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Ludmilka [50]
3 years ago
7

A certain ultrasound device can measure a fetal heart rate as low as 50 beats per minute. This corresponds to the surface of the

heart moving at about 4.0 x 10-4 m/s. If the probe generates ultrasound that has a frequency of 2.0 MHz (1 MHz = 1 megahertz = 106 Hz), what frequency shift must the machine be able to detect
Physics
1 answer:
balu736 [363]3 years ago
7 0

Answer:

The. Machine must detect a shift of

1 Hz

Explanation:

Frequency shift is given as

={ ( Vsound +V/ V sound -V) -1}f emitted

So by substitution we have

= { 1540+4E-4/1540-4E)-1) 2*10^6

= 1Hz

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A kangaroo can jump over an object 2.46 m high. (a) Calculate its vertical speed (in m/s) when it leaves the ground.
Elenna [48]

Answer:

a) 6.95 m/s

b) 1.42 seconds

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.81 m/s²

v^2-u^2=2as\\\Rightarrow -u^2=2as-v^2\\\Rightarrow -u^2=2\times -9.81\times 2.46-0^2\\\Rightarrow u=\sqrt{2\times 9.81\times 2.46}\\\Rightarrow u=6.95\ m/s

a) The vertical speed when it leaves the ground. is 6.95 m/s

v=u+at\\\Rightarrow t=\frac{v-u}{a}\\\Rightarrow t=\frac{0-6.95}{-9.81}\\\Rightarrow t=0.71\ s

Time taken to reach the maximum height is 0.71 seconds

s=ut+\frac{1}{2}at^2\\\Rightarrow 2.46=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{2.46\times 2}{9.81}}\\\Rightarrow t=0.71\ s

Time taken to reach the ground from the maximum height is 0.71 seconds

b) Time it stayed in the air is 0.71+0.71 = 1.42 seconds

3 0
3 years ago
Please help will mark Brainliest the
VLD [36.1K]

Answer: 100 units

Explanation:

4 0
3 years ago
A child goes down the slide,starting from rest. If the length of the slide is 2m and it takes the child 3 seconds to go down the
lapo4ka [179]

Answer:

0.44 m/s^2

Explanation:

We have the following data:

- distance covered by the child: d = 2 m (length of the slide)

- time taken to cover this distance: t = 3 s

- initial velocity of the child: 0 m/s (he starts from rest)

So we can find the acceleration by using the equation:

d=ut+\frac{1}{2}at^2

Where a is the acceleration.

Substituting the values and solving for a,

a=\frac{2d}{t^2}=\frac{2(2)}{3^2}=0.44 m/s^2

3 0
3 years ago
Martha wants to calculate an object's velocity. What will she need to do?​
rjkz [21]

Answer:

you divide the distance by the time it takes to travel that same distance, then you add your direction to it.

4 0
2 years ago
A silver wire 2.6 mm in diameter transfers a charge of 420 C in 80 min. Silver contains 5.8 x 10- free electrons per cubic meter
kifflom [539]

Answer:

a). 87.5 mA or 87.5 x10^{-3}A

b). 1.78 \frac{m}{s}

Explanation:

d=2.6 mm \\Q=420C\\t=80min\\n=5.8x10^{28} \\q=1.6x10^{-19}

n the number of free electrons is 28 in text reference and if they don't give q is take as the charge of electron.

a).

I=\frac{Q}{t}\\ I= \frac{420 C}{80 min}*\frac{1min}{60 s} =\frac{420 C}{4800s}\\  I=87.5 x10^{-3}A

b).

I=n*abs (q)*V_{d}*A

A= \pi * (\frac{d}{2})^{2} \\A=\pi (*\frac{2.6x10^{-3} m}{2})^{2}  \\A=5.309x10^{-6}

V_{d} =\frac{I}{n*abs(q)*A} \\V_{d}=\frac{87.5 x10^{-2} }{5.8x^10{28} *1.6x^{-19} *5.3x^{6} }\\V_{d}=1.78 \frac{m}{s}

8 0
3 years ago
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