To solve this exercise it is necessary to apply the concepts related to the Snells law.
The law defines that,
Incident index
Refracted index
= Incident angle
Refracted angle
Our values are given by
Refractory angle generated when light passes through the fiber.
Replacing we have,
Now for the calculation of the maximum angle we will subtract the minimum value previously found at the angle of 90 degrees which is the maximum. Then,
Therefore the critical angle for the light ray to remain insider the fiber is 28.6°
Answer:
9.8N
Explanation:
Here we can get gravitational acceleration according to the place where object is placed by bellow equation
g = GM/R²
g - Gravitational Acceleration
G - Gravitational constant (6.67×10-11)
R - Distance ( Radius )
g = 6.67 × 10-11 × 1024 /(6.37×106)²
g = 9.8 m/s²
There for
Weight = Mass × Gravitational acceleration
= 1×9.8
= 9.8 N
Answer:
The final kinetic energy of the Helium nucleus (alpha particle) after been scattered through an angle of 120° is
8.00 x 10-13J
Explanation:
In Rutherford Scattering experiment, the collision of the helium nucleus with the gold nucleus is an ELASTIC COLLISION. This means that the kinetic energy is conserved ( The same before and after the collision).
Thus, the final kinetic energy of the helium nucleus is the same as initial kinetic energy (8.00 x 10^-13Joules)
Although, the kinetic energy is converted to potential energy in Coulomb's law equation.
That is,
1/2(mv^2) = (K* q1q2)/r
Where m is the mass of helium nucleus, v is its colliding velocity, k is electrostatic constant, q1 is the charge on helium nucleus, q2 is the charge on gold nucleus, r is impact parameter
Answer:
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Explanation: