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GS-4 Section III Question 95

GS-4 Section III Question 95

Postby admin » Wed Aug 11, 2010 9:38 pm

Be the first to discuss this question!
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Re: GS-4 Section III Question 95

Postby veen » Tue Feb 10, 2015 8:11 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?
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Re: GS-4 Section III Question 95

Postby goldstanda3269 » Wed Feb 11, 2015 5:27 am

'Just wanted to ask if I am on the right track."
- How exactly did you do the calculation? (this is question 94)
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Re: GS-4 Section III Question 95

Postby sabryhassa3517 » Mon Feb 24, 2020 3:53 pm

Hi

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

Thank you
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Re: GS-4 Section III Question 95

Postby goldstanda3269 » Mon Feb 24, 2020 9:24 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.
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Re: GS-4 Section III Question 95

Postby sabryhassa3517 » Tue Feb 25, 2020 8:46 am

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)
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Re: GS-4 Section III Question 95

Postby goldstanda3269 » Tue Feb 25, 2020 11:29 am

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?)
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Re: GS-4 Section III Question 95

Postby sabryhassa3517 » Tue Feb 25, 2020 12:46 pm

Oh okay no worries

Thank you!
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Re: GS-4 Section III Question 95

Postby skogias6733 » Thu Apr 14, 2022 1:34 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
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Re: GS-4 Section III Question 95

Postby goldstanda3269 » Fri Apr 15, 2022 1:18 pm

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.
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Re: GS-4 Section III Question 95

Postby skogias6733 » Sat Apr 16, 2022 9:40 am

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