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Feliz [49]
3 years ago
10

Blood flow rates in the umbilical cord can be found by measuring the Doppler shift of the ultrasound signal reflected by the red

blood cells. If the source emits a frequency f, what is the measured reflected frequency fR? Assume that all of the red blood cells move directly toward the source. Let c be the speed of sound in blood and v be the speed of the red blood cells.
Physics
1 answer:
lesya692 [45]3 years ago
5 0

Answer:

fR = f(c + v)/c

Explanation:

The speed of a wave is its frequency x wavelenght. Therefore,

Frequency is speed of wave over the wavelength.

Since the source (ultrasound machine) is stationary, and the receiver red blood cell is moving towards it. The wavelenght of the wave sent out towards the observer is c/f

The speed of the reflected sound wave is (c + v), so that the reflected frequency fR is given by

fR = f(c + v)/c

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210

Explanation:

A ball rolls horizontally off the cliff at a speed of 30 m/s. It takes 7 seconds for the ball to hit the ground. What is the height of the cliff and the horizontal distance traveled by the ball?

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S = v * t = 30 m/s * 7 s = 210 m for the horizontal distance, again assuming negligible air resistance.

5 0
2 years ago
You have a grindstone (a disk) that is 95.2 kg, has a 0.399 m radius, and is turning at 93 rpm, and you press a steel axe agains
olya-2409 [2.1K]

Answer:

angular acceleration is -0.2063  rad/s²

Explanation:

Given data

mass m = 95.2 kg

radius r = 0.399 m

turning ω = 93 rpm

radial force N  = 19.6 N

kinetic coefficient of friction  μ = 0.2

to find out

angular acceleration

solution

we know frictional force that is = radial force × kinetic coefficient of friction

frictional force = 19.6 × 0.2

frictional force = 3.92 N

and

we know moment of inertia  that is

γ =  I ×α = frictional force × r

so

γ  = 1/2 mr²α

α  = -2f /mr

α  = -2(3.92) /95.2 (0.399)

α  = - 7.84 / 37.9848 = -0.2063

so angular acceleration is -0.2063  rad/s²

3 0
3 years ago
You walk into a darkened room and turn on a flashlight. You see an image of the flashlight reflecting off a plane mirror in fron
user100 [1]

Answer:

4.2 is the answer

Explanation

The image formed in a plane mirror is an equal distance behind the mirror as the object in front of it.

Step 1: the equation to this problem would be: 8.4/2

Step 2: 8.4 ÷ 2 = 4.2

6 0
2 years ago
A charge 3q is at the origin, and a charge −2q is on the positive x axis at x=a. part a where on the x-axis would you place a th
Mila [183]

Answer:

x = a + \frac{a\sqrt2}{\sqrt3 - \sqrt2}

Explanation:

Position of charge 3q is x = 0

position of charge -2q is x = a

so here we know that

when two charges are of opposite nature then the electric field will be zero on the line joining the two charges at the position near to smaller magnitude charge

So here the electric field will be zero if the field due to 3q is counterbalanced by field due to -2q

so here we can say

\frac{k(3q)}{(r+a)^2} + \frac{k(-2q)}{r^2} = 0

\frac{3}{(r + a)^2} = \frac{2}{r^2}

\frac{r+a}{r} = \sqrt{\frac{3}{2}}

1 + \frac{a}{r} =\sqrt{\frac{3}{2}}

\frac{a}{r} = \frac{\sqrt3 - \sqrt2}{\sqrt2}

so we will have

r = \frac{a\sqrt2}{\sqrt3 - \sqrt2}

so the x coordinate of this position is given as

x = a + \frac{a\sqrt2}{\sqrt3 - \sqrt2}

6 0
3 years ago
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