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olga55 [171]
3 years ago
13

Which example does not describe a behavioral adaptation? A. A squirrel buries acorns to dig up and eat later. B. A honeybee does

a dance to tell the other bees where to find flowers. C. Grass stops growing during a drought and begins growing again when it rains. D. The spines of a cactus protect it from animals.
Physics
2 answers:
chubhunter [2.5K]3 years ago
6 0
The answer is B) A honeybee does a dance to tell the other bees where to find flowers.
Honeybees do not dance, first of all.
Second, that is not an adaptation. That is more like a message. An adaptation is changing to the environment to survive. This is not changing to the environment. They are just signaling the other bees.
A, C, and D are all adaptations. They change so that they can survive. They do a different method than usual.
~Deceptiøn
Crazy boy [7]3 years ago
3 0
A I believe is your answer
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If the second harmonic of a certain string is 42 Hz, what is the fundamental frequency of the string?
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Data:
f_{2} = 42 Hz
n (Wave node)
V (Wave belly) 
L (Wave length)
<span>The number of bells is equal to the number of the harmonic emitted by the string.
</span>
f_{n} =  \frac{nV}{2L}

Wire 2 → 2º Harmonic → n = 2

f_{n} = \frac{nV}{2L}
f_{2} = \frac{2V}{2L} &#10;
2V =  f_{2} *2L
V =  \frac{ f_{2}*2L }{2}
V =  \frac{42*2L}{2}
V =  \frac{84L}{2}
V = 42L

Wire 1 → 1º Harmonic or Fundamental rope → n = 1


f_{n} = \frac{nV}{2L}
f_{1} = \frac{1V}{2L}
f_{1} =  \frac{V}{2L}

If, We have:
V = 42L
Soon:
f_{1} = \frac{V}{2L}
f_{1} = \frac{42L}{2L}
\boxed{f_{1} = 21 Hz}

Answer:

<span>The fundamental frequency of the string:
</span>21 Hz

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3 years ago
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Explanation: this might help for

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