Dynamic Planet B/C

UTF-8 U+6211 U+662F
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Re: Dynamic Planet B/C

Post by UTF-8 U+6211 U+662F » April 24th, 2019, 12:55 pm

BennyTheJett wrote:
UTF-8 U+6211 U+662F wrote:
BennyTheJett wrote: They're implying that "box" means cube. that is why they made sure everyone saw it.
I'm not sure that's right... box canyons are a legitimate type of canyon.
I know it is, but the way they accented "box" made me think cube, which would give you a depth measurement. TBH though, I have no idea whether this question is legitimate.
I'm not sure it's solvable even with that assumption, so I guess it's kind of a moot point?

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Re: Dynamic Planet B/C

Post by BennyTheJett » April 24th, 2019, 12:57 pm

UTF-8 U+6211 U+662F wrote:
BennyTheJett wrote:
UTF-8 U+6211 U+662F wrote: I'm not sure that's right... box canyons are a legitimate type of canyon.
I know it is, but the way they accented "box" made me think cube, which would give you a depth measurement. TBH though, I have no idea whether this question is legitimate.
I'm not sure it's solvable even with that assumption, so I guess it's kind of a moot point?
I guess.... frustrating.
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Re: Dynamic Planet B/C

Post by arqto » April 25th, 2019, 9:51 pm

UTF-8 U+6211 U+662F wrote:
sciencegirl03 wrote:Anyone know how to get the answer to Q2?

Q1) An unnamed lake was formed from ice melt. Periodically the ice dam holding back the water would
break, resulting in enormous volumes of water suddenly being released. A typical release might result in
the flow of 10 cubic kilometers of water per hour. At this rate, the lake could be entirely drained in two
days. If the lake covered approximately 800 sq. kilometers, which of the following is closest to the
average depth of the lake. For calculation purposes, assume that the lake had a uniform depth.
Answer: 600 meters (this one is easy)

Q2) If the lake in question 50 were losing water at 10 cubic kilometers per hour, by how much would the
water level in a deep canyon rise if the canyon were 0.5 km across at the bottom and the river was
determined to be moving at 36 m/sec. For calculations, assume the canyon walls are vertical, a true “box”
canyon. Pick the answer that is closest to the calculated amount.
Answer: 100 m?? (how do you get that?)
That's odd, is the original depth of the river given to you?
I think I understand what this question wants you to do, though I'm not getting what the answer is saying. Basically, the depth of the water in the canyon will remain constant if the flow into the canyon and flow out of the canyon are equal to one another.

Essentially 10 km^3/hr (the inflow) = the horizontal flow rate * cross section of the water in the canyon (which is the outflow).
Converting 36 m/s to km/hr gives 129.6 km/hr. From there, 10/129.6 will give the cross sectional area of the canyon in km^2. Since the canyon is 0.5 km wide, dividing by 0.5 km will give the water level since the cross section is a rectangle, giving 0.154 km or 154 m. Maybe it rounds to 100 ?
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Re: Dynamic Planet B/C

Post by BennyTheJett » April 26th, 2019, 3:57 am

arqto wrote:
UTF-8 U+6211 U+662F wrote:
sciencegirl03 wrote:Anyone know how to get the answer to Q2?

Q1) An unnamed lake was formed from ice melt. Periodically the ice dam holding back the water would
break, resulting in enormous volumes of water suddenly being released. A typical release might result in
the flow of 10 cubic kilometers of water per hour. At this rate, the lake could be entirely drained in two
days. If the lake covered approximately 800 sq. kilometers, which of the following is closest to the
average depth of the lake. For calculation purposes, assume that the lake had a uniform depth.
Answer: 600 meters (this one is easy)

Q2) If the lake in question 50 were losing water at 10 cubic kilometers per hour, by how much would the
water level in a deep canyon rise if the canyon were 0.5 km across at the bottom and the river was
determined to be moving at 36 m/sec. For calculations, assume the canyon walls are vertical, a true “box”
canyon. Pick the answer that is closest to the calculated amount.
Answer: 100 m?? (how do you get that?)
That's odd, is the original depth of the river given to you?
I think I understand what this question wants you to do, though I'm not getting what the answer is saying. Basically, the depth of the water in the canyon will remain constant if the flow into the canyon and flow out of the canyon are equal to one another.

