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Gre4nikov [31]
3 years ago
8

What are some of the uses of ultra-sound​

Physics
2 answers:
saw5 [17]3 years ago
7 0

Answer:

Ultrasound can be used for various purposes like diagnosing conditions, including those in the heart, blood vessels, liver and more. Yet most commonly to monitor growth and development of a fetus. A echocardiogram is a ultrasound of the heart.

Explanation:

kirill115 [55]3 years ago
3 0

Answer:

Ultrasou­nd has been used in a variety of clinical settings, including obstetrics and gynecology, cardiology and cancer detection. The main advantage of ultrasound is that certain structures can be observed without using radiation. Ultrasound can also be done much faster than X-rays or other radiographic techniques

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A particular type of automobile storage battery is characterized as "207 - Ampere - hour, 9.4 V."
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Electric Energy = Current x Voltage x time = 207 x 9.4 = 1,945.8 J
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3 years ago
Describe how a drug administered intravenously is absorbed​
Digiron [165]

Answer:

La rapidez con que el fármaco es absorbido en el torrente sanguíneo depende, en parte, del suministro de sangre al músculo: cuanto menor sea el aporte de sangre, más tiempo necesitará el fármaco para ser absorbido.

Para la administración por vía intravenosa se inserta una aguja directamente en una vena

Explanation:

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From the point of view of water when the temperature of water increases from room temperature to 90°C the process of heating the
Andrew [12]

its boiling i believe

5 0
3 years ago
Anything that causes change must have energy?
Makovka662 [10]
This is a true fact.
Example: An object sitting still in place has energy called potential energy so if an object moving, kinetic energy, and pushes the other object not moving, then the potential energy object will now have kinetic energy; Change.

So, that IS a true fact.

I hope this helped!
8 0
3 years ago
Particle-X has a speed of 0.720 c and a momentum of 4.350x1019 kgm/s. What is the mass of the particle? 2.0206 10-27 kg Hints: T
kirill115 [55]

Explanation:

Given that,

Speed of particle = 0.720 c

Momentum = 4.350\times10^{-19}\ kgm/s[/tex]

(I). We need to calculate the mass of the particle

Using formula of momentum

P=mv

m =\dfrac{P}{v}

m=\dfrac{4.350\times10^{-19}}{ 0.720\times3\times10^{8}}

m=2.013\times10^{-27}\ Kg

We need to calculate the rest mass of particle

Using formula of rest mass

m=\dfrac{m_{0}}{\sqrt{1-(\dfrac{v}{c})^2}}

Where, m_{0} = rest mass

Put the value into the formula

m_{0}=2.013\times10^{-27}\times\sqrt{1-(\dfrac{0.720 c}{c})^2}

m_{0}=2.013\times10^{-27}\times\sqrt{1-(0.720)^2}

m_{0}=1.4\times10^{-27}\ kg

(b). We need to calculate the rest energy of the particle

Using formula of energy

E_{0}=m_{0}c^2

Put the value into the formula

E_{0}=1.4\times10^{-27}\times(3\times10^{8})^2

E_{0}=1.26\times10^{-10}\ J

(c).  We need to calculate the kinetic energy of the particle

Using formula of kinetic energy

K.E=mc^2-m_{0}c^2

K.E=(m-m_{0})\timesc^2

K.E=(2.013\times10^{-27}-1.4\times10^{-27})\times3\times10^{8}

K.E=1.84\times10^{-19}\ J

(d). We need to calculate the total energy of the particle

Using formula of energy

E=mc^2

Put the value into the formula

E=2.013\times10^{-27}\times(3\times10^{8})^2

E=1.812\times10^{-10}\ J

Hence, This is the required solution.

8 0
3 years ago
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