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inna [77]
3 years ago
10

A Doppler blood flow unit emits ultrasound at 4.8 MHz. What is the frequency shift of the ultrasound reflected from blood moving

in an artery at a speed of 0.21 m/s?
Physics
1 answer:
butalik [34]3 years ago
6 0

Answer:

1300 Hz

Explanation:

f' = Actual frequency = 4.8 MHz

f = Observed frequency

v_r = Relative velocity of blood = 0.21 m/s

v = Velocity of ultrasound = 1540 m/s

From the Doppler effect formula we have

f=f'\dfrac{v+v_r}{v-v_r}\\\Rightarrow f=4.8\dfrac{1540+0.21}{1540-0.21}\\\Rightarrow f=4.80130\ Hz

Frequency shift is given by

\Delta f=f-f'\\\Rightarrow \Delta f=4.8013-4.8\\\Rightarrow \Delta f=0.0013\ MHz=1300\ Hz

The frequency shift in the sound is 1300 Hz

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Rashid [163]

The frictional force is too much compared to the gravitational force

Explanation:

This is because of the great amount of frictional force between the book and the wall.

Also, the force of gravity on the book is not high enough to cause motion of the book to slide downs.

Frictional force is a force that opposes motion.

  • For the book to slide down, it must overcome the force of friction acting on it opposite to the direction of gravity.
  • According to newton's first law 'a body will remain in its state of rest or uniform motion unless acted upon by an external force'.
  • Assuming the external force is gravitational force, clearly we see that the force is less than the frictional exerted on the book by pinning it to the wall.
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           Fr  >  fg

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fg is the gravitational force

learn more:

Newton laws brainly.com/question/11411375

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3 0
4 years ago
A paper-filled capacitor is charged to a potential difference of V 0 = 2.5 V V0=2.5 V . The dielectric constant of paper is κ =
forsale [732]

Answer:

V_1=9.25 V

Explanation:

We are given that

V_0=2.5 V

k=3.7

We have to find the new potential difference of the capacitor.

When the capacitor is disconnected then the charge stored in capacitor is constant.

When we introduce material of dielectric constant k between the plates of capacitor then the capacitance of capacitor increases k times.

C_0=\frac{Q}{V_0}

C'=kC=\frac{kQ}{V_1}

\frac{Q}{V_0}=\frac{kQ}{V_1}

V_0=\frac{V_1}{k}

Using the formula

V_0=\frac{V_1}{k}

V_1=V_0k=2.5\times 3.7=9.25 V

Hence, the new potential difference V_1=9.25 V

5 0
3 years ago
If you disconnect the wires from the battery and then reconnect them at the opposite ends of the battery, how does that change t
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Explanation:

The current from the battery always flows from the positive terminal of the battery to the negative terminal of the battery.

If we connect a red coloured wire to the positive terminal of the battery and black coloured wire to the negative terminal of the battery, and then reverses the wire to their respective terminals, then there is no change in the direction of flow of current. It does not matter that which wire is connected to the particular terminal. The current always flow from positive to negative terminal of the battery outside the battery.

6 0
3 years ago
The rating of a light bulb at 100 watts indicates
Slav-nsk [51]
That marking means that when the expected voltage is connected
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lubasha [3.4K]

Answer:

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Explanation:

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