To solve this problem, we need to use the formula for pressure in a fluid column, which is P = ρgh, where P is the pressure, ρ is the fluid density, g is the acceleration due to gravity, and h is the height (or depth in this case).

Given:
Absolute pressure P = 5.5 bar = 5.5 * 10^5 Pa (since 1 bar = 10^5 Pa)
Density of ocean water ρ = 1025 kg/m^3
Acceleration due to gravity g = 9.81 m/s^2

We need to find the depth h.

Rearranging the formula for h, we get h = P / (ρg).

Substituting the given values, we get h = (5.5 * 10^5 Pa) / (1025 kg/m^3 * 9.81 m/s^2) = 54.8 m.

So, a diver can descend approximately 54.8 meters deep in the ocean without damaging his watch.

Question

To solve this problem, we need to use the formula for pressure in a fluid column, which is P = ρgh, where P is the pressure, ρ is the fluid density, g is the acceleration due to gravity, and h is the height (or depth in this case).

Given:
Absolute pressure P = 5.5 bar = 5.5 * 10^5 Pa (since 1 bar = 10^5 Pa)
Density of ocean water ρ = 1025 kg/m^3
Acceleration due to gravity g = 9.81 m/s^2

We need to find the depth h.

Rearranging the formula for h, we get h = P / (ρg).

Substituting the given values, we get h = (5.5 * 10^5 Pa) / (1025 kg/m^3 * 9.81 m/s^2) = 54.8 m.

So, a diver can descend approximately 54.8 meters deep in the ocean without damaging his watch.

Knowee AI · Accepted Answer

To solve this problem, we need to use the formula for pressure in a fluid column, which is P = ρgh, where P is the pressure, ρ is the fluid density, g is the acceleration due to gravity, and h is the height (or depth in this case).

Given:
Absolute pressure P = 5.5 bar = 5.5 * 10^5 Pa (since 1 bar = 10^5 Pa)
Density of ocean water ρ = 1025 kg/m^3
Acceleration due to gravity g = 9.81 m/s^2

We need to find the depth h.

Rearranging the formula for h, we get h = P / (ρg).

Substituting the given values, we get h = (5.5 * 10^5 Pa) / (1025 kg/m^3 * 9.81 m/s^2) = 54.8 m.

So, a diver can descend approximately 54.8 meters deep in the ocean without damaging his watch.

How deep can a diver descend in the ocean water without damaging his watch; which an absolute pressure of 5.Sbar? Take the density of ocean water, p ^ 2 = 1025kg / (m ^ 3)

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