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sweet [91]
4 years ago
14

Bats can detect small objects such as insects that are of a size on the order of a wavelength. if bats emit a chirp at a frequen

cy of 69.3 khz and the speed of sound waves in air is 330 m/s, what is the smallest size insect they can detect? give your answer in mm (millimeters)
Physics
1 answer:
crimeas [40]4 years ago
6 0
The smallest size of the insect that the bats can detect corresponds to the wavelength of the chirp they emit.

Their chirp has a frequency of
f=69.3 kHz=69.3 \cdot 10^3 Hz
and the speed of the chirp is equal to the speed of sound in air:
v=330 m/s
Therefore the wavelength of the chirp is
\lambda= \frac{v}{f}= \frac{330 m/s}{69.3 \cdot 10^3 Hz}=4.76 \cdot 10^{-3} m
which corresponds to a size of 4.76 mm.
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If the resultant of two velocity vectors of equal magnitude is also of the same magnitude, then which statement must be correct?
Tamiku [17]

The correct option is C) The angle between the vectors is 120°.

Why?

We can solve the problem and find the correct option using the Law of Cosine.

Let A and B, the given two sides and R the resultant (sum),

Then,

R=A=B

So, using the law of cosines, we have:

R^{2}=A^{2}+B^{2}+2ABCos(\alpha)\\ \\A^{2}=A^{2}+A^{2}+2*A*A*Cos(\alpha)\\\\0=A^{2}+2*A^{2}*Cos(\alpha)\\\\Cos(\alpha)=-\frac{A^{2}}{2*A^{2}}=-\frac{1}{2}\\\\\alpha =Cos(-\frac{1}{2})^{-1}=120\°

Hence, we have that the angle between the vectors is 120°. The correct option is C) The angle between the vectors is 120°

Have a nice day!

4 0
3 years ago
a bubble of air of volume 1cm^3 is released by a deep sea diver at a depth where the pressure is 4.0 atmospheres. assuming its t
lys-0071 [83]

Answer:

hope this helps!

Explanation:

Volume of the air bubble, V1=1.0cm3=1.0×10−6m3

Bubble rises to height, d=40m

Temperature at a depth of 40 m, T1=12oC=285K

Temperature at the surface of the lake, T2=35oC=308K

The pressure on the surface of the lake: P2=1atm=1×1.103×105Pa 

The pressure at the depth of 40 m: P1=1atm+dρg

Where,

ρ is the density of water =103kg/m3

g is the acceleration due to gravity =9.8m/s2

∴P1=1.103×105+40×103×9.8=493300Pa

We have T1P1V1=T2P2V2

Where, V2 is the volume of the air bubble when it reaches the surface.

V2=

8 0
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Answer:

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Explanation:

6 0
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dusya [7]

Answer:

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5 0
4 years ago
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