Solve for Echo Time in Saltwater and Air - Physics Problem

  • Thread starter Thread starter bilalsyed25
  • Start date Start date
  • Tags Tags
    Physics
AI Thread Summary
The discussion revolves around calculating the additional time it takes for an echo to return in air after a sound travels through saltwater. The captain's horn sounds, and the echo in water takes 0.40 seconds to return, leading to a calculated distance of 294 meters. The time for the echo to return in air is then calculated as 0.86 seconds, resulting in an additional time of 0.46 seconds compared to the water echo. Some participants debate the calculations and mention discrepancies found online regarding the distance used. Ultimately, the conversation highlights the importance of accurate distance measurement and time calculation in physics problems.
bilalsyed25
Messages
1
Reaction score
0
The Question:
A boat is floating at rest in dense fog near a large cliff. The captain sounds a horn at water level and the sound travels through the salt water (1470 m/s) and the air (340 m/s) simultaneously. The echo in the water takes 0.40s to return. How much additional time will it take the echo in the air to return?

Relevent equations: v=d/t

My answer:
d = v∆t

d = 1470*0.40

d = 588/2 (I checked online for this question, some ppl got different answers due to this part; 558/2 or just 588?)

d = 294m

294/340 = 0.86s

0.86-0.40 = 0.46

Additional time = 0.46s
 
Physics news on Phys.org
bilalsyed25 said:
d = 588/2 (I checked online for this question, some ppl got different answers due to this part; 558/2 or just 588?)

d = 294m
Realize that the sound must travel to the cliff and back.
 
bilalsyed25 said:
The Question:
A boat is floating at rest in dense fog near a large cliff. The captain sounds a horn at water level and the sound travels through the salt water (1470 m/s) and the air (340 m/s) simultaneously. The echo in the water takes 0.40s to return. How much additional time will it take the echo in the air to return?

Relevent equations: v=d/t

My answer:
d = v∆t

d = 1470*0.40

d = 588/2 (I checked online for this question, some ppl got different answers due to this part; 558/2 or just 588?)

d = 294m

294/340 = 0.86s

0.86-0.40 = 0.46

Additional time = 0.46s

If something takes ##0.4s## to cover a distance and something else takes an extra ##0.46s## to cover the same distance, then the faster thing must be moving approx twice as fast as the slower thing. Is ##1,470m/s## approx twice ##340 m/s##?

Using this logic can you estimate the answer (approx) in your head?
 
Okay, The distance from water is 294 . But for time for echo via air to be heard , t: (2*294)/340 ; 1.73 so addirional time is 1.73-0.4; 1.3seconds
 
bilalsyed25 said:
The Question:
A boat is floating at rest in dense fog near a large cliff. The captain sounds a horn at water level and the sound travels through the salt water (1470 m/s) and the air (340 m/s) simultaneously. The echo in the water takes 0.40s to return. How much additional time will it take the echo in the air to return?

Relevent equations: v=d/t

My answer:
d = v∆t

d = 1470*0.40

d = 588/2 (I checked online for this question, some ppl got different answers due to this part; 558/2 or just 588?)

d = 294m

294/340 = 0.86s

0.86-0.40 = 0.46

Additional time = 0.46s
Why you didnt multiply the distance 294m when finding the time through the air?
 
Nestory said:
Why you didnt multiply the distance 294m when finding the time through the air?
:welcome:

This thread is three years old and the OP has hopefully graduated by now.
 
Thread 'Collision of a bullet on a rod-string system: query'

Thread 'Collision of a bullet on a rod-string system: query'

In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
View full post »
Thread 'A cylinder connected to a hanged mass'

Thread 'A cylinder connected to a hanged mass'

Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
View full post »

Similar threads

Calculate Additional Time for Echo in Air to Return | Physics Problem
Replies
10
Views
5K
Solve Physics Echo Problem: Air Echo Returns 1.33s Later
Replies
3
Views
2K
Calculating Echo Time for Sound in Water and Air
Replies
1
Views
3K
Are the 2nd and 3rd problems in this video correctly solved? (charged dipole and organ pipe problems)
Replies
3
Views
963
Ship and Iceberg - Speed of sound question
Replies
4
Views
2K
Problem set with air conditioners and thermodynamics
Replies
5
Views
2K
Solving the Watermelon Drop Physics Problem: Step-by-Step Guide
Replies
3
Views
3K
Solving a Physics Problem: Calculating Time on the Ground & in Air
Replies
2
Views
2K

Hot Threads

Recent Insights

Back
Top