Solution :
Given :
Rectangular wingspan
Length,L = 17.5 m
Chord, c = 3 m
Free stream velocity of flow,
= 200 m/s
Given that the flow is laminar.
![$Re_L=\frac{\rho V L}{\mu _{\infty}}$](https://tex.z-dn.net/?f=%24Re_L%3D%5Cfrac%7B%5Crho%20V%20L%7D%7B%5Cmu%20_%7B%5Cinfty%7D%7D%24)
![$=\frac{1.225 \times 200 \times 3}{1.789 \times 10^{-5}}$](https://tex.z-dn.net/?f=%24%3D%5Cfrac%7B1.225%20%5Ctimes%20200%20%5Ctimes%203%7D%7B1.789%20%5Ctimes%2010%5E%7B-5%7D%7D%24)
![$= 4.10 \times 10^7$](https://tex.z-dn.net/?f=%24%3D%204.10%20%5Ctimes%2010%5E7%24)
So boundary layer thickness,
![$\delta_{L} = \frac{5.2 L}{\sqrt{Re_L}}$](https://tex.z-dn.net/?f=%24%5Cdelta_%7BL%7D%20%3D%20%5Cfrac%7B5.2%20L%7D%7B%5Csqrt%7BRe_L%7D%7D%24)
![$\delta_{L} = \frac{5.2 \times 3}{\sqrt{4.1 \times 10^7}}$](https://tex.z-dn.net/?f=%24%5Cdelta_%7BL%7D%20%3D%20%5Cfrac%7B5.2%20%5Ctimes%203%7D%7B%5Csqrt%7B4.1%20%5Ctimes%2010%5E7%7D%7D%24)
= 0.0024 m
The dynamic pressure, ![$q_{\infty} =\frac{1}{2} \rho V^2_{\infty}$](https://tex.z-dn.net/?f=%24q_%7B%5Cinfty%7D%20%3D%5Cfrac%7B1%7D%7B2%7D%20%5Crho%20V%5E2_%7B%5Cinfty%7D%24)
![$ =\frac{1}{2} \times 1.225 \times 200^2$](https://tex.z-dn.net/?f=%24%20%3D%5Cfrac%7B1%7D%7B2%7D%20%5Ctimes%201.225%20%20%5Ctimes%20200%5E2%24)
![$=2.45 \times 10^4 \ N/m^2$](https://tex.z-dn.net/?f=%24%3D2.45%20%5Ctimes%2010%5E4%20%5C%20N%2Fm%5E2%24)
The skin friction drag co-efficient is given by
![$C_f = \frac{1.328}{\sqrt{Re_L}}$](https://tex.z-dn.net/?f=%24C_f%20%3D%20%5Cfrac%7B1.328%7D%7B%5Csqrt%7BRe_L%7D%7D%24)
![$=\frac{1.328}{\sqrt{4.1 \times 10^7}}$](https://tex.z-dn.net/?f=%24%3D%5Cfrac%7B1.328%7D%7B%5Csqrt%7B4.1%20%5Ctimes%2010%5E7%7D%7D%24)
= 0.00021
![$D_{skinfriction} = \frac{1}{2} \rho V^2_{\infty}S C_f$](https://tex.z-dn.net/?f=%24D_%7Bskinfriction%7D%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20%5Crho%20V%5E2_%7B%5Cinfty%7DS%20C_f%24)
![$=\frac{1}{2} \times 1.225 \times 200^2 \times 17.5 \times 3 \times 0.00021$](https://tex.z-dn.net/?f=%24%3D%5Cfrac%7B1%7D%7B2%7D%20%5Ctimes%201.225%20%5Ctimes%20200%5E2%20%5Ctimes%2017.5%20%5Ctimes%203%20%5Ctimes%200.00021%24)
= 270 N
Therefore the net drag = 270 x 2
= 540 N
Answer:60 rev/min
Explanation:
Given
angular speed of first shaft ![\omega _1=420\ rev/min](https://tex.z-dn.net/?f=%5Comega%20_1%3D420%5C%20rev%2Fmin)
Moment of inertia of second shaft is seven times times the rotational speed of the first i.e. If I is the moment of inertia of first wheel so moment of inertia of second is 7 I
As there is no external torque therefore angular momentum is conserved
![L_1=L_2](https://tex.z-dn.net/?f=L_1%3DL_2)
![I_1\omega _1=I_2\omega _2](https://tex.z-dn.net/?f=I_1%5Comega%20_1%3DI_2%5Comega%20_2)
![I\times (420)=7 I\times (\omega _2)](https://tex.z-dn.net/?f=I%5Ctimes%20%28420%29%3D7%20I%5Ctimes%20%28%5Comega%20_2%29)
![\omega _2=\frac{420}{7}](https://tex.z-dn.net/?f=%5Comega%20_2%3D%5Cfrac%7B420%7D%7B7%7D)
Answer:
1469 miles /3.25 hours = 452 miles per hours
Explanation:
divide distance by the time it takes to get there
3.5 meters per second
60sec=1min
60min=1 hr
3.5*60*60= meters per hour
=12600
1000meter= 1 km
12600/1000= km per hr
=12.6 km per hour
Lymphocytes are a type of _
<u> white blood cell</u>