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2 years ago

7

A hawk in level flight 135m above the ground drops the fish it caught. If the hawk horizontal speed is 20m/s, how far ahead of t

he drop point will the fish land

1 answer:

Kaylis [27]2 years ago

8 0

Find how long it will take the fish to hit the ground

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How does the position of an object relate to the energy stored in an object?

Kamila [148]

Answer:

Potential Energy

Explanation:

Potential energy is the energy stored in an object due to it's position relative to some zero position. An object possesses gravitational potential energy if it is positioned at a height above (or below) the zero height.

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2 years ago

Frequency is measured in units called?

irina1246 [14]

The frequency, f, of a wave is the number of waves passing a point in a certain time. We normally use a time of one second, so this gives frequency the unit hertz (Hz), since one hertz is equal to one wave per second.

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2 years ago

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klemol [59]

Diagram B .... light shines through at an angle

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2 years ago

Read 2 more answers

Starting from rest, a disk rotates about its central axis with constant angular acceleration. in 6.00 s, it rotates 44.5 rad. du

Klio2033 [76]

a. The disk starts at rest, so its angular displacement at time $t$ is

$\theta=\dfrac\alpha2t^2$

It rotates 44.5 rad in this time, so we have

$44.5\,\mathrm{rad}=\dfrac\alpha2(6.00\,\mathrm s)^2\implies\alpha=2.47\dfrac{\rm rad}{\mathrm s^2}$

b. Since acceleration is constant, the average angular velocity is

$\omega_{\rm avg}=\dfrac{\omega_f+\omega_i}2=\dfrac{\omega_f}2$

where $\omega_f$ is the angular velocity achieved after 6.00 s. The velocity of the disk at time $t$ is

$\omega=\alpha t$

so we have

$\omega_f=\left(2.47\dfrac{\rm rad}{\mathrm s^2}\right)(6.00\,\mathrm s)=14.8\dfrac{\rm rad}{\rm s}$

making the average velocity

$\omega_{\rm avg}=\dfrac{14.8\frac{\rm rad}{\rm s}}2=7.42\dfrac{\rm rad}{\rm s}$

Another way to find the average velocity is to compute it directly via

$\omega_{\rm avg}=\dfrac{\Delta\theta}{\Delta t}=\dfrac{44.5\,\rm rad}{6.00\,\rm s}=7.42\dfrac{\rm rad}{\rm s}$

c. We already found this using the first method in part (b),

$\omega=14.8\dfrac{\rm rad}{\rm s}$

d. We already know

$\theta=\dfrac\alpha2t^2$

so this is just a matter of plugging in $t=12.0\,\mathrm s$ . We get

$\theta=179\,\mathrm{rad}$

Or to make things slightly more interesting, we could have taken the end of the first 6.00 s interval to be the start of the next 6.00 s interval, so that

$\theta=44.5\,\mathrm{rad}+\left(14.8\dfrac{\rm rad}{\rm s}\right)t+\dfrac\alpha2t^2$

Then for $t=6.00\,\rm s$ we would get the same $\theta=179\,\rm rad$ .

7 0

2 years ago

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

Westkost [7]

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

5 0

2 years ago