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Greeley [361]
2 years ago
11

A ship's fathometer (an echo sounder) transmits a sound pulse and records the return of an echo 7.4 seconds later. If the speed

of sound in water is 1500 m/second, what is the water depth in meters
Physics
1 answer:
ira [324]2 years ago
6 0

Based on the velocity of sound in water,  the depth of the water is 5550 m.

<h3>What is the depth of the water if a transmitted sound pulse returns  an echo 7.4 seconds later?</h3>

The velocity of sound in water =  1500 m/s

Time taken to receive the echo = 7.4 seconds

  • Distance travelled by sound = speed * time

Distance = 1500 * 7.4

Distance = 11100 m

Since this is a to and fro distance covered

  • Depth of the water = distance travelled/ 2

Depth of water = 11100/2

Depth of water = 5550 m

Therefore, the depth of the water is 5550 m.

Learn more about echo and sound at: brainly.com/question/14090821

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The magnetic lines of force always travel from north to south true or false
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Explanation:

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Which term best describes the basic unit that makes up all matter?
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2 years ago
An airplane is moving at 350 km/hr. If a bomb is
bogdanovich [222]

Answers:

a) -171.402 m/s  

b) 17.49 s

c) 1700.99 m

Explanation:

We can solve this problem with the following equations:

y=y_{o}+V_{oy}t-\frac{1}{2}gt^{2} (1)

x=V_{ox}t (2)

V_{f}=V_{oy}-gt (3)

Where:

y=0 m is the bomb's final height

y_{o}=1.5 km \frac{1000 m}{1 km}=1500 m is the bomb's initial height

V_{oy}=0 m/s is the bomb's initial vertical velocity, since the airplane was moving horizontally

t is the time

g=9.8 m/s^{2} is the acceleration due gravity

x is the bomb's range

V_{ox}=350 \frac{km}{h} \frac{1000 m}{1 km} \frac{1 h}{3600 s}=97.22 m/s is the bomb's initial horizontal velocity

V_{f} is the bomb's final velocity

Knowing this, let's begin with the answers:

<h3>b) Time </h3>

With the conditions given above, equation (1) is now written as:

y_{o}=\frac{1}{2}gt^{2} (4)

Isolating t:

t=\sqrt{\frac{2 y_{o}}{g}} (5)

t=\sqrt{\frac{2 (1500 m)}{9.8 m/s^{2}}} (6)

t=17.49 s (7)

<h3>a) Final velocity </h3>

Since V_{oy}=0 m/s, equation (3) is written as:

V_{f}=-gt (8)

V_{f}=-(97.22)(17.49 s) (9)

V_{f}=-171.402 m/s (10) The negative sign only indicates the direction is downwards

<h3>c) Range </h3>

Substituting (7) in (2):

x=(97.22 m/s)(17.49 s) (11)

x=1700.99 m (12)

5 0
3 years ago
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