Answer: The atomic mass of a Europium atom is 151.96445 amu.
From the given information:
Percent intensity is 91.61% of Europium atom of molecular weight 150.91986 amu.
Percent intensity is 100.00% of Europium atom of molecular weight 152.92138 amu.
Abundance of Eu-151 atom:
![X_{Eu-151}=\frac{0.9161}{0.9161+1.000}=0.4781](https://tex.z-dn.net/?f=X_%7BEu-151%7D%3D%5Cfrac%7B0.9161%7D%7B0.9161%2B1.000%7D%3D0.4781)
Abundance of Eu-153 atom:
![X_{Eu-153}=\frac{1.000}{0.9161+1.000}=0.5219](https://tex.z-dn.net/?f=X_%7BEu-153%7D%3D%5Cfrac%7B1.000%7D%7B0.9161%2B1.000%7D%3D0.5219)
Atomic mass of Europium atom:
![A=(X_{Eu-151}\times150.91986+X_{Eu-153}\times152.92138)amu\\A=(0.4781\times150.91986+0.5219\times152.92138)amu=151.96445 amu](https://tex.z-dn.net/?f=A%3D%28X_%7BEu-151%7D%5Ctimes150.91986%2BX_%7BEu-153%7D%5Ctimes152.92138%29amu%5C%5CA%3D%280.4781%5Ctimes150.91986%2B0.5219%5Ctimes152.92138%29amu%3D151.96445%20amu)
Therefore, the atomic mass of a Europium atom is 151.96445 amu.
Answer:
Explanation:
heat lost by water will be used to increase the temperature of ice
heat gained by ice
= mass x specific heat x rise in temperature
1 x 2090 x t
heat lost by water in cooling to 0° C
= mcΔt where m is mass of water , s is specific heat of water and Δt is fall in temperature .
= 1 x 2 x 4186
8372
heat lost = heat gained
1 x 2090 x t = 8372
t = 4°C
There will be a rise of 4 degree in the temperature of ice.
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
The conventional signal used by sailboats in conditions of reduced visibility such as heavy fog is one long blast followed by two short blasts.
The blasts help other boat operators locate one another's vessel in a condition where it is not easy to see. This signal is repeated in order to not only let others know of the vessel's position, but also help them know which way it is traveling. For example, if the blasts start to become distant, then the sailboat is travelling away from you.