villarrealc1987

villarrealc1987

29.11.2019 •

Physics

Adoppler flow meter is used to measure the speed of blood flow. transmitting and receiving elements are placed on the skin, as shown in the figure below. typical sound-wave frequencies of about 5.0 mhz are used, which have a reasonable chance of being reflected from red blood cells. by measuring the frequency of the reflected waves, which are doppler-shifted because the red blood cells are moving, the speed of the blood flow can be deduced. "normal" blood flow speed is about 0.1 m/s. suppose that an artery is partly constricted, so that the speed of the blood flow is increased, and the flow meter measures a doppler shift of 780 hz.

what is the speed of blood flow in the constricted region? the effective angle between the sound waves (both transmitted and reflected) and the direction of blood flow is 45o, assume the velocity of sound in tissue is 1540 m/s.

Solved

Show answers

Ответ:

purplefish53

07.06.2020

$-25.2 kg m/s^{2}$

Explanation:

The missing question her is: What is the change in momentum of the baseball?

According to the Impulse-Momentum Theorem, we have:

$F \Delta t= \Delta p$

Where:

is the exerted force

$\Delta t$ is the time interval

$\Delta p=p_{2}-p_{1}$ is the change in momentum

So, in the case of the bat and the ball we have:

For the bat:

$F_{bat} \Delta t=p_{fbat}-p_{obat}$ (1)

For the ball:

$F_{ball} \Delta t=p_{fball}-p_{oball}$ (2)

Since $\Delta t$ is the same for both the bat and the ball, their impulses are equal in magnitude but in opposite direction according to Newton's 3rd law of motion. Hence:

$F_{bat} = -F_{ball}$ (3)

Then:

$p_{fbat}-p_{obat}=-(p_{fball}-p_{oball})$ (4)

Where:

$p_{fbat}=m_{bat} V_{fbat}$

$p_{obat}=m_{bat} V_{obat}$

Being $m_{bat}=1.8 kg$ , $V_{fbat}=28 m/s$ , $V_{obat}=42 m/sm/s$

$p_{fball}=m_{ball} V_{fball}$

$p_{oball}=m_{ball} V_{oball}$

Being $m_{ball}=0.2 kg$

Then (4) is rewritten as:

$m_{bat}(V_{fbat} - V_{obat})=-m_{ball}(V_{fball}-V_{oball})$ (5)

Since $V_{fball}-V_{oball}=\Delta V_{ball}$ :

$m_{bat}(V_{fbat} - V_{obat})=-m_{ball}\Delta V_{ball}$ (6)

Isolating $\Delta V_{ball}$ :

$\Delta V_{ball}=\frac{m_{bat}(V_{fbat} - V_{obat})}{-m_{ball}}$ (7)

$\Delta V_{ball}=\frac{1.8 kg(28 m/s - 42 m/s)}{-0.2 kg}$ (8)

$\Delta V_{ball}=126 m/s$ (9)

Now, the change in momentum of the baseball is:

$\Delta p_{ball}=-m_{ball}\Delta V_{ball}$ (10)

Substituting (9) in (10):

$\Delta p_{ball}=-0.2 kg(126 m/s)$ (11)

Finally:

$\Delta p_{ball}=-25.2 kgm/s^{2}$ This is the change in momentum of the baseball. Note it is negative because its initial velocity is greater than its final velocity

0,0(0 оценок)

Ask your question to the AI advisor

Ai-bot is an expert in any field and is the perfect companion for reliable and useful answers and advice on a variety of topics, including science, history, technology, art, sports, health, culture and more.

Ask your question

More tips

Answers on questions: Physics

Ask an AI advisor a question

I want advice