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galben [10]
3 years ago
6

Please help! 30Points!! An ultrasound machine uses waves to create images. The machine uses sound waves and which wave interacti

on?
A) diffraction
B) interference
C) reflection
D) refraction
Physics
2 answers:
myrzilka [38]3 years ago
8 0

Answer:

C) Reflection

Explanation:

In ultrasound technique we throw a sound of higher frequency towards the cells or the damaged part and then a detector is placed at the same point.

Now the detector is used to detect the reflected sound of the sound which we projected. Now the part from which detector not detect any sound will be termed as defect.

So in this whole process we need to find the part from which reflected sound is not detected.

So here correct answer must be

C) Reflection

Gnesinka [82]3 years ago
5 0
The correct answer should be C
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Answer:

Explanation:

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Natural resources are substances we need and use, which occur naturally. Some come from living things, (example) cotton other are non-living (example)  sand.

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On a straight road, a car speeds up at a constant rate from rest to 20 m/s over a 5 second interval and a truck slows at a const
IceJOKER [234]

Answer:

a)

Explanation:

  • Since the car speeds up at a constant rate, we can use the kinematic equation for distance (assuming that the initial position is x=0, and choosing t₀ =0), as follows:

        x_{fc} = v_{o}*t + \frac{1}{2}*a*t^{2}   (1)

  • Since the car starts from rest, v₀ =0.
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  • Applying the definition of acceleration, as the rate of change of velocity with respect to time, and remembering that v₀ = 0 and t₀ =0, we can solve for a, as follows:

       a_{c} =\frac{v_{fc}}{t} = \frac{20m/s}{5s} = 4 m/s2  (2)

  • Replacing a and t in (1):

       x_{fc} = v_{o}*t + \frac{1}{2}*a*t^{2}  = \frac{1}{2}*a*t^{2} = \frac{1}{2}* 4 m/s2*(5s)^{2} = 50.0 m.  (3)

  • Now, if the truck slows down at a constant rate also, we can use (1) again, noting that v₀ is not equal to zero anymore.
  • Since we have the values of vf (it's zero because the truck stops), v₀, and t, we can find the new value of a, as follows:

       a_{t} =\frac{-v_{to}}{t} = \frac{-20m/s}{10s} = -2 m/s2  (4)

  • Replacing v₀, at and t in (1), we have:

       x_{ft} = 20m/s*10.0s + \frac{1}{2}*(-2 m/s2)*(10.0s)^{2} = 200m -100m = 100.0m   (5)

  • Therefore, as the truck travels twice as far as the car, the right answer is a).
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2 years ago
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Answer:

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Explanation:

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3 years ago
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dangina [55]

Answer:

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Explanation:

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