A submarine is constructed so that it can safely withstand a pressure of 1.6 x 107 Pa. How deep may this submarine descend in the ocean if the average density of seawater is 1025 kg/m3.
I'm sorry I'll stop changing names
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To determine how deep the submarine can descend, we need to consider the pressure it can withstand and the pressure exerted by the water at a certain depth.
The pressure exerted by a fluid (in this case, water) at a specific depth can be calculated using the formula:
P = ρgh
Where:
P is the pressure
ρ (rho) is the density of the fluid (seawater, in this case)
g is the acceleration due to gravity (approximately 9.8 m/s^2)
h is the depth
In this problem, the pressure that the submarine can withstand is 1.6 x 10^7 Pa, and the density of seawater is 1025 kg/m^3.
Let's rearrange the formula to solve for depth (h):
h = P / (ρg)
Substituting the given values:
h = (1.6 x 10^7 Pa) / (1025 kg/m^3 * 9.8 m/s^2)
Now, let's calculate the value of h:
h = 1.6 x 10^7 / (1025 * 9.8) meters
h ≈ 1647.95 meters
Therefore, the submarine can safely descend to a depth of approximately 1647.95 meters in the ocean.
(water density)*g*depth = 1.6 x 10^7 n/m^2
You know the density and g. Solve for depth.
Why do you bother changing names with each post? I consider it annoying, and am less likely to help students who keep doing it.