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PolarNik
1 month ago
7

In which applications are sound waves reflected? Check all that apply.

Physics
2 answers:
inna [3.1K]1 month ago
8 0

used for disease detection, identifying damaged cells, and tracking fetal growth

serg [3.5K]1 month ago
3 0

The applications include identifying illnesses, pinpointing defective components, and tracking fetal growth.

Explanation:

Ultrasound is a type of sound wave that has a frequency above 20,000 Hz.

This technology utilizes high-frequency sound waves for imaging purposes.

Sound waves can be employed to identify issues in large machinery; when the waves bounce back, it indicates the machinery is intact, but failure to do so suggests possible damage or cracks.

Ultrasound serves as a method for medical imaging. The device sends high-frequency sound into the body using a probe to capture images of internal organs. This technique aids in monitoring fetal growth and identifying health conditions like tumors.

Thus, the uses of reflected sound waves encompass identifying diseases, finding defective parts, and monitoring fetal growth.

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Example Response: The technology referenced is fiber optics. Their small size and flexibility allow doctors to visualize areas that would otherwise require surgical intervention to examine.

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1 month ago
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Water, of density 1000 kg/m3, is flowing in a drainage channel of rectangular cross-section. The width of the channel is 15 m, t
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Answer:

The flow rate of water is (300000kg/s) = (300000l/s)

Explanation:

To compute the volume of moving fluid per second in the channel, we consider the channel's section, the water depth, and the fluid velocity:

Volume flow rate = 15m × 8m × (2.5m/s) = 300 m³/s

To find the mass or liters of water flowing per second, multiply the volume of circulating fluid by the water's density:

Flow rate of water = (300m³/s) × (1000kg/m³) = (300000kg/s) = (300000l/s)

It is important to note that 1kg of water is approximately equivalent to 1 liter.

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1 month ago
A 5-kg concrete block is lowered with a downward acceleration of 2.8 m/s2 by means of a rope. The force of the block on the Eart
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The moment the body impacts the ground, two types of Forces are produced: the gravitational pull and the Normal Force. This aligns with Newton's third law, indicating that every action has an equal and opposite reaction. If the downward force of gravity is directed toward the earth, the reactionary force from the block acts upwards, equivalent to its weight:

F = mg

Where,

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Consequently, the answer is E.

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1 month ago
If a steady-state heat transfer rate of 3 kW is conducted through a section of insulating material 1.0 m2 in cross section and 2
Maru [3345]

Answer:

\Delta T = \frac{3000 W *0.025 m}{1 m^2 (0.2 \frac{W}{mK})}= 375 K

Consequently, the temperature difference across the material will be \Delta T = 375 K

Explanation:

In this case, we apply the Fourier Law of heat conduction expressed by the following equation:

Q = -kA \frac{\Delta T}{\Delta x}   (1)

Where k = thermal conductivity = 0.2 W/ mK

A= 1m^2 denotes the cross-sectional area

Q= 3KW signifies the heat transfer rate

\Delta T is the temperature difference we need to determine

represents the thickness of the material\Delta x=2.5 cm =0.025 m

To isolate \Delta T from equation (1), we obtain:

\Delta T =\frac{Q \Delta x}{Ak}

Initially, we convert 3KW to W, resulting in:

Q= 3 KW* \frac{1000W}{1 Kw}= 3000 W

With all variables accounted for, we can substitute and calculate:

\Delta T = \frac{3000 W *0.025 m}{1 m^2 (0.2 \frac{W}{mK})}= 375 K

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5 0
17 days ago
A policeman in a stationary car measures the speed of approaching cars by means of an ultrasonic device that emits a sound with
Ostrovityanka [3204]

Answer:

The beats frequency measures approximately

4.4 kHz

Explanation:

The beat frequency arises from the original ultrasound frequency, f=41.2 kHz, and the frequency of the sound reflected off the car, f':

f_B = f'-f (1)

To calculate the frequency of the reflected sound, we apply the Doppler effect formula:

f'=\frac{v}{v-v_s}f

where

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f=41.2 kHzis the frequency of the sound emitted

By substituting values,

f'=\frac{340 m/s}{340 m/s-33.0 m/s}(41.2 kHz)=45.6 kHz

Thus, the beat frequency (1) is

f_B = 45.6 kHz - 41.2 kHz=4.4 kHz

3 0
1 month ago
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