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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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The emf is induced in the wire will be 1.56 ×10 ⁻³ V. The induced emf is the product of the magnetic field,velocity and length of the wire.

<h3>What is induced emf?</h3>

Emf is the production of a potential difference in a coil as a result of changes in the magnetic flux passing through it.

When the flux coupling with a conductor or coil changes, electromotive Force, or EMF, is said to be induced.

The given data in the problem is;

B is the magnitude of the magnetic field,= 5.0 ×10⁻⁵ T

V(velocity)=125 M/SEC

L(length)=25 cm=0.25 m

The maximum emf is found as;

E=VBLsin90°

E=125 × 5.0 × 10⁻⁵ ×0.25

E=1.56 ×10 ⁻³ V

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brainly.com/question/16764848

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2 years ago
A nasa spacecraft measures the rate r of at which atmospheric pressure on mars decreases with altitude. the result at a certain
Lesechka [4]

Answer:4.21 \times 10^{-10} J/cm^4

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1kPa/km=0.01 Pa/cm

1kPa/km=10^{-8} J/cm^4

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3 0
3 years ago
Read 2 more answers
A stone is dropped into a well. The sound of the splash is heard 3.5 seconds later. What is the depth of the well? Take the spee
Naddika [18.5K]

Answer:

The depth of the well, s = 54.66 m

Given:

time, t = 3.5 s

speed of sound in air, v = 343 m/s

Solution:

By using second equation of motion for the distance traveled by the stone when dropped into a well:

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

Since, the stone is dropped, its initial velocity, u = 0 m/s

and acceleration is due to gravity only, the above eqn can be written as:

s = \frac{1}{2}gt'^{2}

s = \frac{1}{2}9.8t^{2} = 4.9t'^{2}                     (1)

Now, when the sound inside the well travels back, the distance covered,s is given by:

s = v\times t''

s = 343\times t''                                              (2)

Now, total time taken by the sound to travel:

t = t' + t''

t'' = 3.5 - t'                                                                        (3)

Using eqn (2) and (3):

s = 343(3.5 - t')                                                                 (4)

from eqn (1) and (4):

4.9t'^{2} = 343(3.5 - t')

4.9t'^{2} + 343t' - 1200.5 = 0

Solving the above quadratic eqn:

t' = 3.34 s

Now, substituting t' = 3.34 s in eqn (2)

s = 54.66 m

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3 years ago
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d1i1m1o1n [39]

If "0.3 minute" is correct, then it's 9,543,272 Joules.

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telo118 [61]

Answer:

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Explanation:

The orbital velocity of the vessel around the planet can be found by equalizing the force of gravity between the vessel and the planet and the centripetal force:

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where

G is the gravitational constant

m is the mass of the vessel

M is the mass of the planet

r is the distance between the vessel and the centre of the planet

v is the orbital velocity of the vessel

Re-arranging the formula, we find an expression for v:

v=\sqrt{\frac{GM}{r}}

We see that:

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- the bigger the distance between the vessel and the planet, r, the smaller the velocity

So, the correct choices that increase the orbital velocity are:

- Bigger mass of planet B  

- orbiting closer to planet B

6 0
3 years ago
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