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

Answers

Answer 1

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 ( [tex]w_{np}[/tex] ) = 4.27 rad/sec

Explanation:

first we have to state the general form of the equation

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

where :

[tex]w_{np} = Natural frequency of plugiod oscillation[/tex]

[tex]\alpha _{p} = damping ratio of plugiod oscilations[/tex]

comparing the general form with the given equation

[tex]w^{2} _{np}[/tex]  = 18.2329

[tex]w^{2} _{ns} = 0.003969[/tex]

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

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

The Following Is The Longitudinal Characteristic Equation For An F-89 Flying At 20,000 Feet At Mach 0.638.

Related Questions

A hydraulic car jack needs to be designed so it can lift a 2903.57 lb car assuming that a person can exert a force of 24.41 lbs. If the piston the person is pushing on had a radius of 3.26 cm, what should the diameter of the piston be that is used to raise the car?

Answers

Answer:

Diameter of the piston would be 0.71 m (71.1 cm)

Explanation:

From the principle of pressure;

[tex]\frac{F_{1} }{A_{1} }[/tex] = [tex]\frac{F_{2} }{A_{2} }[/tex]

Let [tex]F_{1}[/tex] = 2903.57 lb, [tex]F_{2}[/tex] = 24.41 lbs, [tex]r_{2}[/tex] = 3.26 cm = 0.0326 m.

[tex]A_{2}[/tex] = [tex]\pi r^{2}[/tex]

    = [tex]\frac{22}{7}[/tex] x [tex](0.0326)^{2}[/tex]

    = 0.00334 [tex]m^{2}[/tex]

So that:

[tex]\frac{2903.57}{A_{1} }[/tex] = [tex]\frac{24.41}{0.00334}[/tex]

[tex]A_{1}[/tex] = [tex]\frac{2903.57*0.00334}{24.41}[/tex]

    = 0.3973

[tex]A_{1}[/tex] = 0.4 [tex]m^{2}[/tex]

The radius of the piston can be determined by:

[tex]A_{1}[/tex] = [tex]\pi r^{2}[/tex]

0.3973 = [tex]\frac{22}{7}[/tex] x [tex]r^{2}[/tex]

[tex]r^{2}[/tex] = [tex]\frac{0.3973*7}{22}[/tex]

   = 0.1264

r = [tex]\sqrt{0.1264}[/tex]

 = 0.3555

r = 0.36 m

Diameter of the piston = 2 x r

                                     = 2 x 0.3555

                                     = 0.711

Diameter of the piston would be 0.71 m (71.1 cm).

You have a source of energy containing 21 gj of energy at 600k how much this energy can be converted to work when rejecting heat to the atmosphere at 27°C?

Answers

Answer:

Available energy = 35 x 10⁶ J

Explanation:

Given:

Amount of energy (Q) = 21 gj = 21 x 10⁹ J

Temperature T1 = 600 k

Temperature T0 = 27 + 273 = 300k

Find:

Available energy

Computation:

Available energy = Q[1/T0 - 1/T1]

Available energy = 21 x 10⁹ J[1/300 - 1/600]

Available energy = 35 x 10⁶ J

The image blow shows a certain type of global wind:
What best describes these winds? Polar easterlies caused by air above poles being relatively warmer.
Polar easterlies caused by air above poles being relatively cooler.
Trade winds caused by air above equator being relatively warmer.
Trade winds caused by air above equator being relatively cooler.​

Answers

Answer:

i got u its a

Explanation:

Starting from rest, a coin and a ring roll down a ramp without slipping. Which of the following are true:

A. The ring reaches the bottom first
B. The coin reaches the bottom first
C. The coin and the ring arrive at the same time
D. The one the reaches the bottom first is the one with the largest mass
E. The one that reaches the bottom first is the one with the largest diameter.

