I'm not sure it's solvable even with that assumption, so I guess it's kind of a moot point?BennyTheJett wrote: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.UTF8 U+6211 U+662F wrote:I'm not sure that's right... box canyons are a legitimate type of canyon.BennyTheJett wrote: They're implying that "box" means cube. that is why they made sure everyone saw it.
Dynamic Planet B/C

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Re: Dynamic Planet B/C
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Re: Dynamic Planet B/C
I guess.... frustrating.UTF8 U+6211 U+662F wrote:I'm not sure it's solvable even with that assumption, so I guess it's kind of a moot point?BennyTheJett wrote: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.UTF8 U+6211 U+662F wrote: I'm not sure that's right... box canyons are a legitimate type of canyon.

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Re: Dynamic Planet B/C
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.UTF8 U+6211 U+662F wrote:That's odd, is the original depth of the river given to you?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?)
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
The original depth is not given though?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.UTF8 U+6211 U+662F wrote:That's odd, is the original depth of the river given to you?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?)
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
2020 Events: Astronomy, Dynamic, GeoLogic Mapping

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Re: Dynamic Planet B/C
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.BennyTheJett wrote:The original depth is not given though?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.UTF8 U+6211 U+662F wrote: That's odd, is the original depth of the river given to you?
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
The question asks for the change from the original depth, not the final water level.arqto wrote: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.BennyTheJett wrote:The original depth is not given though?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 ?
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Re: Dynamic Planet B/C

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Re: Dynamic Planet B/C
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.UTF8 U+6211 U+662F wrote:The question asks for the change from the original depth, not the final water level.arqto wrote: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.BennyTheJett wrote:
The original depth is not given though?
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