91影视

1. The speed of air at the end A, $v_A=0m/s$.

2. At point B, the speed of air = the speed of the plane $=v$.

3. The pitot tube is horizontal.

4. The height difference in U-tube,$h=26.0cm$.

5. The density of air,${蚁}_{air}=1.03kg/m^3$.

1. The density of alcohol,$蚁=810kg/m^3$.

Using Bernoulli鈥檚 equation, find the given expression for the airspeed of the plane. Then, using this expression, find the airspeed of the plane for given values. According to Bernoulli鈥檚 equation, as the speed of a moving fluid increases, the pressure within the fluid decreases.

The equation is as follows:

$蚁鈭赌\frac{1}{2}蚁g^{2}h+cons\tan t$

Where, p is pressure, v is velocity, h is height, g is an acceleration due to gravity, h is height, and $蚁$is density.

The airflow obeys Bernoulli鈥檚 principle.

So,

${蚁}_A鈭赌\frac{1}{2}{Pig}_{ir}{g}_{h}hA={蚁}_B鈭赌\frac{1}{2}{蚁}_{ir}g{h}_{air}$

It is given that the pipe is horizontal and$v_A=0m/s$.

Then,

$鈭�p={蚁}_A-p_B=\frac{1}{2}\text{air}\left({}^2\right)$

The difference in pressure is also indicated by the height difference in the liquid columns of the U-tube, i.e.,h.

Thus,

$鈭�pgh$

Hence, by combining these two equations,

$$\begin{gathered} \frac{1}{2}{蚁}_{air}\left(v^2\right)=蚁gh \\ v=\sqrt{\frac{2蚁gh}{{蚁}_{air}}} \end{gathered}$$

Hence, using Bernoulli鈥檚 principle, it is proved that$v=\sqrt{\frac{2蚁gh}{{蚁}_{air}}}$.

Use the equation developed in part (a) to determine the plane鈥檚 speed relative to air as,

$$\begin{gathered} v=\sqrt{\frac{2蚁gh}{{蚁}_{air}}} \\ =\sqrt{\frac{2脳\left(810kg/m^3\right)脳\left(9.8m/s^2\right)脳\left(26.0脳{10}^{-2}m\right)}{1.03kg/m^3}} \\ =63.3m/3 \end{gathered}$$

Hence, the plane鈥檚 speed relative to the air is $63.3m/3$.