The diving bell contains trapped air at the same pressure as the water outside. At the surface, air
pressure is 100 kPa. As the bell descends, the pressure
on it increases by 100 kPa for every 10 m of depth.

A) What is the pressure on the diver at depths of 0m,10m,20m and 30m
B) At the surface, the bell holds 6m^3 of air. If the bell is lowered to a depth of 20m, and no more air is pumped into it, what will be the volume of the trapped air? (assume no change in temperature)

are you like autistic or something?

xd

54 degrees incident angle = reflected angle

angle is from normal turned to line of ray

pythagoras' theorem

square root (-5^2+9^2)

square root (25+81)

square root (106)

angle is tan^-1(9/5)

neg x, pos y

energy stored in the capacitor= 9.72 x 10⁻⁴ j

explanation:

the energy stored in the capacitor is given by

e= 1/2 c v²

e= energy stored

c= capacitance= 1.5 µc= 1.5 x 10⁻⁶f

v= potential difference=36 v

so energy stored= e= 1/2 (1.5 x 10⁻⁶) (36)²

energy stored= 9.72 x 10⁻⁴j