Physics, 02.02.2020 06:43 jerlin1103

Part a - whale communication.
blue whales apparently communicate with each other using sound of frequency 17.0hz , which can be heard nearly 1000km away in the ocean. what is the wavelength of such a sound in seawater, where the speed of sound is 1521m/s ?
? 1 =
m

part b - dolphin clicks.
one type of sound that dolphins emit is a sharp click of wavelength 1.50cm in the ocean. what is the frequency of such clicks?
f =
khz

part c - dog whistles.
one brand of dog whistles claims a frequency of 26.0khz for its product. what is the wavelength of this sound?
? =
cm

part d - bats.
while bats emit a wide variety of sounds, one type emits pulses of sound having a frequency between 39.0khz and 78.0khz . what is the range of wavelengths of this sound?
? min,? max =
mm

part e - sonograms.
ultrasound is used to view the interior of the body, much as x rays are utilized. for sharp imagery, the wavelength of the sound should be around one-fourth (or less) the size of the objects to be viewed. approximately what frequency of sound is needed to produce a clear image of a tumor that is 1.00mm across if the speed of sound in the tissue is 1550m/s ?
f =
mhz

Answers: 3

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Answers

Answer from: mayamcmillan11

A) 89.5 m

The wavelength of a wave is given by:

$\lambda=\frac{v}{f}$

where v is the wave speed and f the frequency. For the sound emitted by the whales,

f=17.0 Hz

v=1521 m/s

Therefore, the wavelength is

$\lambda=\frac{1521 m/s}{17.0 Hz}=89.5 m$

B) 101.4 kHz

The speed of the sound emitted by the dolphins in the water is still

v = 1521 m/s

While the wavelength is

$\lambda=1.50 cm=0.015 m$

So, re-arranging the previous equation we find the frequency:

$f=\frac{v}{\lambda}=\frac{1521 m/s}{0.015 m}=101,400 Hz=101.4 kHz$

C) 1.31 cm

The frequency of the dog whistle is

f = 26.0 kHz = 26,000 Hz

While the speed of sound in air is

v = 340 m/s

Therefore, the wavelength is

$\lambda=\frac{v}{f}=\frac{340 m/s}{26,000 Hz}=0.0131 m=1.31 cm$

D) 4.4 - 8.7 mm

The speed of the sound waves emitted by the bats in air is

v = 340 m/s

The minimum frequency is

f = 39.0 kHz = 39,000 Hz

So the corresponding wavelength is

$\lambda=\frac{v}{f}=\frac{340 m/s}{39,000 Hz}=0.0087 m=8.7 mm$

The maximum frequency is

f = 78.0 kHz = 78,000 Hz

So the corresponding wavelength is

$\lambda=\frac{v}{f}=\frac{340 m/s}{78,000 Hz}=0.0044 m=4.4 mm$

E) 6.2 MHz

The wavelength of the sound must be 1/4 the size of the tumor, so

$\lambda=\frac{1}{4}(1.00 mm)=0.25 mm=2.5\cdot 10^{-4}m$

while the speed of sound across the tissue is

v = 1550 m/s

So the frequency must be

$f=\frac{v}{\lambda}=\frac{1550 m/s}{2.5\cdot 10^{-4} m}=6.2\cdot 10^6 Hz=6.2 MHz$

Answer from: Quest

a solenoid is created by wrapping a l = 90 m long wire around a hollow tube of diameter d = 4.5 cm. the wire diameter is d = 0.9 mm. the solenoid wire is then connected to a power supply so that a current of i = 9 a flows through the wire.

randomized variablesl = 90 m

d = 4.5 cm

d = 0.9 mm

i = 9 a

write an expression for the number of turns, n, in the solenoid. you do not need to take into account the diameter of the wire in this calculation.

calculate the number of turns, n, in the solenoid.

write an expression for the length of the solenoid (l2) in terms of the diameter of the hollow tube d.

calculate the length of the solenoid (l2) in meters.

calculate the magnitude of the magnetic field at the center of the solenoid in teslas.

explanation:

Answer from: Quest

we're given that v₀ = 0 and a = 5.4 m/s². therefore, since v = at + v₀ for constant acceleration the equation for velocity is

$v = 5.4 t$

after 28s,

$v(28) = 5.4(28) = 151.2$ m/s

which i would round to 150 m/s as my final answer

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Part a - whale communication.
blue whales apparently communicate with each other using sound o...