The pressure at a certain depth underwater is:
P = ρgh
P = pressure, ρ = sea water density, g = gravitational acceleration near Earth, h = depth
The pressure exerted on the submarine window is:
P = F/A
P = pressure, F = force, A = area
The area of the circular submarine window is:
A = π(d/2)²
A = area, d = diameter
Set the expressions for the pressure equal to each other:
F/A = ρgh
Substitute A:
F/(π(d/2)²) = ρgh
Isolate h:
h = F/(ρgπ(d/2)²)
Given values:
F = 1.1×10⁶N
ρ = 1030kg/m³ (pulled from a Google search)
g = 9.81m/s²
d = 30×10⁻²m
Plug in and solve for h:
h = 1.1×10⁶/(1030(9.81)π(30×10⁻²/2)²)
h = 1540m
The answere is No pain, no gain
Assuming the gas behaves ideally,
PV/T = constant. P will also be constant in this giving us:
V₁/T₁ = V₂/T₂
40/320 = 20/T₂
T₂ = 160 K
The answer is A.
Answer:
You need to give the options but the formula is p=mv
Explanation:
4 because it’s is the highest one hope it helps
