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3 years ago

Which of the following is not an application of Doppler technology?

2 answers:

8 0

The correct answer to the question above is The third Option: C; ultrasound imaging of the liver. The ultrasound imaging of the liver is definitely not an application of Doppler technology.

Hope this helps! :)

Reil [10]3 years ago

4 0

Answer:

ultrasound imaging of the liver.

Explanation:

Liver ultrasound is not an application of the Doppler effect. The use of ultrasound is more common in medicine, for example, during pregnancy.

Another application in medicine is with the movement of blood. It is a flowmeter which allows the blood flow to be discovered through the Doppler effect, since a sound wave affects the red blood cells that are in motion inside the vessel and collect the echo signal.

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Objects have a tendency to resist changing their motion. This property is called: *

Salsk061 [2.6K]

Answer:

The answer your looking for is option 2 - Inertia

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3 years ago

What do you call the theory that says gravitational attraction caused bits of matter to collapse, begin spinning, and eventually

Semenov [28]

The nebular hypothesis is the name of the theory where gravitational attraction caused bits of matter to collapse, eventually forming a solar system.
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7 0

3 years ago

Halley's comet orbits the sun roughly once every 76 years. It comes very close to the surface of the Sun on its closest approach

Licemer1 [7]

Answer:

r1 = 5*10^10 m , r2 = 6*10^12 m

v1 = 9*10^4 m/s

From conservation of energy

K1 +U1 = K2 +U2

0.5mv1^2 - GMm/r1 = 0.5mv2^2 - GMm/r2

0.5v1^2 - GM/r1 = 0.5v2^2 - GM/r2

M is mass of sun = 1.98*10^30 kg

G = 6.67*10^-11 N.m^2/kg^2

0.5*(9*10^4)^2 - (6.67*10^-11*1.98*10^30/(5*10^10)) = 0.5v2^2 - (6.67*10^-11*1.98*10^30/(6*10^12))

v2 = 5.35*10^4 m/s

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3 years ago

Electrons are accelerated in the picture tube of a television through potential difference of 8.00 * 10 ^ 3 V. (Use the values q

maria [59]

Answer:

$\lambda=1.37\times 10^{-11}\ m$

Explanation:

Give that,

The potential difference of the electrons, $V=8\times 10^{3}\ V$

We need to find the wavelength of the electrons.

Using the conservation of energy,

$2meV=\dfrac{h^2}{\lambda^2}\\\\\lambda=\sqrt{\dfrac{h^2}{2meV}}$

Put all the values,

$\lambda=\sqrt{\dfrac{(6.63\times 10^{-34})^2}{2\times 9.1\times 10^{-31}\times 1.6\times 10^{-19}\times 8\times 10^3}}\\\\\lambda=1.37\times 10^{-11}\ m$

So, the wavelength of the electrons is $1.37\times 10^{-11}\ m$ .

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3 years ago

A student places two books on a table. One book weighs less than the other book

Gnoma [55]

The book that weighs less

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2 years ago