A red rubber ball rolls down a hill from rest with an acceleration of 7.8 m/s 2 . How fast is it moving after it has traveled 5
1 answer:
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Answer:
P /K = 1,997 10⁻³⁶ s⁻¹
Explanation:
For this exercise let's start by finding the radiation emitted from the accelerator
=
the radius of the orbit is the radius of the accelerator a = r = 0.530 m
let's calculate
\frac{dE}{dt} = [(1.6 10⁻¹⁹)² 0.530²] / [6π 8.85 10⁻¹² (3 108)³]
P= \frac{dE}{dt}= 1.597 10⁻⁵⁴ W
Now let's reduce the kinetic energy to SI units
K = 5.0 10⁶ eV (1.6 10⁻¹⁹ J / 1 eV) = 8.0 10⁻¹⁹ J
the fraction of energy emitted is
P / K = 1.597 10⁻⁵⁴ / 8.0 10⁻¹⁹
P /K = 1,997 10⁻³⁶ s⁻¹
Answer:
For electric field inside cylinder, check image 02 attached
For electric field outside cylinder, check image 03 attached
Explanation:
Let's consider the polarized cylinder as superposition of two cylinders with opposite,equal, uniform charge densities in a way shown in the figure in the "image 01"d attached ;
In general, if we have an object with polarization (P¬) , then we have to take two objects with similar shape to the system, with opposite, equal, and uniform charge densities and then we super-impose these two objects in such a way that the total dipole moment of this superimposed system is equal to the total dipole moment of original system.
Now, we can take the super- imposed system as equivalent to the original system for calculating electric field and potential.
Therefore,
For the electric field inside the cylinder, check the solution in "image 02" i attached
For the electric field outside the cylinder, check, "image 03" i attached.
