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goldenfox [79]
3 years ago
5

During an ultrasound, sound waves are sent by a transducer through muscle tissue at a speed of 1300 m/s. Some of the sound waves

are reflected from a metal fragment 3.0 cm into the muscle tissue. How long did it take the transducer to detect the reflected waves from the metal fragment after they were first emitted?
23 seconds
39 seconds
2.3 x 10−5 seconds
3.9 x 10−5 seconds
Physics
1 answer:
erica [24]3 years ago
3 0

Answer: 2.3(10)^{-5} s

Explanation:

The velocity V of the sound wave in the muscle tissue. which is the same as the reflected from a metal fragment is:

V=\frac{d}{t}

Where:

V=1300 m/s

d=3 cm \frac{1m}{100 cm}=0.03 m is the distance traveled by the sound wave

t is the time

Isolating t:

t=\frac{d}{V}

t=\frac{0.03 m}{1300 m/s}

Finally:

t=2.3(10)^{-5} s

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ss7ja [257]

Answer:

223.25 $\text{Jm}^{-1}\text{s}^{-1}^\circ\text{C}^{-1}$

Explanation:

The thermal conductivity of an object is defined as the measure or the ability of the object to transfer heat or conduct heat through its body.

In the context, the thermal conductivity of the material is given as

$=129 \text{ Btu ft}^{-1}\text{h}^{-1}^\circ\text{F}^{-1}$

And it is given that :

1 Btu = 1055 J

1 ft = 0.3048 m

$1^\circ F = \frac{5}{9}^\circ C$

We know that 1 h = 3600 s

So the thermal conductivity of the material in $\text{Jm}^{-1}\text{s}^{-1}^\circ\text{C}^{-1}$ is  :

Thermal conductivity :

$=\frac{129 \text{ Btu}}{1 \text{ ft }\times \text{1 h}\times 1^\circ\text{F}}$

$=\frac{129 \times 1055 \text{ J}}{0.3048 \text{ m} \ \times 3600 \text{ s}\ \times \frac{5}{9}^\circ \text{C}}$

=  223.25 $\text{Jm}^{-1}\text{s}^{-1}^\circ\text{C}^{-1}$

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