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The MGC Chatbox Evo is really easy to use

You can :
  1. Send chats : you just have to enter the text in the dedicated input field and then to validate it by pushing the return key ou by clicking on the OK button.
  2. Format the chats :
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  5. Edit your chats (or others chats) : the edition of a chat can be done by a simple double-click on it.*
  6. Use specific commands : some commands can be accessible in the chatbox depending on the will of the administrator(s) of your forum. These commands let you execute specific actions of formatting, management, etc... You will herebelow the information on the command which are accessible to you. You must notice that you are not forced to use the command prefixes (/command_name) when you use them in the channel they are attached to (except in the general channel).*
* : Depending on the settings chosen by the administrator(s) of the forum, you might not have access to all these features.
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  1. #1
    Junior Member Math Guy's Avatar
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    A boat is floating at rest in dense fog near a large cliff...How much...

    ...additional time will it take? Here's the full 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 and the air (340 m/s) simultaneously. The echo in the water takes 0.40 s to return. How much additional time will it take the echo in the air to return?

    This is my attempt:

    First I find the distance by using the ?d = V(?t) and then divide the distance by 2 since it takes twice the distance for the echo to come back, so I have a distance of 65M. Then I would this distance and the given velocity of the sound-wave to find the time it takes to reach where the sound bounces back and have a value ?t = 0.20 seconds. Then I add both, to get a total value of ?t = 0.60 seconds. Is this the answer? Or am I wrong?

  2. #2
    Unregistered
    Guest
    I am seeking the answer to this question as well, I may have mucked it up a bit though here is my attempt:

    If the speed of sound in salt water is 1470m/s (as stated in the lesson I'm taking), then the answer is that it takes 1.32 seconds longer for the sound to travel through the air than the sea water.

    How I got my answer:

    The time, 0.4s, multiplied by the speed of sound in salt/sea water, 1470m/s, is 588m. Now, divide the distance, 588m, by the velocity, 340m/s, and the answer is 1.72s, now minus the time it takes for the sound to travel through the salt water, 0.4s, and the answer is 1.32 s longer.

    I am trying to make sure my answers are correct and my calculations were done correctly, I hope someone comments and corrects us!


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