Essentially 10 km^3/hr (the inflow) = the horizontal flow rate * cross section of the water in the canyon (which is the outflow).
Converting 36 m/s to km/hr gives 129.6 km/hr. From there, 10/129.6 will give the cross sectional area of the canyon in km^2. Since the canyon is 0.5 km wide, dividing by 0.5 km will give the water level since the cross section is a rectangle, giving 0.154 km or 154 m. Maybe it rounds to 100 ?
The original depth is not given though?
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Re: Dynamic Planet B/C

Post by porphyry » April 26th, 2019, 5:53 am

I've heard rumors of the importance of horns/pyramidal peaks during the Pleistocene epoch being on tests. I haven't come across it myself, but I want to be as prepared as possible. Does anyone know anything about the importance of them? I've tried google but nothing seems to come up.
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Re: Dynamic Planet B/C

Post by arqto » April 26th, 2019, 9:31 am

BennyTheJett wrote:
arqto wrote:
UTF-8 U+6211 U+662F wrote: That's odd, is the original depth of the river given to you?
I think I understand what this question wants you to do, though I'm not getting what the answer is saying. Basically, the depth of the water in the canyon will remain constant if the flow into the canyon and flow out of the canyon are equal to one another.

Essentially 10 km^3/hr (the inflow) = the horizontal flow rate * cross section of the water in the canyon (which is the outflow).
Converting 36 m/s to km/hr gives 129.6 km/hr. From there, 10/129.6 will give the cross sectional area of the canyon in km^2. Since the canyon is 0.5 km wide, dividing by 0.5 km will give the water level since the cross section is a rectangle, giving 0.154 km or 154 m. Maybe it rounds to 100 ?
The original depth is not given though?
The original depth is not needed to find the answer. If 10 km^3/hr of water began flowing into the canyon, the water level in the canyon would rise until the flow out matched the flow in, regardless of how deep the river was originally.
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Re: Dynamic Planet B/C

Post by UTF-8 U+6211 U+662F » April 26th, 2019, 11:42 am

arqto wrote:
BennyTheJett wrote:
arqto wrote:
I think I understand what this question wants you to do, though I'm not getting what the answer is saying. Basically, the depth of the water in the canyon will remain constant if the flow into the canyon and flow out of the canyon are equal to one another.

Essentially 10 km^3/hr (the inflow) = the horizontal flow rate * cross section of the water in the canyon (which is the outflow).
Converting 36 m/s to km/hr gives 129.6 km/hr. From there, 10/129.6 will give the cross sectional area of the canyon in km^2. Since the canyon is 0.5 km wide, dividing by 0.5 km will give the water level since the cross section is a rectangle, giving 0.154 km or 154 m. Maybe it rounds to 100 ?
The original depth is not given though?
The original depth is not needed to find the answer. If 10 km^3/hr of water began flowing into the canyon, the water level in the canyon would rise until the flow out matched the flow in, regardless of how deep the river was originally.
The question asks for the change from the original depth, not the final water level.

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Re: Dynamic Planet B/C

Post by BennyTheJett » April 26th, 2019, 12:17 pm

:(
Last edited by BennyTheJett on April 27th, 2019, 9:32 am, edited 1 time in total.
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Re: Dynamic Planet B/C

Post by arqto » April 27th, 2019, 12:00 am

UTF-8 U+6211 U+662F wrote:
arqto wrote:
BennyTheJett wrote:
The original depth is not given though?
The original depth is not needed to find the answer. If 10 km^3/hr of water began flowing into the canyon, the water level in the canyon would rise until the flow out matched the flow in, regardless of how deep the river was originally.
The question asks for the change from the original depth, not the final water level.
Oops, didn't read the question very carefully. Maybe it just wants you to assume an original depth of 0 but neglected to say it explicitly just because of poor testwriting? Or an original depth of however deep an average canyon river is... I think you would just have to make an assumption about what the question writer intended since there's no way otherwise.
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Re: Dynamic Planet B/C

Post by jz123sst » May 5th, 2019, 12:59 pm

is next year glaciers again for c

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