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goldenfox [79]
3 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]3 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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4 years ago
Pete Zaria works on weekends at Barnaby's Pizza Parlor. His primary responsibility is to fill drink orders for customers. He fil
laila [671]

Answer:

W_n_e_t=7.648512 \approx 7.6J

K.E=0.8J

v=0.7844645406 \approx 0.78m/s

Explanation:

From the question we are told that

Mass of pitcher   M= 2.6kg

Force on pitcher f=8.8N

Distance traveled 48cm=>0.48m

Coefficient of friction \mu=0.28

a)Generally frictional force is mathematically given by

F=\mu N

F=0.28*2.6*9.8

F=7.1344N

Generally work done on the pitcher is mathematically given as

W_n_e_t=W_f+W_F

W_f=8.8*0.48=>4.224N\\W_F=7.1344*0.48=>3.424512N

W_n_e_t=4.224-3.424512

W_n_e_t=0.799488\approx 0.8J

b)Generally K.E can be given mathematically as

K.E= W_n_e_t

Therefore

K.E=0.8J

c)Generally the equation for kinetic energy is mathematically represented by

K.E=1/2mv^2

0.8=1/2mv^2

Velocity as subject

v=\sqrt{\frac{K.E*2}{m} }

v=\sqrt{\frac{0.8*2}{2.6} }

v=0.7844645406 \approx 0.78m/s

6 0
3 years ago
In a lightning bolt, a large amount of charge flows during a time of 1.2 x 10-3 s. Assume that the bolt can be represented as a
ohaa [14]

Answer: 10.58 C has flowed during the lightning bolt

Explanation:

Given that;

Time of flow t = 1.2 × 10⁻³

perpendicular distance r = 21 m

Magnetic field B = 8.4 x 10⁻⁵ T

Now lets consider the expression for magnetic field;

B = u₀I / 2πr

the current flow is;

I = ( B × 2πr ) / u₀

so we substitute

I = ( (8.4 x 10⁻⁵) × 2 × 3.14 × 21 ) / 4π ×10⁻⁷

=  0.01107792 / 0.000001256

= 8820 A

Hence the charge flows during lightning bolt  will be;

q = It

so we substitute

q = 8820 × 1.2 × 10⁻³

q = 10.58 C

therefore 10.58 C has flowed during the lightning bolt

6 0
3 years ago
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