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2 years ago

12

A typical aircraft fuselage structure would be capable of carrying torsion moment. a)True b)- False

1 answer:

Taya2010 [7]2 years ago

8 0

Answer:

True

Explanation:

An aircraft is subject to 3 primary rotations

1) About longitudinal axis known as rolling

2) About lateral axis known as known as pitching

3) About the vertical axis known as yawing

The rolling of the aircraft induces torsion in the body of the aircraft thus the fuselage structure should be capable of carrying torsion

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Answer:

Aluminum

Explanation:

The best material to use when creating an electric fence would be Aluminum. Aluminum wiring is incredibly durable and can be easily obtained. Since aluminum is a non-magnetic metal its conducting capabilities far exceed other metallic options in the market and is also why companies choose aluminum for their high tension cable wiring. Aside from being more expensive than other feasible options its durability and conducting capabilities make it easily the best option.

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2 years ago

Read 2 more answers

An ideal gas initially at 300 K and 1 bar undergoes a three-step mechanically reversible cycle in a closed system. In step 12, p

Veseljchak [2.6K]

Answer:

Ts =Ta E)- 300(

569.5 K

5

Q12-W12 = -4014.26

Mol

AU2s = Q23= 5601.55

Mol

AUs¡ = Ws¡ = -5601.55

Explanation:

A clear details for the question is also attached.

(b) The P,V and T for state 1,2 and 3

P =1 bar Ti = 300 K and Vi from ideal gas Vi=

10

24.9x10 m

=

P-5 bar

Due to step 12 is isothermal: T1 = T2= 300 K and

VVi24.9 x 10x-4.9 x 10-3 *

The values at 3 calclated by Uing step 3l Adiabatic process

B-P ()

Since step 23 is Isochoric: Va =Vs= 4.99 m* and 7=

14

Ps-1x(4.99 x 103

P-1x(29x 10)

9.49 barr

And Ts =Ta E)- 300(

569.5 K

5

(c) For step 12: Isothermal, Since AT = 0 then AH12 = AU12 = 0 and

Work done for Isotermal process define as

8.314 x 300 In =4014.26

Wi2= RTi ln

mol

And fromn first law of thermodynamic

AU12= W12 +Q12

Q12-W12 = -4014.26

Mol

F'or step 23 Isochoric: AV = 0 Since volume change is zero W23= 0 and

Alls = Cp(L3-12)=5 x 8.311 (569.5 - 300) = 7812.18-

AU23= C (13-72) =5 x 8.314 (569.3 - 300) = 5601.53

Inol

Now from first law of thermodynamic the Q23

AU2s = Q23= 5601.55

Mol

For step 3-1 Adiabatic: Since in this process no heat transfer occur Q31= 0

and

AH

C,(T -Ts)=x 8.314 (300- 569.5)= -7842.18

mol

AU=C, (T¡-T)= x 8.314 (300

-5601.55

569.5)

mol

Now from first law of thermodynamie the Ws1

J

mol

AUs¡ = Ws¡ = -5601.55

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2 years ago

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Tools, weapons, hardware, armor

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3 years ago

1: A baseball is hit 4 feet above the ground leaves the bat with an initial speed of 98 ft/sec at an angle of 0 45 is caught by

zvonat [6]

Answer:

299.36 feet

Explanation:

$To \ find \ the \ distance \ of \ the \ ball \ from \ the \ home \ plate. \\ \\ From \ the \ given \ information:$

$Height \ h = 4 \ ft$

$Initial \ speed \ V_o = 98 \ ft/s ec$

$The \ angle \ \theta = 45^0$

$Acceleration \ due \ to \ gravity (g)= 32.2 \ ft/s$

$U_x = V_o \ cos 45 = \dfrac{98}{\sqrt{2}}$

$U_y = V_o \ sin 45 = \dfrac{98}{\sqrt{2}}$

So;

$S_y = u_y t - \dfrac{1}{2}gt^2$

$-1 =\dfrac{98}{\sqrt{2}}t - \dfrac{1}{2}*32*1.85t^2$

By solving:

$t_1 = 4.32 \ sec$

Thus;

$horizontal \ distance = U_x t$

$= \dfrac{98}{\sqrt{2}}\times 4.32$

$\mathbf{=299.36 \ feet}$

$\mathbf{Thus \ , the \ distance \ from \ the \ home \ plate \ = \ 299.36 \ feet}$

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