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3 years ago

7

AN AIR CRAFT HEADS NORTH AT 320KM/H RELATIVE TO THE WIND THE WIND VELOCITY IS 80KM/HFROMNORTH .FIND THE VELOCITY OF THE AIR RELA

TIVE TO THE GROUND

1 answer:

makkiz [27]3 years ago

5 0

Answer:

Explanation:

See attached file for answer.

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Light waves are electromagnetic waves that travel at 3.00 Light waves are electromagnetic waves that travel 108 m/s. The eye is

svlad2 [7]

(a) $5.45 \cdot 10^{14} Hz$

The relationship between frequency and wavelength of an electromagnetic wave is given by

$c=f \lambda$

where

$c=3.00 \cdot 10^8 m/s$ is the speed of light

$f$ is the frequency

$\lambda$ is the wavelength

In this problem, we are considering light with wavelength of

$\lambda=5.50 \cdot 10^{-7} m$

Substituting into the equation and re-arranging it, we can find the corresponding frequency:

$f=\frac{c}{\lambda}=\frac{3.00 \cdot 10^8 m/s}{5.50 \cdot 10^{-7} m}=5.45 \cdot 10^{14} Hz$

(b) $1.83\cdot 10^{-15} s$

The period of a wave is equal to the reciprocal of the frequency:

$T=\frac{1}{f}$

And using $f=5.45 \cdot 10^{14} Hz$ as we found in the previous part, we can find the period of this wave:

$T=\frac{1}{5.45 \cdot 10^{14} Hz}=1.83\cdot 10^{-15} s$

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3 years ago

A skier has an acceleration of 2.5 m/s2. How long does it take her to come to a complete stop from a speed of 18 m/s?

aleksandr82 [10.1K]

(Assuming acceleration is -2.5m/s2) 7.2 seconds.

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3 years ago

Read 2 more answers

A swinging pendulum has a total energy of <img src="https://tex.z-dn.net/?f=E_i" id="TexFormula1" title="E_i" alt="E_i" align="a

Zolol [24]

Answer:

$\frac{E_{2}}{E_{1}} \approx 1 -\frac{3\theta}{1-\theta}$ (for small oscillations)

Explanation:

The total energy of the pendulum is equal to:

$E_{1} = m\cdot g \cdot (1-\cos \theta)\cdot L$

For small oscillations, the equation can be re-arranged into the following form:

$E_{1} \approx m\cdot g \cdot (1-\theta) \cdot L$

Where:

$\theta = \frac{A}{L^{2}}$ , measured in radians.

If the amplitude of pendulum oscillations is increase by a factor of 4, the angle of oscillation is $4\theta$ and the total energy of the pendulum is:

$E_{2} \approx m\cdot g \cdot (1-4\theta)\cdot L$

The factor of change is:

$\frac{E_{2}}{E_{1}} \approx \frac{1 - 4\theta}{1-\theta}$

$\frac{E_{2}}{E_{1}} \approx 1 -\frac{3\theta}{1-\theta}$

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Answer:

A, B Y E

Explanation:

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