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3 years ago

11

AM radio signals have frequencies between 550 kHz and 1600 kHz (kilohertz) and travel with a speed of 3.0Ã108m/s. What are the w

avelengths of these signals?

1 answer:

Westkost [7]3 years ago

5 0

Answer:

The wavelength of these signals is as follow:

Wavelength of 550 kHz is 545.45 m
Wavelength of 1600 kHz is 187.5 m

Explanation:

Given that:

Frequency = 550 kHz & 1600 kHz

Velocity = 3.0 x 10⁸ m/s

As we know that frequency is expressed by the following equation:

Frequency = Velocity / Wavelength ---- (1)

For 550 kHz:

The equation can be rearranged as

Wavelength = Velocity / Frequency

Wavelength = (3.0 x 10⁸ m/s) / (550 x 1000 Hz)

Wavelength = 545.45 m

For 1600 kHz:

Wavelength = Velocity / Frequency

Wavelength = (3.0 x 10⁸ m/s) / (1600 x 1000 Hz)

Wavelength = 187.5 m

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Gretchen runs the first 4.0 km of a race at 5.0 m/s. Then a stiff wind comes up, so she runs the last 1.0 km at only 4.0 m/s.

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Answer:

The velocity is $v = 4.76 \ m/s$

Explanation:

From the question we are told that

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$t_1 = \frac{1000}{4}$

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The total time is mathematically represented as

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=> $t = 800 + 250$

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Generally the constant velocity that would let her finish at the same time is mathematically represented as

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You are standing in a building whose height (40m) you throw a ball downward at a angle of -30 at a speed of (10m/s) acceleration

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Answer: 3.41 s

Explanation:

Assuming the question is to find the time $t$ the ball is in air, we can use the following equation:

$y=y_{o}+V_{o}sin \theta t-\frac{1}{2}gt^{2}$

Where:

$y=0m$ is the final height of the ball

$y_{o}=40 m$ is the initial height of the ball

$V_{o}=10 m/s$ is the initial velocity of the ball

$t$ is the time the ball is in air

$g=9.8 m/s^{2}$ is the acceleration due to gravity

$\theta=30\°$

Then:

$0 m=40 m+(10 m/s)(sin(30\°))t-\frac{1}{2}9.8 m/s^{2}t^{2}$

$0 m=40 m+5m/s t-4.9 m/s^{2}t^{2}$

Multiplying both sides of the equation by -1 and rearranging:

$4.9 m/s^{2}t^{2}-5m/s t-40 m=0$

At this point we have a quadratic equation of the form $at^{2}+bt+c=0$ , which can be solved with the following formula:

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Substituting the known values:

$t=\frac{-(-5) \pm \sqrt{(-5)^{2}-4(4.9)(-40)}}{2(4.9)}$

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