Answer:
Explanation:
The electric field equation of a electromagnetic wave is given by:
(1)
- E(max) is the maximun value of E, it means the amplitude of the wave.
- k is the wave number
- ω is the angular frequency
We know that the wave length is λ = 700 nm and the peak electric field magnitude of 3.5 V/m, this value is correspond a E(max).
By definition:
![k=8.98*10^{6} [rad/m]](https://tex.z-dn.net/?f=k%3D8.98%2A10%5E%7B6%7D%20%5Brad%2Fm%5D)
And the relation between λ and f is:
The angular frequency equation is:
![\omega=2.69*10^{15} [rad/s]](https://tex.z-dn.net/?f=%5Comega%3D2.69%2A10%5E%7B15%7D%20%5Brad%2Fs%5D)
Therefore, the E equation, suing (1), will be:
(2)
For the magnetic field we have the next equation:
(3)
It is the same as E. Here we just need to find B(max).
We can use this equation:
Putting this in (3), finally we will have:
(4)
I hope it helps you!
E=(mV^2)/2
m=1000kg, V=20m/s
then, E=(1000kg*(20m/s)^2)/2
E=(1000*400)/2 J = 200000J
I think the answer you're looking for is -are randomly oriented. if not sorry... i tried.
Answer:
distance between object and image = 18.9 cm
Explanation:
given data
radius of curvature = 18 cm
focal length = 1/2 radius of curvature
magnification = 40%
to find out
distance between object and image
solution
we know lens formula that is
1/f = 1/v + 1/u ....................1
here f = 18 /2 and v and u is object and image distance
and we know m = 40% = 0.40
so 0.40 = -v / u
so here v = - 0.40 u
so from equation 1
1/f = 1/v + 1/u
2/18 = - 1/0.40u + 1/u
u = -13.5 cm ..................2
and
v = -0.40 (- 13.5)
v = 5.4 cm ......................3
so from equation 2 and 3
distance between object and image = 5.4 + 13.5
distance between object and image = 18.9 cm
Answer:
When the object is placed between centre of curvature and principal focus of a concave mirror the image formed is beyond C as shown in the figure and it is real, inverted and magnified.