Answers

Answer:

The answer is D

Explanation:

The object with the heavy mass will reach first because the cause it is heaver so it will go faster

an object of mass 4kg moving with initial velocity if 20m/s accelerates for 10s and attaind a final velocity of 60m/s calculate the acceleration​

Answers

Answer:

given us,

mass= 4×9.8gm m(9.8) formula

= 39.2

final velocity (v)= 60m/s

initial velocity (u)= 20m/s

time(t)= 10s

acleration(a)=?

now,

accleration(a)= v-u/t=60- 20/10

=40/10

=4m/s

:. the acceleration is 4 m/s

Explanation:

first we have to calculate mass and we can use acceleration formula

A block of mass 12.2 kg is sliding at an initial velocity of 3.9 m/s in the positive x-direction. The surface has a coefficient of kinetic friction of 0.173. (indicate the direction with signs of your answer) (a) What is the force of kinetic friction in N acting on the block? (b) What is the block's acceleration in /s^2? (c) How Far will it slide (in m) before coming to rest? Plz answer as soon as possible

Answers

Answer:

Explanation:

a) Force of friction = μ R where μ is coefficient of kinetic friction and R is reaction force

R = mg where m is mass of the block

Force of friction F = μ x mg

= .173 x 12.2 x 9.8

= 20.68 N

b ) Only force of friction is acting on the body so

deceleration = force / mass = 20.68 / 12.2 = 1.7 m /s²

acceleration = - 1.7 m /s²

c )

v² = u² - 2 a s

v = 0 , u = 3.9 m /s

a = 1.7 m /s

0 = 3.9² - 2 x 1.7 x s

s = 4.47  m

A 0.22 caliber handgun fires a 1.9g bullet at a velocity of 765m/s. Calculate the de Broglie wavelength of the bullet. Is the wave nature of matter significant for the bullets?

Answers

Answer:

de Broglie wavelength of the bullet is 4.56 x 10⁻³⁴ mThe value of the wavelength shows that wave nature of matter is insignificant for the bullet because it is larger than particles.

Explanation:

Given;

mass of the bullet, m = 1.9 g = 0.0019 kg

velocity of the bullet, v = 765 m/s

de Broglie wavelength of the bullet is given by;

[tex]\lambda = \frac{h}{mv}[/tex]

where;

h is Planck's constant = 6.626 x 10⁻³⁴ J/s

λ is de Broglie wavelength of the bullet

[tex]\lambda = \frac{h}{mv}\\\\ \lambda =\frac{(6.626*10^{-34})}{(0.0019)(765)}\\\\ \lambda =4.56 *10^{-34} \ m[/tex]

Thus, this value of the wavelength shows that wave nature of matter is insignificant for the bullet because it is larger than particles.

The wavelength will be "[tex]4.56\times 10^{-34} \ m[/tex]".

Given:

Mass, m = 1.9 g or, 0.0019 kgVelocity, v = 765 mPlank's constant, h = 6.626 × 10⁻³⁴ J/s

The De-Broglie wavelength,

→ [tex]\lambda = \frac{h}{mv}[/tex]

By putting the values,

     [tex]= \frac{6.626\times 10^{-34}}{0.0019\times 765}[/tex]

     [tex]= 4.56\times 10^{-34} \ m[/tex]

Thus the response above is right.

Learn more about wavelength here:

https://brainly.com/question/10931065

1. When an object is at rest, not moving, and is crashed into by another
object...which object is experiencing a force and in what direction? Or are
both experiencing a force - what direction?

Answers

Answer:

both experience forces or at least a force

Explanation:

it would go in the direction the other object

(second object, the one that crashed) was going

si if going right then right if left then left

plus or minus

A 760-kg horse whose power output is 1 hp is pulling a sled over the snow at 3.9 m/s. Find the force the horse exerts on the sled. Round your intermediate step to one decimal place and the final answer to two decimal places.

Answers

Answer:

191 N

Explanation:

Power can be regarded as the amount of energy that is been transfered at a unit time and can be calculated using the Express below

Po = F*V

P= power

F= force

V= velocity

From the question, we were given power output as 1 hp and velocity= 3.9 m/s.

But

1hp= 746 Watts = 746 Joules/s.

Then substitute the values

Po = F*V = 746 J/s

F ×3.9 = 746

F= 746/3.9

F = 191 N.

Therefore, the force the horse exerts on the sled iss 191 N

The diagram below shows two bowling balls, A and B, each having a mass of 7.0 kg, placed 2.00 m apart between their centers.

