To find the depth at which the pressure is 511 pounds per square foot, we use the formula for pressure under water:
\[
P = P_0 + \rho g h
\]
where:
- \( P \) is the pressure at depth,
- \( P_0 \) is the atmospheric pressure at the surface (approximately 2116.22 pounds per square foot),
- \( \rho \) is the density of water (approximately 62.4 pounds per cubic foot),
- \( g \) is the acceleration due to gravity (approximately 32.2 feet per second squared),
- \( h \) is the depth in feet.
Rearranging the formula to solve for \( h \):
\[
h = \frac{P - P_0}{\rho g}
\]
Substituting the given values:
\[
h = \frac{511 - 2116.22}{62.4 \times 32.2}
\]
Calculating:
\[
h = \frac{-1605.22}{2008.48} \approx -0.7995
\]
Since the depth cannot be negative, it seems there is an error in the given pressure value or assumptions. Assuming the pressure is gauge pressure (excluding atmospheric pressure), we recalculate:
\[
h = \frac{511}{62.4 \times 32.2} \approx 0.2541
\]
To find the equation of the line through points \((-8, -61)\) and \( (7, 59) \), we first calculate the slope \( m \):
\[
m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{59 - (-61)}{7 - (-8)} = \frac{120}{15} = 8
\]
Using the point-slope form \( y - y_1 = m(x - x_1) \) with point \((-8, -61)\):
\[
y + 61 = 8(x + 8)
\]
Simplifying to slope-intercept form \( y = mx + b \):
\[
y = 8x + 64 - 61
\]
\[
y = 8x + 3
\]
For a line with slope \(-\frac{4}{3}\) passing through point \((10, 7)\), use the point-slope form:
\[
y - 7 = -\frac{4}{3}(x - 10)
\]
Simplifying:
\[
y - 7 = -\frac{4}{3}x + \frac{40}{3}
\]
Converting to standard form \( Ax + By = C \):
\[
3y - 21 = -4x + 40
\]
\[
4x + 3y = 61
\]
- The team is measuring at \(\boxed{0.2541}\) feet below the surface.
- The equation of the line through the points is \(\boxed{y = 8x + 3}\).
- The equation in standard form is \(\boxed{4x + 3y = 61}\).
Answered
