Both objects have the same electrical charge. Opposite charges attract. And if they were neutral they would not do anything.
Beginning when the bottom of the object first touches the water,
and as it descends and more and more of it goes under, the
buoyant force on it increases during that time.
As soon as the object is completely underwater, it doesn't matter
how deep under it is, the buoyant force on it remains the same.
Answer:
the volume is 0.253 cm³
Explanation:
The pressure underwater is related with the pressure in the surface through Pascal's law:
P(h)= Po + ρgh
where Po= pressure at a depth h under the surface (we assume = 1atm=101325 Pa) , ρ= density of water ,g= gravity , h= depth at h meters)
replacing values
P(h)= Po + ρgh = 101325 Pa + 1025 Kg/m³ * 9.8 m/s² * 20 m = 302225 Pa
Also assuming that the bubble behaves as an ideal gas
PV=nRT
where
P= absolute pressure, V= gas volume ,n= number of moles of gas, R= ideal gas constant , T= absolute temperature
therefore assuming that the mass of the bubble is the same ( it does not absorb other bubbles, divides into smaller ones or allow significant diffusion over its surface) we have
at the surface) PoVo=nRTo
at the depth h) PV=nRT
dividing both equations
(P/Po)(V/Vo)=(T/To)
or
V=Vo*(Po/P)(T/To) = 0.80 cm³ * (101325 Pa/302225 Pa)*(277K/293K) = 0.253 cm³
V = 0.253 cm³