A small lead ball with a volume of 0.02 cm ^ 3 falls evenly in the water. At what depth was the ball

A small lead ball with a volume of 0.02 cm ^ 3 falls evenly in the water. At what depth was the ball, if in the process of its movement an amount of heat equal to 12.42 mJ was released?

To find out the immersion depth of the specified lead ball, consider the equality: Q = A = Fr * S = (Ft – FA) * hlim = (mw * g – ρw * g * Vsh) * hlim = (ρw * Vsh * g – ρw * g * Vsh) * hlim = (ρw * – ρw) * Vsh * g * hlim, whence we express: hlim = Q / ((ρwith – ρw) * Vsh * g).

Const: ρw – lead density (ρw = 11340 kg / m3); ρw – water density (assumed ρw (fresh water) = 1000 kg / m3); g – acceleration due to gravity (g ≈ 9.81 m / s2).

Data: Q – released heat (Q = 12.42 mJ = 12.42 * 10-3 J); Vsh – the volume of the specified lead ball (Vsh = 0.02 cm3, in the SI system V = 2 * 10-8 m3) …

Let’s perform the calculation: hlim = Q / ((ρw – ρw) * Vsh * g) = 12.42 * 10-3 / ((11340 – 1000) * 2 * 10-8 * 9.81) ≈ 6.12 m.

Answer: The specified lead ball should be at a depth of 6.12 m.