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spayn [35]
4 years ago
15

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?
Physics
1 answer:
Alinara [238K]4 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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Answer:

The answer to the question is

The object would fall 57.625 m in the first 5 seconds

Explanation:

To solve the question, we note that

the height of fall = 490 ft = ‪149.352‬ m

Time to touch the ground = 7 seconds

We are required to find out how far the object falls in the first 5 seconds

We apply the relation

S = u·t + 0.5×g·t ² = We then have

‪149.352‬ = U×7+0.5*9.81*49 From where u = -13 m/s

Therefore to find how far it falls in the first 5 seconds, we have

-13*5 + 0.5*9.81*25 = 57.625 m

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3 years ago
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svlad2 [7]

A - the objects are too small

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How does an understanding of plate motion help scientists
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advance in the understanding of cellular movement

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6 0
4 years ago
Three identical train cars, coupled together, are rolling east at 1.8 m/s . A fourth car traveling east at 4.5 m/s catches up wi
Vedmedyk [2.9K]

Answer:

v = 1.98\ m/s

Explanation:

consider the mass of each train car be m

m₁ = m₂ = m₃ = m

speed of the three identical train

u₁ = u₂ = u₃ = 1.8 m/s

m₄ = m             u₄ = 4.5 m/s

m₅ = m              u₅ = 0 (initial velocity )

final velocity

v₁ = v₂ = v₃ = v₄ = v₅ = v

using conservation of momentum

m₁u₁ + m₂u₂ + m₃u₃ + m₄u₄ + m₅u₅ = m₁v₁ + m₂v₂ + m₃v₃ + m₄v₄ + m₅v₅

m (1.8 + 1.8 + 1.8 +4.5) = 5 m v

v = \dfrac{9.9}{5}

v = 1.98\ m/s

5 0
3 years ago
A horizontal force of 92.7 N is applied to a 40.5 kg crate on a rough, level surface. If the crate accelerates at 1.13 m/s2, wha
lord [1]

Answer:

The value is F_f =  46.935 \  N

Explanation:

From the question we are told that

    The  magnitude of the horizontal force is F  =  92.7 \  N

     The mass of the crate is  m  =  40.5 \  kg

     The acceleration of the crate is  a =  1.13 \ m/s

Generally the net force acting on the crate is mathematically represented as

       F_{net} =  F -  F_f =  ma

Here F_f is force of kinetic friction (in N) acting on the crate

      So  

            92.7  -  F_f =  40.5 * 1.13

=>         F_f =  46.935 \  N

5 0
3 years ago
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