Given hollow cubical box of side 0.30 m=3 dM
volume V=3^3=27 dm^3 = 27 L
Weight of box
= weight of (fresh) water (1 kg/L) displaced
= V/3 = 9 L of water = 9 kg
When water is filled so that the total weight of water equals the weight of volume of box, the top rim of the box will be at water level.
Remaining volume = 27-9=18L
height of 18L of water when water rushes in
= 18/(3^2) = 2 dm (=0.2m)
Answer:
What kind of problem? Multiplication? Division? What?
Step-by-step explanation:
Step-by-step explanation:
9^2 - 2^2 = 9*9 - 2*2= 81-4= 77
