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dedylja [7]
3 years ago
13

The following is the longitudinal characteristic equation for an F-89 flying at 20,000 feet at Mach 0.638. The Short Period natu

ral frequency is _____.
Physics
1 answer:
BartSMP [9]3 years ago
8 0

Answer:

hello your question is incomplete  attached below is the missing part  

answer : short period oscillations frequency  = 0.063 rad / sec

              phugoid oscillations natural frequency ( w_{np} ) = 4.27 rad/sec

Explanation:

first we have to state the general form of the equation

= ( S^2 + 2\alpha _{p} w_{np} S + w^{2} _{np} ) (S^{2} + 2\alpha _{s} w_{ns}S + w^{2} _{ns}  ) = 0

where :

w_{np}  = Natural frequency of plugiod oscillation

\alpha _{p} = damping ratio of plugiod  oscilations

comparing the general form with the given equation

w^{2} _{np}  = 18.2329

w^{2} _{ns} = 0.003969

hence the short period oscillation frequency ( w_{ns} ) =  0.063 rad/sec

phugoid oscillations natural frequency ( w_{np} ) = 4.27 rad/sec

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Answer:

Acceleration will be -1.620m/sec^2

Explanation:

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We know that 1 km = 1000 m

And 1 hour = 3600 sec

So 70km/hr=70\times \frac{1000m}{3600sec}=19.444m/sec

It is given that car stops in 12 sec

So final speed of the car v = 0 m/sec

Time t = 12 sec

From first equation of motion v = u+at

So 0=19.444+a\times 12

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Use the Bohr model to address this question. When a hydrogen atom makes a transition from the 5 th energy level to the 2nd, coun
iris [78.8K]

Answer:

A. 2.82 eV

B. 439nm

C. 59.5 angstroms

Explanation:

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n2: initial state = 2

Where the energy is electron volts. You replace the values of n1 and n2 in the equation (1):

E_{5,2}=-13.6(\frac{1}{5^2}-\frac{1}{2^2})=2.82eV

B. The energy of the emitted photon is given by the following formula:

E=h\frac{c}{\lambda}   (2)

h: Planck's constant = 6.62*10^{-34} kgm^2/s

c: speed of light = 3*10^8 m/s

λ: wavelength of the photon

You first convert the energy from eV to J:

2.82eV*\frac{1J}{6.242*10^{18}eV}=4.517*10^{-19}J

Next, you use the equation (2) and solve for λ:

\lambda=\frac{hc}{E}=\frac{(6.62*10^{-34} kg m^2/s)(3*10^8m/s)}{4.517*10^{-19}J}=4.39*10^{-7}m=439*10^{-9}m=439nm

C. The radius of the orbit is given by:

r_n=n^2a_o   (3)

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You use the equation (3) with n=5:

r_5=5^2(2.380)=59.5

hence, the radius of the atom in its 5-th state is 59.5 anstrongs

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