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goldenfox [79]
2 years ago
5

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 3.0 cm into the muscle tissue. How long did it take the transducer to detect the reflected waves from the metal fragment after they were first emitted?
23 seconds
39 seconds
2.3 x 10−5 seconds
3.9 x 10−5 seconds
Physics
1 answer:
erica [24]2 years ago
3 0

Answer: 2.3(10)^{-5} s

Explanation:

The velocity V of the sound wave in the muscle tissue. which is the same as the reflected from a metal fragment is:

V=\frac{d}{t}

Where:

V=1300 m/s

d=3 cm \frac{1m}{100 cm}=0.03 m is the distance traveled by the sound wave

t is the time

Isolating t:

t=\frac{d}{V}

t=\frac{0.03 m}{1300 m/s}

Finally:

t=2.3(10)^{-5} s

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A transformer has a primary coil with 375 turns of wire and a secondary coil with 1,875 turns. An AC voltage source connected ac
Sonbull [250]

Answer:

The rms voltage (in V) measured across the secondary coil is 459.62 V

Explanation:

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number of turns in the secondary coil, Ns = 1875 turns

peak voltage across the primary coil, Ep = 130 V

peak voltage across the secondary coil, Es = ?

\frac{N_P}{N_s} = \frac{E_p}{E_s} \\\\E_s = \frac{N_sE_p}{N_p} \\\\E_s = \frac{1875*130}{375} \\\\E_s = 650 \ V

The rms voltage (in V) measured across the secondary coil is calculated as;

V_{rms} = \frac{V_0}{\sqrt{2} } = \frac{E_s}{\sqrt{2} } \\\\V_{rms} = \frac{650}{\sqrt{2} } = 459.62 \ V

Therefore, the rms voltage (in V) measured across the secondary coil is 459.62 V

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