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lidiya [134]
3 years ago
9

A research submarine has a 30-cm-diameter window that is 8.1 cm thick. The manufacturer says the window can withstand forces up

to 1.1×106 N . Part A Part complete What is the submarine's maximum safe depth in sea water? The pressure inside the submarine is maintained at 1.0 atm. Express your answer in meters.
Physics
1 answer:
malfutka [58]3 years ago
5 0

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

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The electrons making the shock come from the women's body.

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7 0
2 years ago
What happens when an object is dropped?
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3 years ago
Metal sphere A is hung from the ceiling by a long, thin string and given a positive charge. An identical sphere B is suspended n
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7 0
3 years ago
An ideal monatomic gas initially has a temperature of 330 K and a pressure of 6.00 atm. It is to expand from volume 500 cm3 to v
Sonbull [250]

a. Final pressure= 5atm

b. Work done = 5000 Joules

<h3>How to determine the parameters</h3>

Given;

  • Temperature = 330K
  • P1 = 6 atm
  • V1 = 500cm^3
  • V2 = 1500cm^2

In adiabatic expansion, temperature is constant

P1V1 = P2V2

Now, let's substitute the values into the formula

6 × 500 = 1500 P2

300 = 1500P2

Make 'p2' subject of formula

P2 = 1500/300

P2 = 5 atm

The formula for work done is given as:

Workndone = P ∆ volume

Work done = 5 × ( 1500 - 500)

Work done = 5 × 1000

Work done = 5000 Joules

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8 0
2 years ago
You have just landed your first job as a structural engineer and you have been asked to design a regional airport for small plan
lidiya [134]

Answer: 197.40\ m

Explanation:

Given

final velocity at takeoff v=28.1\ m/s

Acceleration of the plane can be a=2\ m/s^2

Initial velocity is zero for the plane i.e. u=0

Using the equation of motion

\Rightarrow v^2-u^2=2as\quad [\text{s=displacement}]\\\text{Insert the values}\\\Rightarrow (28.1)^2-0=2\times 2\times s\\\\\Rightarrow s=\dfrac{789.61}{4}\\\\\Rightarrow s=197.40\ m

Thu,s the minimum length must be 197.40\ m

6 0
2 years ago
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