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Gnom [1K]
4 years ago
11

The sound waves used for sonar are _____.

Physics
2 answers:
lakkis [162]4 years ago
6 0

<u>Ultrasound </u>''Although humans cannot hear ultrasound, scientists have developed technology to make use of these high-frequency sounds. For example, sonar technology uses ultrasound.''

Stella [2.4K]4 years ago
4 0

Acoustic Energy


Sonar uses the echo principle by sending out sound waves underwater or through the human body to locate objects

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A client has developed dystrophic calcification as a result of macroscopic deposition of calcium salts. The tissue that would be
Komok [63]

Answer:

Tissues that are damaged or injured.

Explanation:

Dystrophic calcification involves the deposition of calcium in soft tissues despite no disturbance in the calcium metabolism, and this is often seen at damaged tissues.

Examples of areas in the body where dystrophic calcification can occur include atherosclerotic plaques and damaged heart valves.

4 0
3 years ago
A 5.6 MeV (kinetic energy) proton enters a 0.39 T field, in a plane perpendicular to the field. What is the radius of its path?
Kisachek [45]

Answer:

the radius of the protons path is r = 0.85 m.

Explanation:

the force due to magnetic fields lead to the cetripetal force, such that:

F = q×v×B = m×(v^2)/r

         q×B = m×v/r

then:

r = m×v/q×B

r = p/q×B

then, the kinetic energy of the proton:

K = 1/2×m×v^2 = p^2/(2×m)

q×B = \sqrt{2×m×K}/r

    r =  \sqrt{2×m×K}/(q×B)

      =  \sqrt{2×(1.67×10^-27)×(5.3×1.60×10^-13)}/(1.60×10^-19×0.39)

      = 0.85 m  

7 0
4 years ago
3. Two spherical objects at the same altitude move with identical velocities and experience the same drag force at a time t. If
Daniel [21]

Answer:

Object 2 has the larger drag coefficient

Explanation:

The drag force, D, is given by the equation:

D = 0.5 c \rho A v^2

Object 1 has twice the diameter of object 2.

If d_2 = d

d_1 = 2d

Area of object 2, A_2 = \frac{\pi d^2 }{4}

Area of object 1:

A_1 = \frac{\pi (2d)^2 }{4}\\A_1 = \pi d^2

Since all other parameters are still the same except the drag coefficient:

For object 1:

D = 0.5 c_1 \rho A_1 v^2\\D = 0.5 c_1 \rho (\pi d^2) v^2

For object 2:

D = 0.5 c_2 \rho A_2 v^2\\D = 0.5 c_2 \rho (\pi d^2/4) v^2

Since the drag force for the two objects are the same:

0.5 c_1 \rho (\pi d^2) v^2 = 0.5 c_2 \rho (\pi d^2/4) v^2\\4c_1 = c_2

Obviously from the equation above, c₂ is larger than c₁, this means that object 2 has the larger drag coefficient

8 0
3 years ago
A 1.70 m long string has a standing wave with 2 loops at a frequency of 38.4 Hz. What is the speed of the waves in the string? (
Luden [163]

Answer:65.28

Explanation:

7 0
3 years ago
I’ll give points plz and brainliest I know the result I just need the process.
shusha [124]
Shiddd ion even know but can you like my comment?
7 0
3 years ago
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