Answer:
B=9.1397*10^-4 Tesla
Explanation:
To find the velocity first we put kinetic energy og electron is equal to potential energy of electron
K.E=P.E
![\frac{1}{2}*m*v^{2} =e*V](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%2Am%2Av%5E%7B2%7D%20%20%3De%2AV)
where :
m is the mass of electron
v is the velocity
V is the potential difference
eq 1
Radius of electron moving in magnetic field is given by:
eq 2
where:
m is the mass of electron
v is the velocity
q=e=charge of electron
B is the magnitude of magnetic field
Put v from eq 1 into eq 2
![R=\frac{m*\sqrt{\frac{2*e*V}{m} } }{e B}](https://tex.z-dn.net/?f=R%3D%5Cfrac%7Bm%2A%5Csqrt%7B%5Cfrac%7B2%2Ae%2AV%7D%7Bm%7D%20%7D%20%7D%7Be%20B%7D)
![B=\sqrt{\frac{2*m*V}{e*R^{2} } }](https://tex.z-dn.net/?f=B%3D%5Csqrt%7B%5Cfrac%7B2%2Am%2AV%7D%7Be%2AR%5E%7B2%7D%20%7D%20%7D)
![B=\sqrt{\frac{2*(9.31*10^{-31})*(2.12*10^{3}) }{(1.60*10^{-19})*(0.170)^{2} } }](https://tex.z-dn.net/?f=B%3D%5Csqrt%7B%5Cfrac%7B2%2A%289.31%2A10%5E%7B-31%7D%29%2A%282.12%2A10%5E%7B3%7D%29%20%20%7D%7B%281.60%2A10%5E%7B-19%7D%29%2A%280.170%29%5E%7B2%7D%20%20%7D%20%7D)
B=9.1397*10^-4 Tesla
Answer
22.5 m/s
Explanation
We shall use the trigonometric ratio cosine to find the horizontal component.
cos = adjacent/hypotenuse
Adjacent is the horizontal and hypotenuse is the fly speed.
cos 30° = horizontal / 26
horizontal velocity = 26 × cos 30°
= 26 × 0.866
= 22.5166
= 22.5 m/s
Answer:
The inlet velocity is 21.9 m/s.
The mass flow rate at reach exit is 1.7 kg/s.
Explanation:
Given that,
Mass flow rate = 2 kg/s
Diameter of inlet pipe = 5.2 cm
Fifteen percent of the flow leaves through location (2) and the remainder leaves at (3)
The mass flow rate is
![m_{2}=0.15\times2](https://tex.z-dn.net/?f=m_%7B2%7D%3D0.15%5Ctimes2)
We need to calculate the mass flow rate at reach exit
Using formula of mass
![m_{3}=m_{1}-m_{2}](https://tex.z-dn.net/?f=m_%7B3%7D%3Dm_%7B1%7D-m_%7B2%7D)
![m_{3}=2-0.15\times2](https://tex.z-dn.net/?f=m_%7B3%7D%3D2-0.15%5Ctimes2)
![m_{3}=1.7\ kg/s](https://tex.z-dn.net/?f=m_%7B3%7D%3D1.7%5C%20kg%2Fs)
We need to calculate the inlet velocity
Using formula of velocity
![v=\dfrac{m}{\rho A}](https://tex.z-dn.net/?f=v%3D%5Cdfrac%7Bm%7D%7B%5Crho%20A%7D)
Put the value into the formula
![v=\dfrac{2}{42.868\times\dfrac{\pi}{4}\times(5.2\times10^{-2})^2}](https://tex.z-dn.net/?f=v%3D%5Cdfrac%7B2%7D%7B42.868%5Ctimes%5Cdfrac%7B%5Cpi%7D%7B4%7D%5Ctimes%285.2%5Ctimes10%5E%7B-2%7D%29%5E2%7D)
![v=21.9\ m/s](https://tex.z-dn.net/?f=v%3D21.9%5C%20m%2Fs)
Hence, The inlet velocity is 21.9 m/s.
The mass flow rate at reach exit is 1.7 kg/s.
2000÷330=6.06 repatant so the answer would be about 6.06 seconds