Answers

Answer:

F = 1.63 x 10⁻⁹ N

Explanation:

Complete question is as follows:

The diagram below shows two bowling balls, A and B, each having a mass of 7.0 kg, placed 2.00 m apart between their centers. Find the magnitude of Gravitational Force?

Answer:

The gravitational force is given by Newton's Gravitational Law as follows:

F = Gm₁m₂/r²

where,

F = Gravitational Force = ?

G = Universal Gravitational Constant = 6.67 x 10⁻¹¹ N.m²/kg²

m₁ = m₂ = mass of each ball = 7 kg

r = distance between balls = 2 m

Therefore,

F = (6.67 x 10⁻¹¹ N.m²/kg²)(7 kg)(7 kg)/(2 m)²

F = 1.63 x 10⁻⁹ N

Section 4.1- Newton's First Law

Answers

Answer:

Newton's first law states that every object will remain at rest or in uniform motion in a straight line unless compelled to change its state by the action of an external force. This is normally taken as the definition of inertia. ... If that velocity is zero, then the object remains at rest.

Explanation:

Answer:

Newton's First Law is about inertia; objects at rest stay at rest unless acted upon and objects in motion continue that motion in a straight line unless acted upon. The amount of inertia an object has is simply related to the mass of the object.

I NEED THIS ASAP!!
Which formula defines the unit for electrical power?

Answers

C option
Power = voltage x current

Answer:

1 W = 1 V x 1 A

Explanation:

Other dude is wrong this is right

If a penny is dropped from rest from a building takes 2 seconds to hit the ground, calculate the velocity of the penny right before it touches the ground?
a. 19.6 m/s
b 9.8 m/s
c. 0 m/s
d 29.4 m/s​

Answers

Answer:

20m/s

Explanation:

u = 0m/s

t = 2s

a = +g = 10m/s²

t = 2d

v = ?

v = ut + 1/2at²

v = 0(2) + 1/2(10)(2)²

v = 0 + 5(4)

v = 20m/s

A 36.3 kg cart has a velocity of 3 m/s. How much kinetic energy does the object have?

Answers

Answer:

163.35

__________________________________________________________

We are given:

Mass of the object (m) = 36.3 kg

Velocity of the object (v) = 3 m/s

Kinetic Energy of the object:

We know that:

Kinetic Energy = 1/2(mv²)

KE = 1/2(36.3)(3)²            [replacing the variables with the given values]

KE = 18.15 * 9

KE = 163.35 Joules

Hence, the cart has a Kinetic Energy of 163.35 Joules

1. Opposite charges
O repel
attract

Answers

Answer:

attract

Explanation:

that is the answer

Answer:

Attract.

Explanation:

I took the quiz.

If you are pushing 200 kg of textbooks with acceleration of 2m/s2, how much net force are you exerting on the books? (Fnet=ma)
( There is more than one answer)

200N

100N

400Kg

400N

400 kg.m/s2


Answers

Explanation:

m=200kg

a=2m/s2

F=ma

F=200kg×2m/s2

=400kg.m/s2 or 400N

g When the movable mirror of the Michelson interferometer is moved a small distance X while making a measurement, 246 fringes are counted moving into the field of the viewing mirror. What is X if the wavelength of the light entering the interferometer is 562 nm

Answers

Answer:

X = 69.1 x 10⁻⁶ m = 69.1 μm

Explanation:

The relationship between the motion of the moveable mirror and the fringe count of the Michelson's Interferometer is given by the following formula:

d = mλ/2

where,

d = distance moved by the mirror = X = ?

m = No. of Fringes counted = 246

λ = wavelength of light entering interferometer = 562 nm = 5.62 x 10⁻⁷ m

Therefore,

X = (246)(5.62 x 10⁻⁷ m)/2

Therefore,

X = 69.1 x 10⁻⁶ m = 69.1 μm

If it requires 7.0 J of work to stretch a particular spring by 1.8 cm from its equilibrium length, how much more work will be required to stretch it an additional 3.6 cm?

