Answer:
Amplitude = 8 Volts
Frequency = 0.067 kHz
Explanation:
Note: The missing picture in question is attached for your review.
Given:
Volts/Div = 2 V/div
Time/Div = 5 msec/div
Finding Amplitude:
Now, as you can see in the attached picture, there are 4 division between two peaks of the waveform, so,
(Multiplying by 2 V/div because oscilloscope dial is set at 2 V/div)
Finding Frequency:
As can be seen in attached picture, 3 division are there for one complete cycle of waveform,so,
Since,
Answer:
A
Explanation:
There are three basic forces in aerodynamics: acceleration, which moves an airplane forward; drag, which holds it back; and height, which keeps it airborne. Lift is generally explained by three theories: Bernoulli's principle, the Coanda effect, and Newton's third law of motion.
The time lapse between when the bat emits the sound and when it hears the echo is 0.05 s.
From the question given above, the following data were obtained:
Velocity of sound (v) = 343 m/s
Distance (x) = 8.42 m
Time (t) =?
We can obtain obtained the time as illustrated below:
v = 2x / t
343 = 2 × 8.42 / t
343 = 16.84 / t
Cross multiply
343 × t = 16.84
Divide both side by 343
t = 16.84/343
t = 0.05 s
Thus, the time between when the bat emits the sound and when it hears the echo is 0.05 s.
<h3>
How does a bat know how far away something is?</h3>
A bat emits a sound wave and carefully listens to the echoes that return to it. The returning information is processed by the bat's brain in the same way that we processed our shouting sound with a stopwatch and calculator. The bat's brain determines the distance of an object by measuring how long it takes for a noise to return.
Learn more about time elapses between when the bat emits the sound :
<u>brainly.com/question/16931690</u>
#SPJ4
Correction question:
A bat emits a sonar sound wave (343 m/s) that bounces off a mosquito 8.42 m away. How much time elapses between when the bat emits the sound and when it hears the echo? (Unit = s)
No,because they may have more particles
