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2. inspector Lestrade has sent a small piece of very reactive metal to the crime lab. Lab technician Earl LaMayer weighs the soft metal and finds that it weighs 54.3g. It appears to be either lithium,sodium,or potassium, all highly reactive with water. Therefore LaMayer places the metal into a glass graduated cylinder of radius 4 cm that contains a nonreactive liquid. The meatal causes the liquid level to rise 2.0 cm. Which metal is it?
http://www.docstoc.com/docs/80649033/Volume-Problem-Set number 5 and 7 only :D
the soft metal weighs 54.3g. It appears to be either lithium,sodium,or potassium, all highly reactive with water. Therefore LaMayer places the metal into a glass graduated cylinder of radius 4 cm that contains a nonreactive liquid. The meatal causes the liquid level to rise 2.0 cm. Which metal is it? what is the volume of the shape?
i dont see this problem in your link ..
does the metal sink or float?
yup, its separated. so the volume of cylinder V=pi r^2 h
so the radius is 4 and high is 2?
yep
i got the answer of 0.54
D=m/V
how? either you havent given the full problem, or im missing something
D= 54.3g/100.53 cm^3
D= 0.54 g/cm^3
100.53 i get :)
was the metal fully submerged tho? the volume of the piece is only equal to the volume displaced IF the shape is fully submerged
if a cube is floating half way out of the water, the volume of the water displaced is not equal to the volume of the cube, right?
what number are you talking about?
what is the original question? i dont see it asking for anything ... what am i missing?
ok, lets do number 5 lol
gut a large cylindr with a small cylindar
pi R^2 h - pi r^2 h = pi h (R^2-r^2)
can you substitute the given, im dizzy in this problem lol
what our lengths?
200
no, thats the number of pipes
160
number of pipes? length of each pipe? outer radius? inner radius? pi?
200 160 6 5 3.14
radiuses, they arent 6 and 5
3, 2.5
6/2 and 5/2 are fine for me so 160 pi (6^2-5^2)/2^2 is the volume of material for each pipe
the high is 160?
well yeah, stand the pipes on end ....
|dw:1431653341758:dw|
i got 440pi
so that is the total volume
440pi cm^3 and it weighs 7.7 grams per cm^3 giving us?
D=m/V 7.7 = m/440pi m=3388pi
thats the wieght of each pipe in grams .. and we have 200 of them ... so total weight is?
3388pix200?
yeah, unless you want to add it all up instead
2128743.182
might be a little catch here, the pipes are discrete, so we cant haul more than so many at a time ... what is a 1/4 tonne ?
250kg
3388pi * n <= 1/4 tonne is how many she can fit on one load
3.388pi kg is a fair conversion?
n <= 250/(3.388pi)
10.64 kg
23.5
and i assume we aint cutting them in half, so 23 full pipes per load
this is a little aside, it may be that they didnt think of it .... its just precautionry is all
23 divides 200 into how many loads?
8.7
so 8 + 1more trip
9
now assuming this is just over thinking 2128743.182 grams 2128.743182 kg in total, divided by 250 kg per load
still gets us 8.5 loads ... so it was negligble
8.5 loads will transport to deliver the 200length of steel pipe?
no driver makes half a run ... the .5 means that we make 1 last run after 8 are finished.
at least thats my take on it, i dunno whos grading it tho
well, lets move in number 7
im moving elsewhere ...
long posts lag my system
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