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3 years ago

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An echocardiogram uses 4.4 MHz ultrasound to measure blood flow in the aorta. The blood is moving away from the probe at 1.4 m/s

. What is the frequency shift of the reflected ultrasound?

1 answer:

Alinara [238K]3 years ago

7 0

Answer:

$\Delta f=17959.2\ Hz$

Explanation:

Given:

original frequency of the wave, $f=4.4\times 10^6\ Hz$

velocity of the blood stream moving away, $v=1.4\ m.s^{-1}$

<u>Doppler shift is given as:</u>

$\frac{f}{f_o} =\frac{s+v_s}{s-v_o}$

where:

s = velocity of ultrasound in blood = $1570\ m.s^{-1}$

$v_s=$ velocity of sound source towards the observer

$v_o=$ velocity of the observer away from the source

$\frac{4.4\times 10^6}{f_o} =\frac{1570+0}{1570-1.4}$

$f_o=4396076.43\ Hz$

Therefore the shift in frequency:

$\Delta f=f-f_o$

$\Delta f=3923.56\ Hz$

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A sample of helium has a volume of 12.7 m3. The temperature is raised to 323 K at which time the gas occupies 32.5 m3? Assume pr

jasenka [17]

Answer: The original temperature was

$T_{1}=126.51K$

Explanation:

Let's put the information in mathematical form:

$V_{1}=12.7m^{3}$

$T_{1}=?$

$V_{2}=32.5m^{3}$

$T_{2}=323K$

$P_{1}=P_{2}=3atm$

If we consider the helium as an ideal gas, we can use the Ideal Gas Law:

$PV=nRT$

were <em>R</em> is the gas constant. And <em>n</em> is the number of moles (which we don't know yet)

From this, taking $R=0.08205746\frac{atm.l}{mol.K}$ , we have:

$n=\frac{P_{2}V_{2}}{RT_{2}}$

⇒ $n=3.67mol$

Now:

$T_{1}=\frac{P_{1}V_{1}}{nR}$

⇒ $T_{1}=126.51K$

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2 years ago

Read 2 more answers

lanet R47A is a spherical planet where the gravitational acceleration on the surface is 3.45 m/s2. A satellite orbitsPlanet R47A

qaws [65]

$2.6×10^6\:\text{m}$

Explanation:

The acceleration due to gravity g is defined as

$g = G\dfrac{M}{R^2}$

and solving for R, we find that

$R = \sqrt{\dfrac{GM}{g}}\:\:\:\:\:\:\:(1)$

We need the mass M of the planet first and we can do that by noting that the centripetal acceleration $F_c$ experienced by the satellite is equal to the gravitational force $F_G$ or

$F_c = F_G \Rightarrow m\dfrac{v^2}{r} = G\dfrac{mM}{r^2}\:\:\:\:\:(2)$

The orbital velocity <em>v</em> is the velocity of the satellite around the planet defined as

$v = \dfrac{2\pi r}{T}$

where <em>r</em><em> </em>is the radius of the satellite's orbit in meters and <em>T</em> is the period or the time it takes for the satellite to circle the planet in seconds. We can then rewrite Eqn(2) as

$\dfrac{4\pi^2 r}{T^2} = G\dfrac{M}{r^2}$

Solving for <em>M</em>, we get

$M = \dfrac{4\pi^2 r^3}{GT^2}$

Putting this expression back into Eqn(1), we get

$R = \sqrt{\dfrac{G}{g}\left(\dfrac{4\pi^2 r^3}{GT^2}\right)}$

$\:\:\:\:=\dfrac{2\pi}{T}\sqrt{\dfrac{r^3}{g}}$

$\:\:\:\:=\dfrac{2\pi}{(1.44×10^4\:\text{s})}\sqrt{\dfrac{(5×10^6\:\text{m})^3}{(3.45\:\text{m/s}^2)}}$

$\:\:\:\:= 2.6×10^6\:\text{m}$

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2 years ago

The head of a rattlesnake can accelerate at 49 m/s2 in striking a victim. If a car could do as well, how long would it take to r

Anettt [7]

<h2>Time taken is 0.459 seconds</h2>

Explanation:

We have equation of motion v = u + at

Initial velocity, u = 0 m/s

Final velocity, v = 81 km/hr = 22.5 m/s

Time, t = ?

Acceleration, a = 49 m/s²

Substituting

v = u + at

22.5 = 0 + 49 x t

t = 0.459 seconds

Time taken is 0.459 seconds

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3 years ago

Does most psychologists believe that ESP exist

vodomira [7]

Answer:

no

Explanation:

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2 years ago

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AysviL [449]

I think the answer is D

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