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Dmitriy789 [7]
3 years ago
15

A coyote is looking for some food for its young. It spots a pocket mouse near a cactus and starts to track and hunt the mouse fo

r food.
In this scenario, the cactus is considered:
a. predator
b. prey
c. neither
Physics
2 answers:
Dmitrij [34]3 years ago
8 0

Answer:

c neither

Explanation:

the cactus is just kinda chillin there

makkiz [27]3 years ago
7 0

Answer:

Neither

Explanation: <em>The cactus isn't really used for anything. The coyote only saw the mouse nearby it. The coyote was only hunting for the mouse, not the cactus.</em>

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To calculate we use the formula for a magnetic force in a current-carrying wire expressed as the product of the current, magnetic field and the length of the wire.
 F = I x L x B

where F is the force on the wire, I is the current flowing on the wire, L is the length of the wire and B is the magnetic field.

F = 10.0 A x 1.2 m x 0.050 T
F = 0.60 N
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Calculate the wavelength of each frequency of electromagnetic radiation: a. 100.2 MHz (typical frequency for FM radio broadcasti
Natalka [10]

Answer:

a). 100.2 MHz (typical frequency for FM radio broadcasting)

The wavelength of a frequency of 100.2 Mhz is 2.99m.

b. 1070 kHz (typical frequency for AM radio broadcasting) (assume four significant figures)

The wavelength of a frequency of 1070 khz is 280.3 m.

c. 835.6 MHz (common frequency used for cell phone communication)

The wavelength of a frequency of 835.6 Mhz is 0.35m.

Explanation:

The wavelength can be determined by the following equation:

c = \lambda \cdot \nu  (1)

Where c is the speed of light, \lambda is the wavelength and \nu is the frequency.  

Notice that since it is electromagnetic radiation, equation 1 can be used. Remember that light propagates in the form of an electromagnetic wave.

<em>a). 100.2 MHz (typical frequency for FM radio broadcasting)</em>

Then, \lambda can be isolated from equation 1:

\lambda = \frac{c}{\nu} (2)

since the value of c is 3x10^{8}m/s. It is necessary to express the frequency in units of hertz.

\nu = 100.2 MHz . \frac{1x10^{6}Hz}{1MHz} ⇒ 100200000Hz

But 1Hz = s^{-1}

\nu = 100200000s^{-1}

Finally, equation 2 can be used:

\lambda = \frac{3x10^{8}m/s}{100200000s^{-1}}

\lambda = 2.99 m

Hence, the wavelength of a frequency of 100.2 Mhz is 2.99m.

<em>b. 1070 kHz (typical frequency for AM radio broadcasting) (assume four significant figures)</em>

<em> </em>

\nu = 1070kHz . \frac{1000Hz}{1kHz} ⇒ 1070000Hz

But  1Hz = s^{-1}

\nu = 1070000s^{-1}

Finally, equation 2 can be used:

\lambda = \frac{3x10^{8}m/s}{1070000s^{-1}}

\lambda = 280.3 m

Hence, the wavelength of a frequency of 1070 khz is 280.3 m.

<em>c. 835.6 MHz (common frequency used for cell phone communication) </em>

\nu = 835.6MHz . \frac{1x10^{6}Hz}{1MHz} ⇒ 835600000Hz

But  1Hz = s^{-1}

\nu = 835600000s^{-1}

Finally, equation 2 can be used:

\lambda = \frac{3x10^{8}m/s}{835600000s^{-1}}

\lambda = 0.35 m

Hence, the wavelength of a frequency of 835.6 Mhz is 0.35m.

6 0
3 years ago
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Svetach [21]

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So distance between node and its nearest antinode will be

d = \frac{\lambda}{4}

d = \frac{0.26}{4} = 0.065 m = 6.5 cm

So the distance will be 6.5 cm

3 0
3 years ago
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