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nydimaria [60]
3 years ago
13

Sperm whales, just like bats, use echolocation to find prey. A sperm whale’s vocal system creates a single sharp click, but the

emitted sound consists of several equally spaced clicks of decreasing intensity. Researchers use the time interval between the clicks to estimate the size of the whale that created them. Explain how this might be done.
Hint: The head of a sperm whale is complex, with air pockets at either end.
Physics
1 answer:
Tamiku [17]3 years ago
3 0

Answer: The size of the whale can be estimated by using echo Sounder or Sonar

Explanation:The researchers and physicst uses the knowledge of reflection of sound in water or echo to solve a lot of puzzle under water , including calclulating the distance and size of objects like fish,sunk ship,etc

The sonar is able to receive the signals from the whales and the succesive echoes that follows when placed in a strategic position in the water in the path of the whale .

The sonar has a traducers which has both a transmitter and a receiver,so it receives pulses from the whale and transmit back same echo back to the whale and to the bottom of the sea.

Using the speed of sound underwater and time taken for succesive echoes of pulses ,it can calculate with the precision the location of the whale while the image of the whale is also capture by the traducers to determine the size.

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If the error in the angle is 0.50, the error in sin 90o is​
Sindrei [870]

At the given erro in angle, the error in the measurement of sin 90 degrees would be 0.001.

<h3>Percentage error</h3>

The percentage error of any measurement is obtained from the ratio of the error to the actual measurement.

The error of sin 90 degrees is calculated as follows;

sin 90 = 1

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<h3>Error in sin 90 degrees</h3>

Error in sin 90 degrees = 1 - 0.999

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A parallel-plate capacitor stores charge Q. The capacitor is then disconnected from its voltage source, and the space between th
Stells [14]

Answer:

The relationship between the initial stored energy PE_{i} and the stored energy after the dielectric is inserted PE_{f} is:

c) PE_{f} =0.5\ PE_{i}

Explanation:

A parallel plate capacitor with C_{o} that is connected to a voltage source V_{o} holds a charge of Q_{o} =C_{o} V_{o}. Then we disconnect the voltage source and keep the charge Q_{o} constant . If we insert a dielectric of \kappa=2 between the plates while we keep the charge constant, we found that the potential decreases as:

                                                     V=\frac{V_{o}}{\kappa}

The capacitance is modified as:

                                              C=\frac{Q}{V} =\kappa\frac{Q_{o}}{V_{o}}=\kappa\ C_{o}

The stored energy without the dielectric is

                                               PE_{i}=\frac{1}{2}\frac{Q_{o}^{2}}{C_{o}}=\frac{1}{2}C_{o}V_{o}^{2}

The stored energy after the dielectric is inserted is:

                                               PE_{f}=\frac{1}{2}\frac{Q^{2}}{C}=\frac{1}{2}CV^{2}

If we replace in the above equation the values of V and C we get that

                                         PE_{f}=\frac{1}{2}\kappa\ C_{o}(\frac{V_{o}}{\kappa})^{2}=\frac{1}{\kappa}(\frac{1}{2}C_{o}V_{o}^{2})

                                                   PE_{f} =\frac{PE_{i}}{\kappa}

Finally

                                                  PE_{f} =0.5\ PE_{i}

                                               

                                     

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