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yan [13]
2 years ago
10

A helicopter is hovering in a steady cross wind at a gross weight of 3,000 lb (1,360.8 kg). This helicopter has 275 hp (205 kW)

delivered to the main rotor shaft. The tail rotor is 2.3 ft (0.701 m) and has an induced power factor of 1.15. The tail rotor is located 15.3 ft (4.66 m) from the main rotor shaft. Determine the crosswind conditions (velocity and direction) in which the tail rotor effectiveness may be reduced or lost. If the center of gravity is assumed to lie on the rotor shaft axis, determine the feasible yawing angular velocity that the pilot can demand that may also result in a loss of tail rotor effectiveness
Engineering
1 answer:
damaskus [11]2 years ago
6 0

Answer: Answer Bellow

Explanation:The helicopter has 275 hp (205 kW) delivered to the main rotor shaft. The tail rotor radius is 2.3 ft (0.701 m) and has an induced power factor of 1.15.

(I would appreciate brainiest)

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Determine (with justification) whether the following systems are (i) memoryless, (ii) causal, (iii) invertible, (iv) stable, and
lina2011 [118]

Answer:

a.

y[n] = x[n] x[n-1]  x[n+1]

(i) Memory-less - It is not memory-less because the given system is depend on past or future values.

(ii) Causal - It is non-casual because the present value of output depend on the future value of input.

(iii) Invertible - It is invertible and the inverse of the given system is \frac{1}{x[n] . x[n-1] x[n+1]}

(iv) Stable - It is stable because for all the bounded input, output is bounded.

(v) Time invariant - It is not time invariant because the system is multiplying with a time varying function.

b.

y[n] = cos(x[n])

(i) Memory-less - It is memory-less because the given system is not depend on past or future values.

(ii) Causal - It is casual because the present value of output does not depend on the future value of input.

(iii) Invertible - It is not invertible because two or more than two input values can generate same output values .

For example - for x[n] = 0 , y[n] = cos(0) = 1

                       for x[n] = 2\pi , y[n] = cos(2\pi) = 1

(iv) Stable - It is stable because for all the bounded input, output is bounded.

(v) Time invariant - It is time invariant because the system is not multiplying with a time varying function.

3 0
2 years ago
A satellite orbits the Earth every 2 hours at an average distance from the Earth's centre of 8000km. (i) What is the average ang
AlexFokin [52]

Answer:

i)ω=3600 rad/s

ii)V=7059.44 m/s

iii)F=1245.8 N

Explanation:

i)

We know that angular speed given as

\omega =\dfrac{d\theta}{dt}

We know that for one revolution

θ=2π

Given that time t= 2 hr

So

ω=θ/t

ω=2π/2 = π rad/hr

ω=3600 rad/s

ii)

Average speed V

V=\sqrt{\dfrac{GM}{R}}

Where M is the mass of earth.

R is the distance

G is the constant.

Now by putting the values

V=\sqrt{\dfrac{GM}{R}}

V=\sqrt{\dfrac{6.667\times 10^{-11}\times 5.98\times 10^{24}}{8000\times 10^3}}

V=7059.44 m/s

iii)

We know that centripetal fore given as

F=\dfrac{mV^2}{R}

Here given that m= 200 kg

R= 8000 km

so now by putting the values

F=\dfrac{mV^2}{R}

F=\dfrac{200\times 7059.44^2}{8000\times 10^3}

F=1245.8 N

3 0
2 years ago
In a production turning operation, the foreman has decreed that a single pass must be completed on the cylindrical workpiece in
stellarik [79]

Answer:

V = 125.7m/min

Explanation:

Given:

L = 400 mm ≈ 0.4m

D = 150 mm ≈ 0.15m

T = 5 minutes

F = 0.30mm ≈ 0.0003m

To calculate the cutting speed, let's use the formula :

T = \frac{pi* D * L}{V*F}

We are to find the speed, V. Let's make it the subject.

V = \frac{pi* D * L}{F*T}

Substituting values we have:

V = \frac{pi* 0.4 * 0.15}{0.0003*5}

V = 125.68 m/min ≈ 125.7 m/min

Therefore, V = 125.7m/min

7 0
3 years ago
Implement the following Matlab code:
vagabundo [1.1K]
28384 *x soít cos estematema
3 0
3 years ago
Four kilograms of carbon monoxide (CO) is contained in a rigid tank with a volume of 1 m3. The tank is fitted with a paddle whee
Juli2301 [7.4K]

Answer:

a) 1 m^3/Kg  

b) 504 kJ

c) 514 kJ

Explanation:

<u>Given  </u>

-The mass of C_o2 = 1 kg  

-The volume of the tank V_tank = 1 m^3  

-The added energy E = 14 W  

-The time of adding energy t = 10 s  

-The increase in specific internal energy Δu = +10 kJ/kg  

-The change in kinetic energy ΔKE = 0 and The change in potential energy  

ΔPE =0  

<u>Required  </u>

(a)Specific volume at the final state v_2

(b)The energy transferred by the work W in kJ.  

(c)The energy transferred by the heat transfer W in kJ and the direction of  

the heat transfer.  

Assumption  

-Quasi-equilibrium process.  

<u>Solution</u>  

(a) The volume and the mass doesn't change then, the specific volume is constant.

 v= V_tank/m ---> 1/1= 1 m^3/Kg  

(b) The added work is defined by.  

W =E * t --->  14 x 10 x 3600 x 10^-3 = 504 kJ  

(c) From the first law of thermodynamics.  

Q - W = m * Δu

Q = (m * Δu) + W--> (1 x 10) + 504 = 514 kJ

The heat have (+) sign the n it is added to the system.

7 0
2 years ago
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