How Far Does a Bat's Echo Travel in a Cave?

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To determine how far a bat's echo travels in a cave, the speed of sound in air is approximately 340 m/s. The total time for the echo to return is 2.9 seconds, meaning the sound travels to the cave wall and back. Using the formula v = d/t, the distance can be calculated by rearranging the equation to d = vt. Since the echo travels to the wall and back, the distance to the wall is half of the total distance calculated. The final distance to the cave wall is 493 meters.
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Need help with a SOUND question~!><

A bat flying in a cave emits a sound pulse and receives its echo in 2.9 s. How far away is the cave wall?
 
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Do you know of an equation relating speed, distance and time?

What is the speed of sound waves in a 'cave' (i.e in air)?

Don't forget that the sound needs to travel to the cave wall AND back
 


v=d/t
340m/s
 


And you can't solve it from here?
 


So, do you take 2.9 times by 2??
 


re-arrange your velocity equation, plug in the numbers and solve for distance. Just know, that the distance you end up with will be the TOTAL distance the sound waves traveled, from the bat to the wall and back to the bat.
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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