Answer
Given,
Air enter density, ρ₁ = 2.21 kg/m³
speed of entry, v₁ = 40 m/s
Air exit density , ρ₂ = 0.752 kg/m³
speed of exit, v₂ = 180 m/s
Inlet area = 90 cm²
a) mass flow rate through nozzle.
![m = \rho_1 A_1 v_1](https://tex.z-dn.net/?f=m%20%3D%20%5Crho_1%20A_1%20v_1)
![m = 2.21\times 90\times 10^{-4}\times 40](https://tex.z-dn.net/?f=m%20%3D%202.21%5Ctimes%2090%5Ctimes%2010%5E%7B-4%7D%5Ctimes%2040)
m = 0.796 kg/s
b) exit area
Using continuity equation
![\rho_1 A_1 v_1=\rho_2A_2 v_2](https://tex.z-dn.net/?f=%5Crho_1%20A_1%20v_1%3D%5Crho_2A_2%20v_2)
![A_2 = \dfrac{\rho_1 A_1 v_1}{\rho_2 v_2}](https://tex.z-dn.net/?f=A_2%20%3D%20%5Cdfrac%7B%5Crho_1%20A_1%20v_1%7D%7B%5Crho_2%20v_2%7D)
![A_2 = \dfrac{2.21\times 90\times 40}{0.752\times 180}](https://tex.z-dn.net/?f=A_2%20%3D%20%5Cdfrac%7B2.21%5Ctimes%2090%5Ctimes%2040%7D%7B0.752%5Ctimes%20180%7D)
![A_2 = 58\ cm^2](https://tex.z-dn.net/?f=%20A_2%20%3D%2058%5C%20cm%5E2)
Exit area of nozzle is equal to ![58\ cm^2](https://tex.z-dn.net/?f=58%5C%20cm%5E2)
Answer:
194,22g/mol
Explanation:
8*12,01+10*1,01+4*14,01+2*16=194,22g/mol
Because the light colors will only work on a dark background, as dark colors will only work on a light background. If light color were on the white background not only will it be nearly impossible to read, the letters with have to be much larger for older people to read or see. Therefore, Black letters on a white background is most common. But a Black background with white letters could also work, kind of like a chalkboard.
Answer:
The force is ![3.2\times10^{-20}\ N](https://tex.z-dn.net/?f=3.2%5Ctimes10%5E%7B-20%7D%5C%20N)
Explanation:
Given that,
A long, straight wire carries a current.
Suppose, a long, straight wire carries a current 0.86 A. An electron is traveling in the vicinity of the wire. At the instant when the electron is 4.50 cm from the wire and traveling at speed of
directly toward the wire. What are the magnitude of the force that the magnetic field of the current exerts on the electron?
We need to calculate the magnetic field
Using formula of magnetic field
![B=\dfrac{\mu_{0}I}{4\pi r}](https://tex.z-dn.net/?f=B%3D%5Cdfrac%7B%5Cmu_%7B0%7DI%7D%7B4%5Cpi%20r%7D)
Where, I = current
r = distance
Put the value into the formula
![B=\dfrac{4\pi\times10^{-7}\times0.86}{2\pi\times4.5\times10^{-2}}](https://tex.z-dn.net/?f=B%3D%5Cdfrac%7B4%5Cpi%5Ctimes10%5E%7B-7%7D%5Ctimes0.86%7D%7B2%5Cpi%5Ctimes4.5%5Ctimes10%5E%7B-2%7D%7D)
![B=0.0000038\ T](https://tex.z-dn.net/?f=B%3D0.0000038%5C%20T)
![B=3.8\times10^{-6}\ T](https://tex.z-dn.net/?f=B%3D3.8%5Ctimes10%5E%7B-6%7D%5C%20T)
We need to calculate the force
Using formula of force
![F=qvB\sin\theta](https://tex.z-dn.net/?f=F%3DqvB%5Csin%5Ctheta)
Here, ![\theta=90^{\circ}](https://tex.z-dn.net/?f=%5Ctheta%3D90%5E%7B%5Ccirc%7D)
Where, q = charge of electron
v = velocity of electron
B = magnetic field
Put the value into the formula
![F=1.6\times10^{-19}\times6\times10^{4}\times3.8\times10^{-6}\sin90](https://tex.z-dn.net/?f=F%3D1.6%5Ctimes10%5E%7B-19%7D%5Ctimes6%5Ctimes10%5E%7B4%7D%5Ctimes3.8%5Ctimes10%5E%7B-6%7D%5Csin90)
![F= 3.2\times10^{-20}\ N](https://tex.z-dn.net/?f=F%3D%203.2%5Ctimes10%5E%7B-20%7D%5C%20N)
Hence, The force is ![3.2\times10^{-20}\ N](https://tex.z-dn.net/?f=3.2%5Ctimes10%5E%7B-20%7D%5C%20N)
Answer:
<h2>
The force is ![36*10^-^9 N](https://tex.z-dn.net/?f=36%2A10%5E-%5E9%20N)
</h2>
Explanation:
According to newtons law on gravitation formula, which we are going to apply,
![F= \frac{Gm1m2}{d^2}](https://tex.z-dn.net/?f=F%3D%20%5Cfrac%7BGm1m2%7D%7Bd%5E2%7D)
Where G represents universal gravitation constant, ![6.67*10^-^1^1 N.m^2/kg^2](https://tex.z-dn.net/?f=6.67%2A10%5E-%5E1%5E1%20N.m%5E2%2Fkg%5E2)
Given data
m1= 90 kg
m2= 60 kg
d= 1 m
Substituting our given data into the formula we have
![F= \frac{6.67*10^-^1^1*90*60}{1^2}](https://tex.z-dn.net/?f=F%3D%20%5Cfrac%7B6.67%2A10%5E-%5E1%5E1%2A90%2A60%7D%7B1%5E2%7D)
![F= \frac{6.67*10^-^1^1*5.4*10^3}{1^2}\\\\F= 36*10^-^9 N](https://tex.z-dn.net/?f=F%3D%20%5Cfrac%7B6.67%2A10%5E-%5E1%5E1%2A5.4%2A10%5E3%7D%7B1%5E2%7D%5C%5C%5C%5CF%3D%2036%2A10%5E-%5E9%20N)