Answer:
(a) [tex]p = 0.43d[/tex]
(b) [tex]p = 21.5[/tex]
(c) [tex]d = 151.163[/tex]
Step-by-step explanation:
Given
Variation: Direct
Represent ear pressure with p and depth with d.
So, we have:
[tex]p\ \alpha\ d[/tex]
[tex]d =10[/tex] when [tex]p =4.3[/tex]
Solving (a): Write a variation
We have:
[tex]p\ \alpha\ d[/tex]
Write as an equation
[tex]p = kd[/tex]
Where: k is the constant of variation
[tex]d =10[/tex] when [tex]p =4.3[/tex]
So, we have:
[tex]4.3 = k * 10[/tex]
Make k the subject
[tex]k = \frac{4.3}{10}[/tex]
[tex]k = 0.43[/tex]
Substitute 0.43 for k in [tex]p = kd[/tex]
[tex]p = 0.43d[/tex]
Solving (b): Pressure at 50 feet
Substitute 50 for d in [tex]p = 0.43d[/tex]
[tex]p = 0.43 * 50[/tex]
[tex]p = 21.5[/tex]
The pressure is 21.5 pounds per square
Solving (c): The safe depth
The safe depth stops at a pressure of 65psi.
Substitute 65 for p in [tex]p = 0.43d[/tex]
[tex]65 = 0.43 * d[/tex]
Make d the subject
[tex]d = \frac{65}{0.43}[/tex]
[tex]d = 151.162790698[/tex]
[tex]d = 151.163[/tex] -- approximated
The safe distance for an amateur is 151.163ft