Consider the field E1 in the dielectric on the left of this interface. E1= 3ax+4ay+5az The interface lies in the z=0 plane. Find
the total E2 field in medium #2. Also find the angles that the fields in each dielectric make with the normal to the interface. ET2 Ert = 2 Er2 = 5 E2 = EN Do EN2 Z0
1 answer:
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Answer:
The velocity of the photo electron is .
Explanation:
Given that,
Supplied energy,
Minimum energy of the electron to escape from the metal,
We need to find the velocity of the photo electron. The energy supplied by the photon is equal to the sum of minimum escape energy and the kinetic energy of the escaping electron. So,
The formula of kinetic energy is given by :
So, the velocity of the photo electron is .
Answer:
α = 0.0135 rad/s²
Explanation:
given,
t = 133 min = 133 x 60 = 7980 s
angular speed varies from 570 rpm to 1600 rpm
now,
570 rpm =
= 59.69 rad/s
1600 rpm = =
= 167.6 rad/s
using equation of rotational motion
ωf = ωi + αt
167.6 = 59.7 + α x 7980
α x 7980 = 107.9
α = 0.0135 rad/s²