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Step 1 of 3(a)Write the equation for average velocity as follows:\begin{aligned}\tilde{F} & =\frac{1}{S} \int_{T} u d S \\& =\frac{1}{\pi r_{0}^{2}} \int_{0}^{2} u d\left(\pi r^{2}\right) \\& =\frac{\pi}{\pi r_{0}^{2}} \int_{0} u 2 \pi r d r\end{aligned}The average velocity expression it,P=\frac{2}{r_{r}^{5}} \int_{0}^{e} \omega rNow, calculate the values of ur as follows:Draw a plot between or vi r and calculate the area under the curve as follows:r Vs ruArea under the curve \left(\int_{0}^{\text {ur }}\right. ur \left.d\right)= no. of boxes \times \mathrm{X} axis units \times \mathrm{Y} axis units\begin{array}{l}=23 \times 0.005 \mathrm{~m} \times 0.005 \mathrm{~m}^{2} / \mathrm{s} \\=5.75 \times 10^{-4} \mathrm{~m}^{2} / \mathrm{s}\end{array}Substitute the corresponding values in equation (1) to ealculate the average velocity.\begin{aligned}\tilde{F}^{F} & =\frac{2}{r_{s}^{T}} \int_{i}^{r} u r d r \\& =\frac{2}{(0.0375 \mathrm{~m})^{2}} \times 5.75 \times 10^{-\mathrm{At}} \mathrm{m}^{3} / \mathrm{s} \\& =0.81 \frac{\mathrm{m}^{3} / \mathrm{s}}{\mathrm{m}^{7}} \\& =0.81 \frac{\mathrm{m}}{\mathrm{s}}\end{aligned}Therefore, the average velocity is 0.81 \mathrm{~m} / \mathrm{s}.Step 2 of 3(b)Write the equation for kirietic energy correction factor as follows:\begin{aligned}a & =\frac{\int w^{3} d S}{p^{3} S} \\& =\frac{\int_{u^{3}} d\left(\pi r^{2}\right)}{F^{3} \pi r_{w}^{2}} \\& =\frac{\pi u^{3} 2 r d r}{P^{3} \pi r_{*}^{2}}\end{aligned}The kirietic energy correction foctor is,\alpha=\frac{2}{p^{3} r_{-}^{2}} \int_{i}^{\alpha} u^{3} r d rCalculate the values of u^{3} r from the data as follows:Draw a plot between u^{3} r vs r and calculate the atea under the curve as follows:r vs rus\text { Area under the curve } \begin{aligned}\left(\int_{0} u^{\prime} r d r\right) & =\text { no. of boxs } \times X \text { axis units } \times Y \text { axis units } \\& =43 \times 0.005 \mathrm{~m} \times 0.002 \mathrm{~m}^{2} / \mathrm{s}^{3} \\& =4.3 \times 10^{-4} \mathrm{~m}^{4} / \mathrm{s}^{3}\end{aligned}Substitute the corresponding values in equation (2) to calculate the kinetic friction factor.\begin{array}{l}\alpha=\frac{2}{P^{3} r^{2}} \int_{-}^{r} u^{3} r d r \\=\frac{2}{(0.81 \mathrm{~m} / \mathrm{s})^{1}(0.0375 \mathrm{~m})^{2}} \times 4.3 \times 10^{-64} \mathrm{~m}^{2} / \mathrm{s}^{2} \\=1.15 \\\end{array}Therefore, the kinetic energy correction factor is \mathbf{1 . 1 5}Step 3 of 3(C)Write the expression of momentum correction factor as follows:\beta=\frac{i / S}{\left.\rho\right|^{i}}Substitute the value of M / S in the above equation.\begin{array}{l}\beta=\frac{\left(\frac{\rho \int_{0}^{2} d S}{S}\right)}{\rho v^{n}} \\=\frac{\int u^{2} d\left(\pi r^{2}\right)}{\Gamma^{3} \pi r^{2}} \\=\frac{\pi \int v^{2} \cdot 2 r d r}{F r r_{w}^{2}} \\=\frac{2}{\sqrt{2}^{2} r^{2}} \int_{0}^{2} u^{2} r d r \\\end{array}The momentum correction factor expression is,\rho=\frac{2}{r^{2} r^{2}} \int_{0}^{2} u^{2} r d-(3)Calculate the values of u^{2} r from the dota as follows:Draw a plot between u^{2} r vs r and calculate the area under the curve as follows:r vs run 2Area under the curve \left(\int_{0}^{r} s^{2} r d r\right)= No. of boxes \times x- axis units \times y- axis units\begin{array}{l}=19 \times 0.005 \mathrm{~m} \times 0.005 \mathrm{~m}^{3} / \mathrm{s}^{\mathrm{t}} \\=4.75 \times 10^{-24} \mathrm{~m}^{1} / \mathrm{s}^{2}\end{array}Substitute the corresponding values in equation (3) to calculate the momentum correction factor as follows:\begin{aligned}\beta & =\frac{2}{\bar{V}^{2} r_{w}^{2}} \int_{0}^{r_{\tilde{2}}} u^{2} r d r \\& =\frac{2}{(0.81 \mathrm{~m} / \mathrm{s})^{2}(0.0375 \mathrm{~m})^{2}}\left(4.75 \times 10^{-04} \mathrm{~m}^{3} / \mathrm{s}^{2}\right) \\& =1.02\end{aligned}Thus, the momentum correction factor is 1.02 .I HOPE THAT YOU HAVE UNDERSTOOD THE PROBLEM. IF NOT, FEEL FREE TO ASK ME. BUT PLEASE GIVE ME A LIKE(UPVOTE) ud83dudc4d   * STAY HOME - STAY SAFE * ...