A hot-air balloon has a volume of 440 × 10^3 ^3. Calculate the buoyant force that the surrounding cold air exerts on the balloon? Given that the density of the surrounding air is 2.58 kg/^3.

A hot air balloon has a volume of 2.2 times 10^3 m^3. What is the buoyant force that the surrounding cold air exerts on the balloon? Assume that the density of the surrounding air is 1.29 kg/m^3 (a) 2.79 times 10^2 N (b) 2.79 times 10^3 N (c) 2.79 times 10^4 N (d) 2.79 times 10^5 N When water flows vertically downward out of the cross sectional area of the stream of water gradually narrows This phenomenon can be explained .

graph slopes whilst gas cools, then it is flat whilst conensation to liquid happens, the coolong liquid then graphs as a slope when it solidifies graph is flat

latent heat

answer: 2×10^-2kg

explanation: density= mass per volume

vol=mass/density

converting to s.i units we have;

3÷1000kg = 0.003

1.5÷100m = 0.15

0.003/0.15 = 0.02kg