Answer:
The magnetic energy density is 
Explanation:
From the question we are told that
The earths magnetic field is 
Now
=> 
So
Generally the magnetic energy density is mathematically represented as
Here 
is the permeability of free space with a constant value
substituting values for equation above
Kinetic energy of the horse and rider = 1/2 (mass) (speed)²
At 4.0 m/s : KE = 1/2 (55kg) (4 m/s)² = 440 joules
At 6.0 m/s : KE = 1/2 (55kg) (6 m/s)² = 990 joules
She increased the kinetic energy of herself and her vehicle by 550 joules,
so she must have put at least that much work into it.
Displacement is your answer :)
Answer:Answer:
They are inhaled or eaten
Explanation:
Radioactivity can be defined as the process in which an unstable atomic nucleus spontaneously emits ionizing radiation and charge particles. This eventually results in the formation of an energetically stable atomic nucleus.
Examples of radioactive elements are Uranium, Polonium, Thorium, Radon, etc. The radiation emitted during this process is classified as;
1. Alpha radiation (α).
2. Beta radiation (β).
3. Gamma radiation (G).
Alpha-emitting substances, such as radon gas, can be a serious health hazard only if they are inhaled or eaten. Alpha-emitting substances compared to other radiation has very short-range particle and as such cannot penetrate the human skin or body.
Answer:
A = 13000K has a maximum at lam = 1,9984 10⁻⁷ m = 199.84 nm
, this star is visually separated from the other two by its constant emission spectrum and is not affected by the other two.
we have a fluctuation of the intensity emitted by the stars. Consequently by this fluctuation the amateur astronomer can conclude that this system is made up of two stars.
Explanation:
The radiation of a black body is characterized by its temperature, with Wien's law of displacement we can find the maximum wavelength emitted by each star.
λ T = 2,898 10⁻³
therefore the emission the star of A = 13000K has a maximum at lam = 1,9984 10⁻⁷ m = 199.84 nm
The emission of the premiere is in the ultraviolet light range, as this star is visually separated from the other two by its constant emission spectrum and is not affected by the other two.
The burst with A = 4300K has a bad emission maximum = 6.7395 10⁻⁷ m = 673.95 nm, which corresponds to an emission in the visible in the orange range, giving a blackbody spectrum of this range, but since the emission is formed by two stars, we see that when the two are placed one in front of the other the intensity of the emission must increase significantly and when they are placed next to each other the intensity reaches its minimum, consequently we have a fluctuation of the intensity emitted by the stars.
Consequently by this fluctuation the amateur astronomer can conclude that this system is made up of two stars.