Answers

Answer:

56 J

Explanation:

The following data were obtained from the question:

Energy 1 (E₁) = 7 J

Extention 1 (e₁) = 1.8 cm

Extention 2 (e₂) = 1.8 + 3.6 = 5.4 cm

Energy 2 (E₂) =?

Energy stored in a spring is given by the following equation:

E = ke²

Where E is the energy.

K is the spring constant.

e is the extension.

E = ke²

Divide both side by e²

K = E/e²

Thus,

E₁/e₁² = E₂/e₂²

7/ 1.8² = E₂/ 5.4²

7 / 3.24 = E₂/ 29.16

Cross multiply

3.24 × E₂ = 7 × 29.16

3.24 × E₂ = 204.12

Divide both side by 3.24

E₂ = 204.12 / 3.24

E₂ = 63 J

Thus, the additional energy required can be obtained as follow:

Energy 1 (E₁) = 7 J

Energy 2 (E₂) = 63 J

Additional energy = 63 – 7

Additional energy = 56 J

A light truck with manual transmission has a circular brake drum with a diameter of d = 310 millimeters. Each brake pad, which presses against the drum, is s = 306 millimeters long. What central angle is subtended by one of the brake pads? Write your answer in both radians and degrees.

Answers

Answer:

113 degrees

1.974 radians

Explanation:

The angle can be calculated using below expression

Angle = arc/r

Where arc= diameter of d = 310 millimeters

r= s = 306 millimeters long.

Since both have the same unit, no need for conversion

Angle = arc/r = 306/(0.5×310 )

= 306/155

= 1.974 radians

Then to find the angle In degrees we have

R x 180/pi = D

Where R= angle in radians= 1.974 radians

1.974 x 180/π

Where π= 22/7

= 113 degrees( approximately)

Therefore, central angle is subtended by one of the brake pads is 1.974 radians

can vectors be strung together?

Answers

Answer:

The head-to-tail method is a graphical way to add vectors, The tail of the vector is the starting point of the vector, and the head (or tip) of a vector is the final, pointed end of the arrow.

Explanation:

3. Which object has more inertia?
A. A tractor trailer rig moving at 2 m/s
B. A pingpong ball rolling a 2 m/s
C. A bowling ball rolling at 1m/s
D. A car rolling at 5 m/s

Answers

Answer:

A. A tractor trailer rig moving at 2 m/s

Explanation:

Inertia can be defined as the tendency of an object or a body to continue in its state of motion or remain at rest unless acted upon by an external force.

In physics, Sir Isaac Newton's first law of motion is known as law of inertia and it states that, an object or a physical body in motion will continue in its state of motion at continuous velocity (the same speed and direction) or, if at rest, will remain at rest unless acted upon by an external force.

The inertia of an object such as a tractor trailer rig is greatly dependent or influenced by its mass; the higher quantity of matter in a tractor trailer rig, the greater will be its tendency to continuously remain at rest.

Hence, the object that has more inertia is a tractor trailer rig moving at 2 m/s because it has more mass than all the other objects in the category. Also, the mass of an object is directly proportional to its inertia.

In a simulation on earth, an astronaut in his space suit climbs up a vertical ladder. On the moon, the same astronaut makes the same climb. In which case does the gravitational potential energy of the astronaut change by a greater amount?

Answers

Answer:

Gravitational potential energy of the astronaut will change by a greater amount on the earth

Explanation:

Gravitational potential energy is expressed by the formula;

GPE = mgh

This means that the gravitational potential energy is directly proportional to the gravity(g)

Now, from constant values, gravity of moon is 1.62 m/s² while gravity of the earth is 9.81 m/s².

This means that if we plug in the values of g on the earth and g on the moon, the potential energy on the earth would be greater than that of the moon

Thus, gravitational potential energy of the astronaut will change by a greater amount on the earth

Use the above picture to fill in the blanks for the following statement.

One of the element carbon combines with one of the element oxygen to form one of the compound carbon dioxide.

Answers

Answer:

C + 2O ------ CO2

Explanation:

"One" element of Carbon combines with "Two" elements of Oxygen, to form "One" compound of Carbon dioxide.

