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steposvetlana [31]
2 years ago
5

Calculate the acceleration of a 2000kg single engine airlane just before take off when the thrust of the engine is 500n?

Engineering
1 answer:
pshichka [43]2 years ago
4 0

Answer:

0.25 m/s²

Explanation:

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Can someone help me plz!!
pogonyaev
It has to do with mechanical engineering
6 0
2 years ago
A 500-km, 500-kV, 60-Hz, uncompensated three-phase line has a positivesequence series impedance. z = 5 0.03 1 + j 0.35 V/km and
Anni [7]

Answer:

A) 282.34 - j 12.08 Ω

B) 0.0266 + j 0.621 / unit

C)

A = 0.812 < 1.09° per unit

B =  164.6 < 85.42°Ω  

C =  2.061 * 10^-3 < 90.32° s

D =  0.812 < 1.09° per unit

Explanation:

Given data :

Z ( impedance ) = 0.03 i  + j 0.35 Ω/km

positive sequence shunt admittance ( Y ) = j4.4*10^-6 S/km

A) calculate Zc

Zc = \sqrt{\frac{z}{y} }  =  \sqrt{\frac{0.03 i  + j 0.35}{j4.4*10^-6 } }    

    = \sqrt{79837.128< 4.899^o}   =  282.6 < -2.45°

hence Zc = 282.34 - j 12.08 Ω

B) Calculate  gl

gl = \sqrt{zy} * d  

 d = 500

 z = 0.03 i  + j 0.35

 y = j4.4*10^-6 S/km

gl =  \sqrt{0.03 i  + j 0.35*  j4.4*10^-6}  * 500

   = \sqrt{1.5456*10^{-6} < 175.1^0} * 500

   = 0.622 < 87.55 °

gl = 0.0266 + j 0.621 / unit

C) exact ABCD parameters for this line

A = cos h (gl) . per unit  =  0.812 < 1.09° per unit ( as calculated )

B = Zc sin h (gl) Ω  = 164.6 < 85.42°Ω  ( as calculated )

C = 1/Zc  sin h (gl) s  =  2.061 * 10^-3 < 90.32° s ( as calculated )

D = cos h (gl) . per unit = 0.812 < 1.09° per unit ( as calculated )

where :  cos h (gl)  = \frac{e^{gl} + e^{-gl}  }{2}

             sin h (gl) = \frac{e^{gl}-e^{-gl}  }{2}

     

7 0
2 years ago
An inductor has a 50.0-Ω reactance when connected to a 60.0-Hz source. The inductor is removed and then connected to a 45.0-Hz s
nignag [31]

Given:

X_{L} = 50.0 \ohm

frequency, f = 60.0 Hz

frequency, f' = 45.0 Hz

V_rms} = 85.0 V

Solution:

To calculate max current in inductor, I_{L(max):

At f = 60.0 Hz

X_{L} = 2\pi fL

50.0 = 2\pi\times 60.0\times L

L = 0.1326 H

Now, reactance X_{L} at f' = 45.0 Hz:

X'_{L} = 2\pi f'L

X'_{L} = 2\pi\times 45.0\times 0.13263 = 37.5\ohm

Now, I_{L(max) is given by:

I_{L(max) = \sqrt {\frac{2V_{rms}}{X'_{L}}}

I_{L(max) = \sqrt {\frac{2\times 85.0}{37.5}} = 2.13 A

Therefore,  max current in the inductor, I_{L(max) = 2.13 A

7 0
3 years ago
I need a thesis statement about Engineers as Leaders.
algol [13]

Answer:

Engineers are a very beneficial contribution in which offers great solutions to national problems.

5 0
2 years ago
Calculate the areas under the stress-strain curve (toughness) for the materials shown in Fig. below, (a) plot them as a
defon

Answer:

Explanation:

Wow

5 0
2 years ago
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