see how long this 1 takes u

when a boat is at anchor 5 of the rungs on the rope ladder over its side r underwater. if the rungs r 30cm apart r 3 inches thick how many rungs will be underwater in 4 hours time if the tide rises at 35cm per hour

Originally posted by orkyboy
when a boat is at anchor 5 of the rungs on the rope ladder over its side r underwater. if the rungs r 30cm apart r 3 inches thick how many rungs will be underwater in 4 hours time if the tide rises at 35cm per hour

the same amount as the boat rises with the water as it is floating on it

Andrew

Does the tide rise at a constant rate, or does it rise at that rate only once 4 hours have passed?

Originally posted by waldorf
Does the tide rise at a constant rate, or does it rise at that rate only once 4 hours have passed?

does it matter? bishops right-isn't he?...

Originally posted by orkyboy
when a boat is at anchor 5 of the rungs on the rope ladder over its side r underwater. if the rungs r 30cm apart r 3 inches thick how many rungs will be underwater in 4 hours time if the tide rises at 35cm per hour

here is another along the same lines....

"A ship loaded with iron ore is in a canal lock, with the gates closed at both ends. Suddenly the ship dumps its entire load of ore into the lock. Does the water level rise, stay the same, or drop?"

Andrew

i dunno bout the ore 1 but latex bishop got the oringinal puzzle right. your 2 good at these latex bishop

Originally posted by latex bishop
"A ship loaded with iron ore is in a canal lock, with the gates closed at both ends. Suddenly the ship dumps its entire load of ore into the lock. Does the water level rise, stay the same, or drop?"

Andrew

Assuming that the density of the ore is greater than that of the water, and that the ore is not shaped in such a way that it is able to float, I would have to guess that the water level relative to the canal lock would drop. When the ore is floating in the ship, it is displacing a volume of water X of equal weight to the ore. When it is submerged, the ore displaces a volume of water Y of equal volume to the ore. Since the ore is more dense than the water, X is larger than Y, and the submerged ore displaces less water. Less displaced water means lower water level in the lock. The water level relative to the ship should drop, too, since the ship is now carrying less weight and will bob up a bit.

Originally posted by godzillion
Assuming that the density of the ore is greater than that of the water, and that the ore is not shaped in such a way that it is able to float, I would have to guess that the water level relative to the canal lock would drop. When the ore is floating in the ship, it is displacing a volume of water X of equal weight to the ore. When it is submerged, the o ...[text shortened]... to the ship should drop, too, since the ship is now carrying less weight and will bob up a bit.

I do not know the correct answer, but I would go along with your logic

Andrew

Originally posted by godzillion
Assuming that the density of the ore is greater than that of the water, and that the ore is not shaped in such a way that it is able to float, I would have to guess that the water level relative to the canal lock would drop. When the ore is floating in the ship, it is displacing a volume of water X of equal weight to the ore. When it is submerged, the o ...[text shortened]... to the ship should drop, too, since the ship is now carrying less weight and will bob up a bit.

Seems plausible