I didn't really get what you meant but this is my guess of what you meant

A mass of 100 g is tied to the end of an 80.0-cm string and swings in a vertical circle about a fixed center under the influence of gravity. The speed of the mass at the top of the swing is 3.50 m/s. What is the speed of the mass at the bottom of its swing?

Answers

Answer:

the speed of the mass at the bottom of its swing is 6.61m/s

Explanation:

Applying energy conservation

[tex]\frac{1}{2}m(Vlowest)^2 = mg(2R) + \frac{1}{2}m(Vtop)^2[/tex]

There is no potential energy at the bottom as the body will have a kinetic energy there.

h= 2R = 1.6m as the diameter of the circle will represent the height in the circle.

g = 9.8m/s^2

m will cancel out, so the net equation becomes.

[tex]\frac{(Vbottom)^2}{2} = 2gR + \frac{(Vtop)^2}{2}[/tex]

                 = [tex]2*9.8*0.8 + \frac{(3.5)^2}{2}[/tex]

                  =  15.68+ 6.125

         [tex]\frac{(Vbottom)^2 }{2}[/tex]     =  21.805

(Vb)^2 = 2*21.805

     = 43.64

Vb = 6.61m/s

A parallel plate capacitor is made up of two metal squares with sides of length 8.8 cm, separated by a distance 5.0 mm. When a voltage 187 V is set up across the terminals of the capacitor, the charge stored on the positive plate is equal to __________ nC. g

Answers

Answer:

2.56 nC

Explanation:

By definition, the capacitance is expressed by the following relationship between the charge stored on one of the plates of the capacitor and the potential difference between them, as follows:

       [tex]C =\frac{Q}{V} (1)[/tex]

For a parallel-plate capacitor, assuming a uniform surface charge density σ, if the area of the plates is A, the charge on one of the plates can be written as follows:

       [tex]Q = \sigma * A (2)[/tex]

Assuming an uniform electric field E, the potential difference V can be expressed as follows:

        [tex]V = E*d (3)[/tex]

        where d is the distance between plates.

Applying Gauss 'Law to a closed surface half within one plate, half outside it, we find that E can be written as follows:

       [tex]E =\frac{\sigma}{\epsilon_{0}} (4)[/tex]

Replacing (4) in (3), and (2) in (1), we can express the capacitance C as follows:

       [tex]C= \frac{\epsilon_{0}*A}{d} (5)[/tex]

Taking (1) and (5), as both left sides are equal each other, the right sides are also equal, so we can write the following equality:

        [tex]\frac{Q}{V} = \frac{\epsilon_{0}*A}{d} (6)[/tex]

Solving for Q, we get:

       [tex]Q = \frac{\epsilon_{0}*A*V}{d} = \frac{8.85e-12F/m*(0.088m)^{2}*187 V}{5.0e-3m} = 2.56 nC[/tex]

Find analytically the velocity of the object at the end point of the inclined plane for a certain angle Ө

Answers

I don't know if there is other given information that's missing here, so I'll try to fill in the gaps as best I can.

Let m be the mass of the object and v₀ its initial velocity at some distance x up the plane. Then the velocity v of the object at the bottom of the plane can be determined via the equation

v² - v₀² = 2 a x

where a is the acceleration.

At any point during its motion down the plane, the net force acting on the object points in the same direction. If friction is negligible, the only forces acting on the object are due to its weight (magnitude w) and the normal force (mag. n); if there is friction, let f denote its magnitude and let µ denote the coefficient of kinetic friction.

Recall Newton's second law,

F = m a

where the symbols in boldface are vectors.

Split up the forces into their horizontal and vertical components. Then by Newton's second law,

• net horizontal force:

F = n cos(θ + 90º) = m a cos(θ + 180º)

→  - n sin(θ) = - m a cos(θ)

→  n sin(θ) = m a cos(θ) ……… [1]

• net vertical force:

F = n sin(θ + 90º) - w = m a sin(θ + 180º)

→   n cos(θ) - m g = - m a sin(θ)

→   n cos(θ) = m (g - a sin(θ)) ……… [2]

where in both equations, a is the magnitude of acceleration, g = 9.80 m/s², and friction is ignored.

