What did you get from your calculations?
hi, um the problem is that i didnt know how to start. i was just thinking.. V which is 1.3L/ the area would give me Depth but im in sure and i dont get conversions properly
The amount of rainfall that fell is equal to the depth of water collected in a rectangular water-tight box over the period of time.
A rain gauge is an instrument by which even minute amount of rainfall can be measureed accurately by using an oversized funnel and collecting the water in a container that can measure the quantity accurately. By dividing the amount of water collected divided by the area of the funnel (collecting area), we can obtain the amount of rainfall in mm or any other unit.
So the depth required is the quantities of water divided by the area, both in consistent units.
Let's work in centimetres.
1.3 litres = 1300 cm³
We know that 1 inch = 2.54 cm, so
1 in² = 2.54² cm²
= 30.50*2.54^2 cm²
= 196.645 cm²
Amount of rainfall
= volume of water / area of collector area
= 1300 cm³ / 196.645 cm²
= 6.61 cm
= 66.1 mm
This measn that to do these questions, you will need to understand the definition of each term, make necessary converions of units to arrive at the correct answer.
Give a try for the remaining questions, start by
1. reading up and understanding the definitions of the terms, then
2. decide on the units you should be using,
3. do the conversions, then
4. do the calculations.
Start with Question 2 and post the steps that you have done if you have difficulties.
Umm why do you have to convert it to CM? So this is not right:
30.50 inches squared= 774.6999mm squared
1.3 litres= 1 300 000 mm cubed
Depth= 1 300 000/774.699
Depth = 1678.069147 mm
Im so confused with this stuff! Im in management so science is definitely not my area of expertise. Please help!
would you use the depth in part a) to calculate part b)?
I started with cm so that the numbers remain reasonable and can be "visualized" in size. You could have very well used mm all along, then there would be no conversion in the end.
For part 2, yes, exactly, you would first read up your teacher's notes to find out what is the definition of intensity. I believe it is the amount of rainfall (in mm) that fell in one hour, that is why the units are mm hr-1.
Then you use the amount calculated in part 1, in mm that fell in 26 minutes and 45 seconds, to calculate what would have fallen in 1 hour, if the intensity did not change. Can you proceed now with #2?
I understand now! Just to clarify my way of calculating depth isn't right? And also when I calculate part b), I have to convert 24 minutes and 45 secs into hours right? When I do that, do I convert it as 24.45 or 24mins seperately 45 secs seperately....sorry just confused!
Since there are 60 seconds in a minute, so
26 minutes 45 seconds is converted to
26+45/60 minutes, or 26.75 minutes.
In the same way, since there are 60 minutes in an hour, so 26.75 minutes converts to
Now for the intensity of rainfall, which is measured by mm/hour, we divide the amount fallen in 26 minutes 45 seconds by the duration (in hours), so the intensity would be:
______ / 0.44583 = ______ mm/hour.
Thank you! I was wondering if the person who originally started this forum had any input ("pleasehelp").
Sorry, I assumed you're the original poster. Sometimes they change names along the way.
I hope you got something out of it!
Do you have a similar problem or was it by curiosity that you joined in?
Similar question! I think we are both in the same course!
So for the second part i got 148.26 mm/hr
how do i write that as mm h-1?
148.26 mm-hr-1 or
148.26 mm/hr will both be correct.
I would go with the former if this is the unit the question suggested.
On the other hand, the volume collected was 1.30 litres, meaning that it is accurate to at most 3 significant figures. So I would give the answer as 148 mm-hr-1.
Are you OK with the rest of the questions?
I really feel bad asking but im honestly not...its taking me a while to figure it out...do you mind helping me with the rest of the questions :$...sorrryy....this is definitely not my area of expertise:(....thank you so much i really appreciate it:)
If its a problem please let me know...thank you:)
Not a problem at all!
The volume of water would be calculated like a big flat rectangular slab of area 1 acre (4840 yd²) with a height of 66.1 mm.
You need conversions:
= 4840 (3 ft)²
= 4840 (3*.3048m)²
= 4840 *(3*.3048)^sup2; m²
Now multiply by the height of 66.1 mm = 0.0661 m. to get
4840 *(3*.3048)² m² * 0.0661 m
4840 *(3*.3048)²* 0.0661 m³
= _______ m³ (= V)
Discharge, D is measured in m³/s, so
D = V m/ 3hr
= V m / 3*3600 s
= V/(3*3600) m/s
5. Today at closing, according to the Bank of Canada, 1 US$ = 1.0661 C$
and Google gives 1 US gallon = 3.78541178 litres
= 4*1.0661C$ / 3.78541178 litres
= 4*1.0661 / 3.7854 C$/litre
I will leave it to you for the numerical calculations.
Post if you need a check. However, DO make sure you understand every step that I have done. It is more important to know how to do it than to get the right answer.
You are truely a savior mathmate, but i honestly dont understand part 3)...where are you getting all these numbers from?
I'm sorry that I did not explain in detail:
These are standard conversions
1 acre = 4840 sq. yds = 43560 sq.ft
1 inch = 2.54 cm (exactly)
so 1 foot = 12*2.54 cm = 30.48 cm = 0.3048 m.
and 1 sq. ft = 0.3048² m² = 0.09290304 m²
Hope that clears up a little.
Ahhhhhhhhhhhhhhhhhhhhh okkk :S but the question is asking for 7 acres
I apologize for the oversight, so the volume will have to be multiplied by 7.
It will affect #4 also, although it said "...all the water ran off the area in part c. in 3.00 hours."
I suppose that part c was converted to #3 when it was posted.
Also, I can continue tomorrow. I need some sleep before I don't know what I'm writing!
Ok thank you...if you can explain it to me more tomorrow i will greatly appreciate it! good night!!
In a measurement of precipitation (rainfall), a raingauge with an orifice (collector) area of 30.50 in2 collects 1.30 litres of water over a period of 26 minutes and 45 seconds.
3) The volume (m3) of water that would have fallen on an area of 7.00 acres.
4)The discharge in m3 s-1 that would occur if all the water ran off the area in part c. in 3.00 hours.
Thank you for your help thus far. Would be able to explain it to me in steps tomorrow? GOOD NIGHT MATH MATE!!
Thank you good night
Would the fact
1 (american) gallon = 3.785L
1 (canadian) gallon = 4.55L
come into affect for # 5?
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