As the earth is an oblate spheroid, its radius near the equator is more than its radius near poles. In the present case, the earth is under rotation with a constant angular velocity ω, then the test mass moves in a circular path of radius ‘r’ with an angular velocity ω.This is the case of a non-inertial frame of reference so there exists a As both these forces are acting from the same point these are known as co-initial forces and as they lie along the same plane they are termed as co-planar forces.We know from parallelogram law of vectors, if there are two coplanar vectors forming two sides of a parallelogram then the resultant of those two vectors will always along the diagonal of the parallelogram.We know ‘r’ is the radius of the circular path and ‘R’ is the radius of the earth, then r = Rcosθ.Where g′ is the apparent value of acceleration due to gravity at the latitude due to the rotation of the earth and g is the true value of gravity at the latitude without considering the rotation of the earth.Acceleration due to Gravity – Formula, Unit and ValuesAcceleration due to Gravity on the Surface of EarthImportant Conclusions on Acceleration due to Gravity :

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So if a person moves from the equator to poles his weight decreases as the value of g decreases.Consider a test mass (m) is on a latitude making an angle with the equator. The standard deviation about the regression, therefore, is\[s_r = \sqrt{\frac {1.596 \times 10^{-5}} {6 - 2}} = 1.997 \times 10^{-3} \nonumber\]Next, we need to calculate the standard deviations for the slope and the \[s_{b_1} = \sqrt{\frac {6 \times (1.997 \times 10^{-3})^2} {6 \times (1.378 \times 10^{-4}) - (2.371 \times 10^{-2})^2}} = 0.3007 \nonumber\]\[s_{b_0} = \sqrt{\frac {(1.997 \times 10^{-3})^2 \times (1.378 \times 10^{-4})} {6 \times (1.378 \times 10^{-4}) - (2.371 \times 10^{-2})^2}} = 1.441 \times 10^{-3} \nonumber\]and use them to calculate the 95% confidence intervals for the slope and the \[\beta_1 = b_1 \pm ts_{b_1} = 29.57 \pm (2.78 \times 0.3007) = 29.57 \text{ M}^{-1} \pm 0.84 \text{ M}^{-1} \nonumber\]\[\beta_0 = b_0 \pm ts_{b_0} = 0.0015 \pm (2.78 \times 1.441 \times 10^{-3}) = 0.0015 \pm 0.0040 \nonumber\]\[C_A = \frac {S_{samp} - b_0} {b_1} = \frac {0.114 - 0.0015} {29.57 \text{ M}^{-1}} = 3.80 \times 10^{-3} \text{ M} \nonumber\]\[s_{C_A} = \frac {1.997 \times 10^{-3}} {29.57} \sqrt{\frac {1} {3} + \frac {1} {6} + \frac {(0.114 - 0.1183)^2} {(29.57)^2 \times (4.408 \times 10^{-5})}} = 4.778 \times 10^{-5} \nonumber\]\[\mu = C_A \pm t s_{C_A} = 3.80 \times 10^{-3} \pm \{2.78 \times (4.778 \times 10^{-5})\} \nonumber\]\[\mu = 3.80 \times 10^{-3} \text{ M} \pm 0.13 \times 10^{-3} \text{ M} \nonumber\]You should never accept the result of a linear regression analysis without evaluating the validity of the model. As we saw earlier, the residual error for a single calibration standard, If the regression model is valid, then the residual errors should be distributed randomly about an average residual error of zero, with no apparent trend toward either smaller or larger residual errors (Figure \(\PageIndex{6}\)a). sketches of equation derivations may be useful in making certain points. Rotameter Equations and Derivations. The Rotameters are one of the most common flow meters. With this in mind, many simple factors influencing As you can see from these equations, a pressure increase of 50 psig will more than double the In addition, based on Dr. Roger Gilmont's extensive 30 year research we are able to correlate flows using a unique data analysis program which can correct for specific operating conditions. The equation for this line isFigure \(\PageIndex{3}\) shows the residual errors for the three data points. Bernoulli’s Equation. The input of the unit is …

A more general form of the equation, written in terms of \[s_{x} = \frac {s_r} {b_1} \sqrt{\frac {1} {m} + \frac {1} {n} + \frac {\left( \overline{Y} - \overline{y} \right)^2} {(b_1)^2 \sum_{i = 1}^{n} \left( x_i - \overline{x} \right)^2}} \nonumber\]A close examination of Equation \ref{5.12} should convince you that the uncertainty in Three replicate analyses for a sample that contains an unknown concentration of analyte, yield values for The average signal, \(\overline{S}_{samp}\), is 29.33, which, using Equation \ref{5.11} and the slope and the \[C_A = \frac {\overline{S}_{samp} - b_0} {b_1} = \frac {29.33 - 0.209} {120.706} = 0.241 \nonumber\]To calculate the standard deviation for the analyte’s concentration we must determine the values for \(\overline{S}_{std}\) and for \(\sum_{i = 1}^{2} (C_{std_i} - \overline{C}_{std})^2\).

Now, the force acting on the test mass due to gravity is;Where M is the mass of earth and R is the radius of the earth. Rotameter Equations and Derivations. The Subnet Training Guide V2.5 by Brendan Choi . The bottom of the tube is narrow and gets wider as the. UNIT- III FLUID MECHANICS The former is just the average signal for the calibration standards, which, using the data in \[\sum_{i = 1}^{n} (C_{std_i} - \overline{C}_{std})^2 = (s_{C_{std}})^2 \times (n - 1) \nonumber\]where \(s_{C_{std}}\) is the standard deviation for the concentration of analyte in the calibration standards.

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