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- admin
**Posts:**1065**Joined:**Tue Jun 29, 2010 7:47 pm

Hi,

I solved it like this

velocity = wavelength x Frequency

= wavelength x (1/period)

I got the right answer. Just wanted to ask if I am on the right track.

Does that mean length of one wave is 13720m?

I solved it like this

velocity = wavelength x Frequency

= wavelength x (1/period)

I got the right answer. Just wanted to ask if I am on the right track.

Does that mean length of one wave is 13720m?

- veen
**Posts:**6**Joined:**Thu Dec 11, 2014 10:01 am

'Just wanted to ask if I am on the right track."

- How exactly did you do the calculation? (this is question 94)

- How exactly did you do the calculation? (this is question 94)

- goldstanda3269
**Posts:**1713**Joined:**Wed Aug 25, 2010 10:59 pm

Hi

I'm just wondering if you can please go through the equation that was used.

Thank you

I'm just wondering if you can please go through the equation that was used.

Thank you

- sabryhassa3517
**Posts:**21**Joined:**Fri Jan 31, 2020 2:09 pm

sabryhassa3517 wrote:Hi

I'm just wondering if you can please go through the equation that was used.

Thank you

GS-4 Section III Question 95 does not use any equation. Perhaps you are referring to a different exam or question? Please copy/paste the question. Thanks.

- goldstanda3269
**Posts:**1713**Joined:**Wed Aug 25, 2010 10:59 pm

Hi,

I think there is a glitch in the system. Because when I press on this Q it comes up with Question 95 but the question number is 101.

Question: "Sonar is one of the applications of sound used extensively in undersea mapping. A transmitter emits a pulse of sound and the time it takes to return is used to determine the depth. If the sound wave takes 20 seconds to return, and sound travels four times as quickly in water as in air, how deep is that region?"

Explanation:

use 10 sec. for dist.

Speed of sound in water = (343 x 4) m s-1

Time taken to return = 20 s

Therefore, time taken to reach = 1/2 x time taken to return = 10 s

Depth of region = (343 x 4) m s-1 x 10 s = 13 720 m

Just wondering if we can have the explanation with equation that was used i.e. c= f x (wavelength)

I think there is a glitch in the system. Because when I press on this Q it comes up with Question 95 but the question number is 101.

Question: "Sonar is one of the applications of sound used extensively in undersea mapping. A transmitter emits a pulse of sound and the time it takes to return is used to determine the depth. If the sound wave takes 20 seconds to return, and sound travels four times as quickly in water as in air, how deep is that region?"

Explanation:

use 10 sec. for dist.

Speed of sound in water = (343 x 4) m s-1

Time taken to return = 20 s

Therefore, time taken to reach = 1/2 x time taken to return = 10 s

Depth of region = (343 x 4) m s-1 x 10 s = 13 720 m

Just wondering if we can have the explanation with equation that was used i.e. c= f x (wavelength)

- sabryhassa3517
**Posts:**21**Joined:**Fri Jan 31, 2020 2:09 pm

This is classic GAMSAT: No equation, only dimensional analysis (see Chapter 2, GAMSAT Maths in the GS Book). Pay only attention to the units and that leads you to the next step.

Is there a step in the sequence that did not seem to follow? (for your Section 2: non sequitur?)

Is there a step in the sequence that did not seem to follow? (for your Section 2: non sequitur?)

- goldstanda3269
**Posts:**1713**Joined:**Wed Aug 25, 2010 10:59 pm

Oh okay no worries

Thank you!

Thank you!

- sabryhassa3517
**Posts:**21**Joined:**Fri Jan 31, 2020 2:09 pm

HEAPS 3 - Q95

Hi,

Just wondering why we half the 20 (to give 10 seconds) to get the depth, if it says the sound takes 20 seconds to return?

This is the worked solution

use 10 sec. for dist.

Speed of sound in water = (343 x 4) m s-1

Time taken to return = 20 s

Therefore, time taken to reach = 1/2 x time taken to return = 10 s

Depth of region = (343 x 4) m s-1 x 10 s = 13 720 m

Thanks

Hi,

Just wondering why we half the 20 (to give 10 seconds) to get the depth, if it says the sound takes 20 seconds to return?

This is the worked solution

use 10 sec. for dist.

Speed of sound in water = (343 x 4) m s-1

Time taken to return = 20 s

Therefore, time taken to reach = 1/2 x time taken to return = 10 s

Depth of region = (343 x 4) m s-1 x 10 s = 13 720 m

Thanks

- skogias6733
**Posts:**32**Joined:**Sat Feb 06, 2021 11:17 am

skogias6733 wrote:HEAPS 3 - Q95

Just wondering why we half the 20 (to give 10 seconds) to get the depth, if it says the sound takes 20 seconds to return?

It is possible, but not 100% certain, that if such a question was on the real GAMSAT then ACER would ask the question a little differently in order to remove the ambiguity that you encountered.

Here is the ambiguity: The time to return is from the moment that the sound was reflected (which is your calculation) or from the time that the sound was emitted (which is the basis of this question)?

Possible alternate wordings: "the time for just the return segment of the travel" vs "the time to return since the sound was emitted"

The great majority of GAMSAT questions weed out such ambiguities; however, such situations may rarely present on the real exam.

- goldstanda3269
**Posts:**1713**Joined:**Wed Aug 25, 2010 10:59 pm

Ohhh so it takes 20 seconds all together from start to finish. Thanks for clearing that up

- skogias6733
**Posts:**32**Joined:**Sat Feb 06, 2021 11:17 am

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