Answer:
308,000 or 30.8×10^3
Explanation:
v=f×lamda
v is ?, f is 875Hz, lamda is 352m
v=875×352
v=308,000
v=30.8×10^3 m/s
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)
Answer:
, it flows through your community's sanitary sewer system to a wastewater treatment facility.
Answer:
C) 1.0 m
Explanation:
The component of the velocity parallel to the sidewalk is:
vₓ = v cos θ
vₓ = 0.1 m/s cos 45°
vₓ = 0.0707 m/s
The distance traveled after 14 seconds is:
d = vₓ t
d = (0.0707 m/s) (14 s)
d = 0.99 m
Closest answer is C) 1.0 m.
