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4 years ago

7

The Doppler effect using ultrasonic waves of frequency 2.25 × 106 Hz is used to monitor the heartbeat of a fetus. A (maximum) be

at frequency of 240 Hz is observed. Assuming that the speed of sound in tissue is 1540 m/s, calculate the maximum velocity of the surface of the beating heart.

1 answer:

miss Akunina [59]4 years ago

4 0

Answer:

$v = 2.88 \times 10^7 m/s$

Explanation:

As per Doppler's effect we know that the frequency of the sound that is observed by the detector is the reflected sound

This reflected sound is given as

$f' = f (\frac{v - v_h}{v + v_h})$

$f' = 2.25\times 10^6(\frac{v - 1540}{v + 1540})$

so we know that the beat frequency is

$\Delta f = 240 Hz$

so we will have

$f - f' = \Delta f$

$2.25 \times 10^6 - 2.25\times 10^6(\frac{v - 1540}{v + 1540}) = 240$

$1 - (\frac{v - 1540}{v + 1540}) = 1.07 \times 10^{-4}$

so we have

$0.99989 = (\frac{v - 1540}{v + 1540})$

$1.99989\times 1540 = 1.067 \times 10^{-4} v$

$v = 2.88 \times 10^7 m/s$

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t = 3.0 s is the time taken

a is the acceleration

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6. 62.5 m

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and the mass of the car

$m=1000 kg$

so we can find its acceleration:

$a=\frac{F}{m}=\frac{-5000 N}{1000 kg}=-5 m/s^2$

So now we can find the minimum distance to stop by using the equation

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where in this case we have

v = 0 m/s is the final speed

u = 25 m/s is the initial speed

a = -5 m/s^2 is the acceleration

d is the distance

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$d=\frac{v^2-u^2}{2a}=\frac{0^2-(25 m/s)^2}{2(-5 m/s^2)}=62.5 m$

7a. 14 m/s

We can solve the problem by using the law of conservation of energy: in fact, the initial gravitational potential energy of the person is all converted into kinetic energy as she hits the water below

$mgh=\frac{1}{2}mv^2$

where

m = 65 kg is the mass of the person

g = 9.8 m/s^2

h = 10 m is the initial height of the diver

v is the final speed as she enters the water

Solving for v, we find

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7b. -3185 N

We need to find the acceleration of the diver during the motion of 2.0 m below the water:

$v^2-u^2 = 2ad$

where

v = 0 is the final speed

u = 14 m/s is the initial speed as she enters the water

a is the acceleration

d = 2.0 m is the distance covered

Solving for a,

$a=\frac{v^2-u^2}{2d}=\frac{0^2-(14 m/s)^2}{2(2.0 m)}=-49 m/s^2$

And so now we can find the net force acting on the diver

$F=ma=(65 kg)(-49 m/s^2)=-3185 N$

8a. -133 m/s^2

The acceleration of the passenger is given by

$a=\frac{v-u}{t}$

where

v = 0 m/s is the final speed

u = 13.3 m/s is the initial speed

t = 0.10 s is the time interval

Solving for a, we find

$a=\frac{0-13.3 m/s}{0.10 s}=-133 m/s^2$

8b. 3325 N

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So, the magnitude of the force will be

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8c. 245 N

The weight of the child is given by

$W=mg$

where

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g = 9.8 m/s^2 is the acceleration due to gravity

Solving the equation,

$W=(25 kg)(9.8 m/s^2)=245 N$

8d. 55.1 lb

Since we know that

$1 lb = 4.45 N$

We can find the weight in pounds by setting the following proportion

$1 lb : 4.45 N = x : 245 N$

Solving for x,

$x=\frac{(1 lb)(245 N)}{4.45 N}=55.1 lb$

8e. Chances are very low

The force that the driver should exert on the child is

F = 3325 N

This force is equivalent to the force required to lift an object of mass m:

$F=mg\\m=\frac{F}{g}=\frac{3325 N}{9.8 m/s^2}=339.3 kg$

So, it is equivalent to the force required to lift an object of 339.3 kg, which is quite a lot. therefore, the changes are very low.

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