
3 years ago
A balanced three phase load is supplied over a three-phase , 60 hz, transmission line with each line have a series impedance of
posledela

Answer:

Explanation:

Given a three-phase system

Frequency f=60Hz

Line impedance Z= 12.84 + j72.76 Ω

Then,

The resistance is R=12.84Ω

And reactance is X=72.76Ω

Z=√(12.84²+72.76²)

Z=73.88

Angle = arctan(X/R)

Angle = arctan(72.76/12.84)

Angle=80°

Then, Z=73.88 < 80° ohms

Load voltage is 132 kV

Load power P=55 MWA

Power factor =0.8lagging

the relation between the sending and receiving end specifications are given using ABCD parameters by the equations below.

Vs = AVr + BIr

Is = CVr + DIr

Where

Vs is sending Voltage

Vr is receiving Voltage

Is is sending current

Ir is receiving current

A is ratio of source voltage to received voltage A=Vs/Vr when Ir=0

B is short circuit resistance

B= Vs/Ir when Vr=0

C is ratio of source current to received voltage C=Is/Vr when Ir=0

D is ratio of source current to received current D=Is/Ir when Vr=0

Now,

The load at 55MVA at 132kV (line to line)

Therefore, load current is

Ir= P/V√3

Ir=55×10^6/(132×10^3×√3)

Ir=240.56 Amps

It has a power factor 0.8 lagging

PF=Cosθ

0.8=Cosθ

θ=arcCos(0.8)

θ=36.87°

Therefore, Ir=240.56 <-36.87°

Vr=V/√3

Vr=132/√3

Vr=76.21 kV. Phase voltage

Vr= 76210 < 0° V

For series impedance,

Using short line approximation

Vs = Vr + IrZ

Vs = 76210 < 0° + (240.56 <-36.87° × 73.88 < 80°)

Using calculator

Vs=76210<0° + 17772.5728<(-36.87°+80°)

Vs=76210<0° + 17772.5728<43.13°

Vs=89970.67<7.7°

Also

Is = Ir = 240.56 <-36.87° Amps

Therefore, the ABCD parameters is

A=Vs/Vr

A= 89970.67 <7.7° / 76210 <0°

A=1.181 <7.7-0

A=1.18 <7.7° no unit

B = Vs/Ir

B = 89970.67 < 7.7° / 240.56 <-36.87°

B = 347.01 < 7.7+36.87

B= 347.01 < 44.57° Ω

C= Is/Vr = 240.56 <-36.87° / 76210 < 0°

C= 0.003157 <-36.87-0

C= 3.157 ×10^-3 < -36.87° /Ω

C= 3.157 ×10^-3 < -36.87° Ω~¹

D= Is/Ir

Since Is=Ir

Then, D = 1 no unit.

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