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slamgirl [31]
3 years ago
7

A medical ultrasound imaging system sends out a steady stream of very short pulses. To simplify analysis, the reflection of one

pulse should be received before the next is transmitted.
If the system is being used to create an image of tissue 15 cm below the skin, what is the minimum time between pulses?

Also how many pulses per second does this correspond to?
Physics
1 answer:
arlik [135]3 years ago
5 0

Answer:

Minimum time = 1.95x10^-4 s

Number of pulses = 5128.21 pulses/s

Explanation:

We have the speed of sound waves through human tissue with a value of 1540 m/s, to calculate the time it takes for the pulse to travel a distance of 30 cm (since the pulse will first travel a distance of 15 cm and then it will return another 15 cm to be detected by the equipment), therefore, the time between the two pulses will be equal to:

tminimum = 0.30 m/1540 m/s = 1.95x10^-4 s

To calculate the number of pulses, one second must be divided over the minimum time between the two pulses, as follows:

npulses = 1 s/1.95x10^-4 s = 5128.21 pulses/s

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A projectile is launched at an angle of 36.7 degrees above the horizontal with an initial speed of 175 m/s and lands at the same
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Answer:

a) The maximum height reached by the projectile is 558 m.

b) The projectile was 21.3 s in the air.

Explanation:

The position and velocity of the projectile at any time "t" is given by the following vectors:

r = (x0 + v0 · t · cos α, y0 + v0 · t · sin α + 1/2 · g · t²)

v = (v0 · cos α, v0 · sin α + g · t)

Where:

r = position vector at time "t"

x0 = initial horizontal position

v0 = initial velocity

t = time

α = launching angle

y0 = initial vertical position

g = acceleration due to gravity (-9.80 m/s² considering the upward direction as positive).

v = velocity vector at time t

a) Notice in the figure that at maximum height the velocity vector is horizontal. That means that the y-component of the velocity (vy) at that time is 0. Using this, we can find the time at which the projectile is at maximum height:

vy = v0 · sin α + g · t

0 = 175 m/s · sin 36.7° - 9.80 m/s² · t

-  175 m/s · sin 36.7° /  - 9.80 m/s² = t

t = 10.7 s

Now, we have to find the magnitude of the y-component of the vector position at that time to obtain the maximum height (In the figure, the vector position at t = 10.7 s is r1 and its y-component is r1y).

Notice in the figure that the frame of reference is located at the launching point, so that y0 = 0.

y = y0 + v0 · t · sin α + 1/2 · g · t²

y = 175 m/s · 10.7 s · sin 36.7° - 1/2 · 9.8 m/s² · (10.7 s)²

y = 558 m

The maximum height reached by the projectile is 558 m

b) Since the motion of the projectile is parabolic and the acceleration is the same during all the trajectory, the time of flight will be twice the time it takes the projectile to reach the maximum height. Then, the time of flight of the projectile will be (2 · 10.7 s) 21.4 s. However, let´s calculate it using the equation for the position of the projectile.

We know that at final time the y-component of the vector position (r final in the figure) is 0 (because the vector is horizontal, see figure). Then:

y = y0 + v0 · t · sin α + 1/2 · g · t²

0 = 175 m/s · t · sin 36.7° - 1/2 · 9.8 m/s² · t²

0 = t (175 m/s ·  sin 36.7 - 1/2 · 9.8 m/s² · t)

0 = 175 m/s ·  sin 36.7 - 1/2 · 9.8 m/s² · t

-  175 m/s ·  sin 36.7 / -(1/2 · 9.8 m/s²) = t

t = 21.3 s

The projectile was 21.3 s in the air.

7 0
3 years ago
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