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:
d = 90 ft
Explanation:
Here in each swing the distance sweeps by the swing is half of the initial distance that it will move
So here we can say that total distance in whole motion is given as
![d = 45 + \frac{45}{2} + \frac{45}{4} + \frac{45}{8}..........](https://tex.z-dn.net/?f=d%20%3D%2045%20%2B%20%5Cfrac%7B45%7D%7B2%7D%20%2B%20%5Cfrac%7B45%7D%7B4%7D%20%2B%20%5Cfrac%7B45%7D%7B8%7D..........)
since it is half of the distance that it will move in each swing so it would be a geometric progression with common ratio of 1/2
so sum of such GP is given by the formula
![S = \frac{a}{1 - r}](https://tex.z-dn.net/?f=S%20%3D%20%5Cfrac%7Ba%7D%7B1%20-%20r%7D)
![d = \frac{45}{1 - \frac{1}{2}}](https://tex.z-dn.net/?f=d%20%3D%20%5Cfrac%7B45%7D%7B1%20-%20%5Cfrac%7B1%7D%7B2%7D%7D)
![d = 90 ft](https://tex.z-dn.net/?f=d%20%3D%2090%20ft)
Answer:
Use a faster than normal approach and landing speed.
Explanation
For pilots, it is one of the critical moments of the flight that concentrates 12% of fatal accidents. The main difficulty lies in reaching enough speed to take flight within the space of the runway. At present, it ceased to be a challenge for the aircraft, since the engine power improved, so the takeoff ceased to be the most dangerous moment of the flight.
One of the risks that aircraft face today is that some of the engines fail while the plane accelerates. In that case, the pilot must decide in an instant whether it is better to take flight and solve the problem in the air or if it is preferable not to take off.
Although for many staying on the ground might seem the most sensible option, it is not as simple as it seems: to suddenly decelerate an aircraft, with the weight it has and the speed it reaches can cause accidents. However, today a special cement was designed that runs around the runways of the airports, which when coming into contact with the wheels of the aircraft the ground breaks and helps to slow down.
Friction occurs when the surfaces and heat of two surfaces grind against each other.
I think these are the answers
Answer:
3.6 m
Explanation:
![\lambda_R = 650 \ nm\\\\\lambda_R = 650*10^{-9} m\\\\L \ should \ be \ minimum \\\\i.e \ 0.25 \ mm\\\\= 0.25 *10^{-3} m](https://tex.z-dn.net/?f=%5Clambda_R%20%3D%20650%20%5C%20nm%5C%5C%5C%5C%5Clambda_R%20%3D%20650%2A10%5E%7B-9%7D%20m%5C%5C%5C%5CL%20%5C%20should%20%5C%20be%20%5C%20minimum%20%5C%5C%5C%5Ci.e%20%5C%20%200.25%20%5C%20mm%5C%5C%5C%5C%3D%200.25%20%2A10%5E%7B-3%7D%20m)
![\lambda_R = 700 \ nm\\\\\lambda_R = 700*10^{-9} m\\\\](https://tex.z-dn.net/?f=%5Clambda_R%20%3D%20700%20%5C%20nm%5C%5C%5C%5C%5Clambda_R%20%3D%20700%2A10%5E%7B-9%7D%20m%5C%5C%5C%5C)
Also
![\beta = 1 \ mm \ fringe \ width](https://tex.z-dn.net/?f=%5Cbeta%20%3D%201%20%5C%20mm%20%5C%20fringe%20%5C%20%20width)
![D_{min} = \frac{\beta d}{\lambda}\\\\D_{min} = \frac{10^{-3}*0.25*10^{-3}}{700*10^{-9}}\\\\D_{min} = 3.57 \\D_{min} = 3.6 m](https://tex.z-dn.net/?f=D_%7Bmin%7D%20%3D%20%5Cfrac%7B%5Cbeta%20d%7D%7B%5Clambda%7D%5C%5C%5C%5CD_%7Bmin%7D%20%3D%20%5Cfrac%7B10%5E%7B-3%7D%2A0.25%2A10%5E%7B-3%7D%7D%7B700%2A10%5E%7B-9%7D%7D%5C%5C%5C%5CD_%7Bmin%7D%20%3D%203.57%20%5C%5CD_%7Bmin%7D%20%3D%20%203.6%20m)
Therefore, the minimum distance L you can place a screen from the double slit that will give you an interference pattern on the screen that you can accurately measure using an ordinary 30 cm (12 in) ruler. = 3.6 m