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snow_tiger [21]
3 years ago
8

During an ultrasound, sound waves are sent by a transducer through muscle tissue at a speed of 1300 m/s. Some of the sound waves

are reflected from a metal fragment in the muscle and are detected by the transducer 1.5 x 10−4 seconds after they were first emitted. What is the depth of the metal fragment in the muscle tissue?
Physics
1 answer:
Simora [160]3 years ago
4 0

Answer:

0.195 m

Explanation:

Speed is distance moved per unit time, expressed as s=d/t and making d the subject of the formula then d=st

Where d is distance/depth moved, s is rhe speed of waves and t is time in seconds.

Substituting s with 1300 m/s and t with 0.00015 s then the depth of metal segment will be

D=1300*0.00015=0.195 m

Therefore, the depth is equivalent to 0.195 m

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If the sound level at a point is 98.0 dB, what is the intensity of sound at that point (Unit=W/m^2)? please help will give brain
Firlakuza [10]

Answer: The decibel scale is a logarithmic scale where each bel or 10 decibels correspondents to a factor of ten. A power intensity of 10^(-12) watts per square meter is the standard reference for a SPL of 0 db. So an SPL of 98 db corresponds to a power intensity of 10^(9.8)*10^(-12) or 10^(9.8–12) w/m^2.

0.006309573 w/m^2.

You can also readily find the value for any given SPL using the online calculator at: http://www.sengpielaudio.com/calculator-soundlevel.htm

Explanation:

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3 years ago
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vlabodo [156]
You earn swag points.
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A football wide receiver rushes 16 m straight down the playing field in 2.9 s (in the positive direction). He is then hit and pu
Bumek [7]

Answer:

a) v1 = 5.52m/s

b) v2 = -1.52m/s

c) v3 = 4.62m/s

d) vt = 3.85m/s

Explanation:

The velocity of the football wide receiver is his displacement per unit time.

Velocity v = (displacement d)/time t

v = d/t .....1

For each of the cases, equation 1 would be used to calculate the velocity.

a) v1 = d1/t1

d1= 16m

t1 = 2.9s

v1 = 16m/2.9s

v1 = 5.52m/s

b) v2 = d2/t2

d2 = -2.5m

t2 = 1.65s

v2 = -2.5/1.65

v2 = -1.52m/s

c) v3 = d3/t3

d3 = 24m

t3 = 5.2s

v3 = 24/5.2

v3 = 4.62m/s

d) vt = dt/tt

dt = 16m - 2.5m + 24m = 37.5m

tt = 2.9 + 1.65 + 5.2 = 9.75s

vt = 37.5/9.75

vt = 3.85m/s

5 0
4 years ago
A train whistle has a sound intensity level of 70. dB, and a library has a sound intensity level of about 40. dB. How many times
kodGreya [7K]

Answer:

The sound intensity of train is 1000 times greater than that of the library.

Explanation:

We have expression for sound intensity level,

            L=10log_{10}\left ( \frac{I}{I_0}\right )

A train whistle has a sound intensity level of 70 dB

We have

           70=10log_{10}\left ( \frac{I_1}{I_0}\right )

A library has a sound intensity level of about 40 dB

We also have

           40=10log_{10}\left ( \frac{I_2}{I_0}\right )

Dividing both equations

           \frac{70}{40}=\frac{10log_{10}\left ( \frac{I_1}{I_0}\right )}{10log_{10}\left ( \frac{I_2}{I_0}\right )}\\\\\frac{7}{4}=\frac{log_{10}\left ( \frac{I_1}{I_0}\right )}{log_{10}\left ( \frac{I_2}{I_0}\right )}\\\\10^7\frac{I_2}{I_0}=10^4\frac{I_1}{I_0}\\\\\frac{I_1}{I_2}=10^3=1000

The sound intensity of train is 1000 times greater than that of the library.

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3 years ago
Consider the following statement:
vovangra [49]

Answer:

The statement "The magnetic field of a magnet comes out of the north pole and goes into the south pole" is imprecise

Explanation:

This is because the zero divergence equation (∇ · B = 0 ) is valid for any magnetic field, even if it is time dependent rather than static. Physically, it means that there are no magnetic charges otherwise we would have ∇ · B ∝ ρmag instead of ∇ · B = 0. Consequently, the magnetic field lines never begin or end anywhere in space; instead they form closed loops or run from infinity to infinity.

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