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Assoli18 [71]
3 years ago
13

The time delay between transmission and the arrival of the reflected wave of a signal using ultrasound traveling through a piece

of fat tissue was 0.13 ms. At what depth did this reflection occur? (The average propagation speed for sound in body tissue is 1540 m/s)
Physics
1 answer:
Brilliant_brown [7]3 years ago
8 0

Answer:

10.01 cm

Explanation:

Given that,

The time delay between transmission and the arrival of the reflected wave of a signal using ultrasound traveling through a piece of fat tissue was 0.13 ms.

The average propagation speed for sound in body tissue is 1540 m/s.

We need to find the depth when the reflection occur. We know that, the distance is double when transmitting and arriving. So,

v=\dfrac{2d}{t}\\\\d=\dfrac{vt}{2}\\\\d=\dfrac{1540\times 0.13\times 10^{-3}}{2}\\\\d= $$0.1001\ m

or

d = 10.01 cm

So, the reflection will occur at 10.01 cm.

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baherus [9]

Answer:

The speed of the baseball is approximately 19.855 m/s

Explanation:

From the question, we have;

The frequency of the microwave beam emitted by the speed gun, f = 2.41 × 10¹⁰ Hz

The change in the frequency of the returning wave, Δf = +3190 Hz higher

The Doppler shift for the microwave frequency emitted by the speed gun which is then reflected back to the gun by the moving baseball is given by 2 shifts as follows;

 \dfrac{\Delta f}{f} = \dfrac{2 \cdot v_{baseball}}{c}

\therefore{\Delta f}{} = \dfrac{2 \cdot v_{baseball}}{c} \times f

Where;

Δf = The change in frequency observed, known as the beat frequency = 3190 Hz

v_{baseball} = The speed of the baseball

c = The speed of light = 3.0 × 10⁸ m/s

f = The frequency of the microwave beam = 2.41 × 10¹⁰ Hz

By plugging in the values, we have;

\therefore{\Delta f} = 3190 \ Hz =  \dfrac{2 \cdot v_{baseball}}{3.0 \times 10^8 \ m/s} \times 2.41 \times 10^{10} \ Hz

v_{baseball} = \dfrac{3190 \ Hz \times 3.0 \times 10^8 \ m/s }{2.41 \times 10^{10} \ Hz \times 2} \approx 19.855 \ m/s

The speed of the baseball, v_{baseball} ≈ 19.855 m/s

3 0
3 years ago
In an adiabatc process, what happens when gases in a system are compressed?
likoan [24]

Answer:work is done, and temperature increases

Explanation:

In an adiabatic process, when gases are compressed, work is done on the liquid and the temperature increases

8 0
4 years ago
How are galaxies organized and distributed within the universe
den301095 [7]
The galaxies aren't distributed randomly throughout the universe but they are grouped in gravitationally bound clusters.
3 0
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Answer:

D) 735 J(oules)

Explanation:

Work is defined as force * distance

Force is defined as mass * acceleration

Given a mass of 15 kg and a gravitational acceleration of 9.8 m/s² since the box is being lifted up, the force being applied to the box is 15 kg * 9.8 m/s² = 147 N

Since the distance is 5 meters, the work done is 147 N * 5 m = 735 N/m = 735 J, making D the correct answer.

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3 years ago
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TEA [102]

Answer:

Friction, normal force, and weight

Explanation:

If the book slows down, it means that there must be friction acting in the opposite direction of the direction the book is moving in.

Weight is caused by the gravitational pull of the Earth on the book, and normal force is the table pushing the book up because the book is pushing down on the table (3rd law.)

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