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3 years ago

9

I will Brainlist "Burning and Inch". Describe this measurement.

2 answers:

MAVERICK [17]3 years ago

8 0

Answer:

what measurement

Explanation:

Paladinen [302]3 years ago

4 0

Answer:

You measured from the 1" mark (“burning” the first inch), then forgot to add an inch when marking.

Explanation:

Another way to say it is "Burn an Inch."

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Consider a simply supported rectangular beam with a span length of 10 ft, a width of 8 in, and an effective depth of 20 in. Norm

yulyashka [42]

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beam with a span length of 10 ft, a width of 8 in, and an effective depth of 20 in. Normal weight concrete is used for the beam. This beam carries a total factored load of 9.4 kips. The beam is reinforced with tensile steel, which continues uninterrupted into the support. The concrete has a strength of 4000 psi, and the yield strength of the steel is 60,000 psi. Using No. 3 bars and 60,000 psi steel for stirrups, do the followings:

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3 years ago

Select the best answer for the question.

dalvyx [7]

I think the Acid level

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3 years ago

A 1.00 liter solution contains 0.46 M hydrocyanic acid and 0.35 M potassium cyanide If 25.0 mL of water are added to this system

victus00 [196]

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3 years ago

A three-phase wye-connected synchronous generator supplies a network through a transmission line. The network can absorb or deli

Amanda [17]

Answer:

the graph and the answer can be found in the explanation section

Explanation:

Given:

Network rated voltage = 24 kV

Impedance of network = 0.07 + j0.5 Ω/mi, 8 mi

Rn = 0.07 * 8 = 0.56 Ω

Xn = 0.5 * 8 = 4 Ω

If the alternator terminal voltage is equal to network rated voltage will have

Vt = 24 kV/√3 = 13.85 kV/phase

The alternative current is

$I_{a} =\frac{40x10^{6} }{\sqrt{3}*24x10^{3} } =926.2A$

$X_{s} =0.85\frac{13.85}{926.2} =12.7ohm$

The impedance Zn is

$\sqrt{0.56^{2}+4^{2} } =4.03ohm$

The voltage drop is

$I_{a} *Z_{n} =926.2*4.03=3732.58V$

$r_{dc} =\frac{voltage}{2*current} =\frac{13.85}{2*926.2} =7.476ohm$

rac = 1.2rdc = 1.2 * 7.476 = 8.97 Ω

The effective armature resistance is

$Z_{s} =\sqrt{R_{a}^{2}+X_{s}^{2} } =\sqrt{8.97^{2}+12.7^{2} } =15.55ohm$

The induced voltage for leading power factor is

$E_{F} ^{2} =OB^{2} +(BC-CD)^{2}$

if cosθ = 0.5

$E_{F} =\sqrt{(13850*0.5)^{2}+(\frac{3741}{2}-926.2*12.7)^{2} } =11937.51V$

if cosθ= 0.6

EF = 12790.8 V

if cosθ = 0.7

EF = 13731.05 V

if cosθ = 0.8

EF = 14741.6 V

if cosθ = 0.9

EF = 15809.02 V

if cosθ = 1

EF = 13975.6 V

The voltage regulation is

$\frac{E_{F}-V_{t} }{V_{t} } *100$

For each value:

if cosθ = 0.5

voltage regulation = -13.8%

if cosθ = 0.6

voltage regulation = -7.6%

if cosθ = 0.7

voltage regulation = -0.85%

if cosθ = 0.8

voltage regulation = 6.4%

if cosθ = 0.9

voltage regulation = 14%

if cosθ = 1

voltage regulation = 0.9%

the graph is shown in the attached image

for 10% of regulation the power factor is 0.81

8 0

3 years ago

Ok there........................................................................

Juliette [100K]

Answer:

ok THERE

Explanation:

4 0

3 years ago