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Which part of the wave has the highest frequency?

trapecia [35]

Answer:

The last part on the right side of the diagram

Explanation:

Im on plato and just got it right :)

6 0

3 years ago

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

BartSMP [9]

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

8 0

3 years ago

Monochromatic light with a wavelength of 384 nm passes through a single slit and falls on a screen 86 cm away. If the distance o

valkas [14]

This is a Fraunhofer single slit experiment, where the light passing through the slit produces an interference pattern on the screen, and where the dark bands (minima of diffraction) are located at a distance of
$y= \frac{m\lambda D}{a}$
from the center of the pattern. In the formula, m is the order of the minimum, $\lambda$ the wavelenght, $D$ the distance of the screen from the slit and $a$ the width of the slit.

In our problem, the distance of the first-order band (m=1) is $y=0.22 cm$ . The distance of the screen is D=86 cm while the wavelength is $\lambda = 384 nm=384 \cdot 10^{-7}cm$ . Using these data and re-arranging the formula, we can find a, the width of the slit:
$a= \frac{m \lambda D}{y}= \frac{1 \cdot 384 \cdot 10^{-7}cm \cdot 86 cm}{0.22 cm}=0.015 cm$

3 0

3 years ago

2. What is the water pressure at a depth of 24 m in a lake? [Density of water, p = 1 000 kg m- and gravitational acceleration, g

Andrei [34K]

Answer:

Pressure = ρgh

pressure (p) is measured in pascals (Pa)

density (ρ) is measured in kilograms per metre cubed (kg/m3)

The fore of gravitational field strength (g) is measured in N/kg or m/s 2

height of column (h) is measured in metres (m)

Answer = 235,200 Pa

Explanation:

Pressure = ρgh

Pressure = 1,000 x 9.8 x 24

Pressure = 235,200 Pa

4 0

3 years ago

A spy satellite uses a telescope with a 1.7-m-diameter mirror. It orbits the earth at a height of 180 km.

WINSTONCH [101]

Answer: the minimum spacing that must be there between two objects on the earth's surface if they are to be resolved as distinct objects by this telescope 6.45 cm

Explanation:

Given that;

diameter of the mirror d = 1.7 m

height h = 180 km = 180 × 10³ m

wavelength λ = 500 nm = 5 × 10⁻⁹ m

Now Angular separation from the peak of the central maximum is expressed as;

sin∅= 1.22 λ / d

sin∅ = (1.22 × 5 × 10⁻⁹) / 1.7

sin∅ = 3.588 × 10⁻⁷

we know that;

sin∅ = object separation / distance from telescope

object separation = sin∅ × distance from telescope

object separation = 3.588 × 10⁻⁷ × 180 × 10³

object separation =6.45 × 10⁻² m

then we convert to centimeter

object separation = 6.45 cm

Therefore the minimum spacing that must be there between two objects on the earth's surface if they are to be resolved as distinct objects by this telescope 6.45 cm

5 0

3 years ago