Then by multiplying [1] by cos(θ) and [2] by sin(θ), we have

n sin(θ) cos(θ) = m a cos²(θ)

n cos(θ) sin(θ) = m (g sin(θ) - a sin²(θ))

m a cos²(θ) = m (g sin(θ) - a sin²(θ))

a cos²(θ) + a sin²(θ) = g sin(θ)

a = g sin(θ)

and so the object attains a velocity of

v = √(v₀² + 2 g x sin(θ))

If there is friction to consider, then f = µ n, and Newton's second law instead gives

• net horizontal force:

F = n cos(θ + 90º) + f cos(θ) = m a cos(θ + 180º)

→   - n sin(θ) + µ n cos(θ) = - m a cos(θ)

→   n sin(θ) - µ n cos(θ) = m a cos(θ) ……… [3]

• net vertical force:

F = n sin(θ + 90º) + f sin(θ) - w = m a sin(θ + 180º)

→   n cos(θ) + µ n sin(θ) - m g = - m a sin(θ)

→   n cos(θ) + µ n sin(θ) = m g - m a sin(θ) ……… [4]

Then multiply [3] by cos(θ) and [4] by sin(θ) to get

- n sin(θ) cos(θ) + µ n cos²(θ) = - m a cos²(θ)

n cos(θ) sin(θ) + µ n sin²(θ) = m g sin(θ) - m a sin²(θ)

and adding these together gives

µ n (cos²(θ) + sin²(θ)) = m g sin(θ) - m a (cos²(θ) + sin²(θ))

µ n = m g sin(θ) - m a

m a = m g sin(θ) - µ n

m a = m g sin(θ) - µ m g cos (θ)

a = g (sin(θ) - µ cos (θ))

and so the object would instead attain a velocity of

v = √(v₀² + 2 g x (sin(θ) - µ cos (θ)))

A typical elevator car with people has a mass of 1500.0 kg. Elevators are currently approaching speeds of 20.0 m/s - faster than the speed.

Required:
What is the upward force required if the elevator moves upward 200.0 meters before reaching 20.0 m/s?

Answers

Answer:

1500N

Explanation:

Force = mass * acceleration

Given

Mass = 1500kg

Get the acceleration using the equation of motion;

v² = u²+2aS

20² = 0+2s(200)

400 = 400a

a = 400/400

a = 1m/s²

Get the upward force required

F = 1500 * 1

F = 1500N

Hence the upward force required if the elevator moves upward 200.0 meters before reaching 20.0 m/s is 1500N

A sample with a path length of 1 cm absorbs 99.0% of the incident light at a wavelength of 274 nm, measured with respect to an appropriate solvent blank. Tyrosine is known to be the only chromophore present in the sample that has significant absorption at 274 nm. Calculate the molar concentration of tyrosine in the sample.

Answers

Answer:

17. NADH has a molar extinction coefficient of 6200 M2 cm at 340 nm. Calculate the molar concentration of NADH required to obtain an absorbance of 0.1 at 340 nm in a 1-cm path length cuvette. 18. A sample with a path length of 1 cm absorbs 99.0% of the incident light at a wavelength of 274 nm, measured with respect to an appropriate solvent blank. Tyrosine is known to be the only chromophore present in the sample that has significant absorption at 274 nm. Calculate the molar concentration of tyrosine in the sample.

Explanation:

gravities limit is under which sphere as the perimeter?​

Answers

Ndndjxncmcmmxmxmxx I hope this hiked

A car starts from rest and accelerates uniformly over a time of 7 seconds for a distance of 190m. Find the the acceleration of the car.

Answers

Answer:

a = 7.75 [m/²]

Explanation:

To solve this problem we must use the following equation of kinematics.

[tex]x=x_{0} +v_{o} *t + (\frac{1}{2})*a*t^{2}[/tex]

where:

x = final distance = 190 [m]

Xo =  initial distance = 0

Vo = initial velocity = 0 (car starts from the rest)

a = acceleration [m/s²]

t = time = 7 [s]

190 = 0 + (0*7) + 0.5*a*(7²)

190 = 0.5*49*a

a = 7.75 [m/²]

Other Questions
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