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Lina20 [59]
3 years ago
12

How many types of arcs do we have in AutoCad? O a.9 O b. 10 Oc. 12 O d. 11​

Engineering
2 answers:
Dmitry [639]3 years ago
8 0

Answer:

Letter D

Explanation:

AutoCAD provides eleven different ways to create arcs. The different options are used based on the geometry conditions of the design. To create an arc, you can specify various combinations of center, endpoint, start point, radius, angle, chord length, and direction values.

zzz [600]3 years ago
3 0

<em>Answ</em><em>er</em><em> </em><em>:</em><em>-</em><em> </em>

<em><u>corr</u></em><em><u>ect</u></em><em><u> answer</u></em><em><u> is</u></em><em><u> </u></em><em><u>=</u></em><em><u>1</u></em><em><u>1</u></em>

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thermodynamics A nuclear power plant based on the Rankine cycle operates with a boiling-water reactor to develop net cycle power
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(a) the percent thermal efficiency is 27.94%

(b) the temperature of the cooling water exiting the condenser is 31.118°C

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3 years ago
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3 years ago
In a wind-turbine, the generator in the nacelle is rated at 690 V and 2.3 MW. It operates at a power factor of 0.85 (lagging) at
Juli2301 [7.4K]

To solve this problem we will apply the concepts related to real power in 3 phases, which is defined as the product between the phase voltage, the phase current and the power factor (Specifically given by the cosine of the phase angle). First we will find the phase voltage from the given voltage and proceed to find the current by clearing it from the previously mentioned formula. Our values are

V = 690V

P_{real} = 2.3MW

Real power in 3 phase

P_{real} = 3V_{ph}I_{ph} Cos\theta

Now the Phase Voltage is,

V_{ph} = \frac{V}{\sqrt{3}}

V_{ph} = \frac{690}{\sqrt{3}}

V_{ph} = 398.37V

The current phase would be,

P_{real} = 3V_{ph}I_{ph} Cos\theta

Rearranging,

I_{ph}=\frac{P_{real}}{3V_{ph}Cos\theta}

Replacing,

I_{ph}=\frac{2.3MW}{3( 398.37V)(0.85)}

I_{ph}= 2.26kA/phase

Therefore the current per phase is 2.26kA

6 0
3 years ago
A vehicle of 1 200 kg is moving at a speed of 40 km/h on an incline of 1 in 50. The total constant rolling and wind resistance i
Readme [11.4K]

Answer:11.602 KW

Explanation:

mass of vehicle\left ( m\right )=1200 kg

speed=40Km/h

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\eta =80%

Gear ratio\left ( G\right )=4:1

D_{effective}=500mm

Net force to overcome by engine is

F=Resistance + sin component of weight

F=600+mgsin\theta

Where tan\theta =[tex]\frac{1}{50}

\theta =1.1457^{\circ}

F=600+1200\times 9.81\times sin\left ( 1.1457\right )

F=600+235.38=835.38 N

power=F.v=835.38\frac{100}{9}

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4 0
4 years ago
Is a street the same as a avenue
-BARSIC- [3]

they're essentially the same thing so i'd say yes

5 0
3 years ago
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