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Ira Lisetskai [31]
3 years ago
11

An ultrasound machine is being used to try to identify potential kidney stones. The machine is working properly and no kidney st

ones are found. Which statement best describes this situation? The generated sound wave will not be able to reflect off of the object. The sound wave will not be generated, so there will be no echoes received. The echoes will be made but will not be able to be received by the machine. The machine will not be able to process the received echoes into images.
Physics
2 answers:
Kazeer [188]3 years ago
7 0

Answer:

The generated sound wave will not be able to reflect off of the object.

Explanation:

bonufazy [111]3 years ago
4 0

Answer:

Explanation:

e d g e n u i t y 2020

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A 1.60-kg object is held 1.05 m above a relaxed, massless vertical spring with a force constant of 330 N/m. The object is droppe
pentagon [3]

Answer:

(A) l = 0.39 m      

(B)  l =0.38 m  

(C) l = 0.14 m

Explanation:

Answer:

Explanation:

Answer:

Explanation:

from the question we are given the following values:

mass (m) = 1.6 kg

height (h) = 1.05 m

compression of spring (l) = ?

spring constant (k) = 330 N/m

acceleration due to gravity (g) = 9.8 m/s^{2}

(A) initial potential energy of the object = final potential energy of the spring

         potential energy of the object = mg(1.05 + l)  

         potential energy of the spring = 0.5 x k x l^{2}  (k= spring constant)

 therefore we now have

              mg(1.05 + l)  = 0.5 x k x l^{2}

              1.6 x 9.8 x (1.05 + l)  = 0.5 x 300 x l^{2}

               15.68 (1.05 + l) = 150 x l^{2}

                   16.5 + 15.68l = 150l^{2}

l = 0.39 m        

(B)   with constant air resistance the equation applied in part A above becomes

initial P.E of the object - air resistance = final P.E of the spring

mg(1.05 + l) - 0.750(1.05 + l) = 0.5 x k x l^{2}        

     1.6 x 9.8 x (1.05 + l) - 0.750(1.05 + l)  = 0.5 x 300 x l^{2}

         (16.5 + 15.68l) - (0.788 + 0.75l) = 150l^{2}        

          16.5 + 15.68l - 0.788 - 0.75l = 150l^{2}

            15.71 + 14.93l = 150^{2}

            l =0.38 m  

(C)   where g = 1.63 m/s^{2} and neglecting air resistance

      the equation mg(1.05 + l)  = 0.5 x k x l^{2} now becomes

        1.6 x 1.63 x (1.05 + l)  = 0.5 x 300 x l^{2}

        2.608 (1.05 +l) = 0.5 x 300 x l^{2}

        2.74 + 2.608l = 150 x l^{2}

l = 0.14 m

6 0
3 years ago
Effect of Solvent:
Naddik [55]

1. H2O- The water completely dissolved the salt.

Alcohol- The alcohol dissolved the salt slightly.

Glycerin-The salt has not dissolved at all.

2. Water

3. Like dissolve like basically works on the principle of polarity. It means the substances which possess similar chemical properties may dissolve in each other. For example, ethanol can be dissolved in water because both are polar in nature whereas non-polar molecules can be dissolved in non-polar solvents only.

4. Generally, a solute dissolves faster in a warmer solvent than it does in a cooler solvent because particles have more energy of movement. For example, if you add the same amount of sugar to a cup of hot tea and a cup of iced tea, the sugar will dissolve faster in the hot tea.

Learn more about The choice of solvent here:-brainly.com/question/14918321

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6 0
1 year ago
What is the speed that the Earth orbits the Sun?
iris [78.8K]

Answer:

30 Kilometers per second

4 0
3 years ago
You have probably noticed that carrying a person in a pool of water is much easier than carrying a person through air. To unders
Kruka [31]

Answer:

The answer is "0.91238 and 744.8"

Explanation:

In this scenario it is easier to take a person to the water-pool than to transport the people in the air, as the person's strength is increased by water upwards:

f_b \to m \to mg =person \\\\F_B \ in\  air = v\ & air\  g \\\\

               =0.076 \times 1.225 \times 9.8 \\\\ =0.91238 \ N\\\\

F_B \ in \ water = v  \& water \ g \\\\

                    =0.076 \times 1000 \times 9.8\\\\= 744.8 \ N\\

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3 years ago
Match the speed to the section that describes.
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a=2 ok do it and ........

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