Question

\begin{itemize} \item Two columns of data; (i) a depth data starting with 0 m (at the water's surface) and ending at 50 m below the water's surface and (ii) a calculated time data for the allowable time that the diver can stay submerged \item A graph of Time (min) versus Depth (meters) \end{itemize} ICN Scuba Diver Problem Monday, January 31, 2022 11:25 AM $R = \frac{[L]}{[s]}$ $T = const$ $P_t = P_a + \rho_{w}gh$ $P_iV_i = P_tV_t$ $(P_a + \rho_wgh)(Rt) = \frac{P_tV_t}{t}$ Solve for time $\begin{cases} P_a = 1.01 \times 10^5 Pa \\ R = \frac{0.7 L}{s} \\ P_t = 1.2 \times 10^7 Pa \\ V_t = 0.010 m^3 \end{cases}$

          \begin{itemize}
\item Two columns of data; (i) a depth data starting with 0 m (at the water's surface) and ending at 50 m below the water's surface and (ii) a calculated time data for the allowable time that the diver can stay submerged
\item A graph of Time (min) versus Depth (meters)
\end{itemize}
ICN Scuba Diver Problem
Monday, January 31, 2022  11:25 AM
$R = \frac{[L]}{[s]}$
$T = const$
$P_t = P_a + \rho_{w}gh$
$P_iV_i = P_tV_t$
$(P_a + \rho_wgh)(Rt) = \frac{P_tV_t}{t}$
Solve for time
$\begin{cases}
    P_a = 1.01 \times 10^5 Pa \\
    R = \frac{0.7 L}{s} \\
    P_t = 1.2 \times 10^7 Pa \\
    V_t = 0.010 m^3
\end{cases}$

* Two columns of data; (i) a depth data starting with 0 m (at the water's surface) and ending at 50 m below the water's surface and (ii) a calculated time data for the allowable time that the diver can stay submerged

* A graph of Time (min) versus Depth (meters)

ICN Scuba Diver Problem
Monday, January 31, 2022 11:25 AM
R = ([L])/([s])
T = const
Pt = Pa + ρwgh
PiVi = PtVt
(Pa + )(Rt) = (PtVt)/(t)
Solve for time
Pa = 1.01 × 10^5 Pa

R = (0.7 L)/(s)

Pt = 1.2 × 10^7 Pa

Vt = 0.010 m^3

Added by Victor N.

Question

Please give Ace some feedback