Answer: 14.52*10^6 m/s
Explanation: In order to explain this problem we have to consider the energy conservation for the electron within the coaxial cylidrical wire.
the change in potential energy for the electron; e*ΔV is equal to energy kinetic gained for the electron so:
e*ΔV=1/2*m*v^2 v^=(2*e*ΔV/m)^1/2= (2*1.6*10^-19*600/9.1*10^-31)^1/2=14.52 *10^6 m/s
It is definitely letter D. <span>A1 and B1 are like poles, but there is not enough information to tell whether they are north poles or south poles.
A1 and B1 is either both north poles or both south poles. Repulsion of both magnets says it all--like poles always repel while opposite poles always attract. Thus, the best conclusion to this would be choice D.</span>
Answer:
a) 600nm
b) 300nm
Explanation:
the path difference = 2t
t = thickness of the film
L' = wavelength of light in film = L/n
L = wavength of light in air
n = refractive index of glass
(a)
for destructive interference 2t = L'/2 = L/2n
L = 4*t*n
= 4*120*10^-9*1.25
L = 600 nm
(b)
for constructive interference 2t = L' = L/1.25
L = 2tn
= 2 × 1.25 × 120nm
= 300 nm
To solve this problem we will use the concepts related to angular motion equations. Therefore we will have that the angular acceleration will be equivalent to the change in the angular velocity per unit of time.
Later we will use the relationship between linear velocity, radius and angular velocity to find said angular velocity and use it in the mathematical expression of angular acceleration.
The average angular acceleration
Here
= Angular acceleration
Initial and final angular velocity
There is not initial angular velocity,then
We know that the relation between the tangential velocity with the angular velocity is given by,
Here,
r = Radius
= Angular velocity,
Rearranging to find the angular velocity
Remember that the radius is half te diameter.
Now replacing this expression at the first equation we have,
Therefore teh average angular acceleration of each wheel is 
