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>
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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)
Answer:
At time 10.28 s after A is fired bullet B passes A.
Passing of B occurs at 4108.31 height.
Explanation:
Let h be the height at which this occurs and t be the time after second bullet fires.
Distance traveled by first bullet can be calculated using equation of motion
Here s = h,u = 450m/s a = -g and t = t+3
Substituting
Distance traveled by second bullet
Here s = h,u = 600m/s a = -g and t = t
Substituting
Solving both equations
So at time 10.28 s after A is fired bullet B passes A.
Height at t = 7.28 s
Passing of B occurs at 4108.31 height.
Answer:
wavelength = v/f or wavelength equals to velocity over frequency
frequency= v/w or velocity over wavelength
frequency= 1/p or one over period or time
Answer:
f = 2858.33 Hz
Explanation:
given,
distance between speaker (A) and the person = 2.34 m
Distance between speaker (B) and the person is AB =
=
= 2.46 m
path difference d = BP - AP
= 2.46 - 2.34 m
= 0.12 m
now, λ = 0.12
speed of sound = 343 m/s
f = 2858.33 Hz
the lowest frequency that will produce constructive interference is equal to
f = 2858.33 Hz