Answer: 586.60N/m
Explanation:
In this scenario, the elastic potential energy of the spring is converted into potential energy.
0.5*K*x^2 = mgh
Thus K = 2mgh/x^2
=(2*2.90*10^-2*9.8*7.23)/(8.37*10^-2)^2
=586.599
Therefore K = 586.60N/m
Part 1
When the solar atmosphere accumulates a lot of magnetic energy
to a point that cannot accumulate more, all that magnetic energy is suddenly released,
and with it, a lot of radiation. So much, that in fact it covers all of the
electromagnetic spectrum; from radio waves to gamma rays. That burst of
radiation is called a solar flare. In a single solar flare the amount of
radiation released is millions of times greater than all the nuclear bombs in
the face if the earth exploding together. Lucky for us, most of the high-energy
radiation dissipates before reaching the Earth, and the radiation that do reach
us, is deflected by the Earth’s magnetic field.
Part 2
1. Not all the radiation
of solar flares that reach the Earth is deflected by its magnetic field; some
of them reach us and charges the upper atmosphere with ionized particles. Those
particles react with the gases in the atmosphere and produce a light; that
light is what we call Auroras borealis or southern nights; One the most beautiful
natural spectacles in earth, who thought Auroras begin their lives as deadly
solar flares.
2. Solar flares
contain a lot of high-energy radiation that is extremely dangerous for our
electronic devices; when they reach the Earth, they can damage sensible
electronics like satellites. A very powerful solar flare could even damage all
the electronic devices on the surface of the Earth.
Answer: It frees up valuable portions of the broadcast spectrum, it has better audio and picture quality, and there are more options on digital broadcasting
Explanation:
Answer:
The correct answer is B
Explanation:
Let's calculate the electric field using Gauss's law, which states that the electric field flow is equal to the charge faced by the dielectric permittivity
Φ
= ∫ E. dA = 
/ ε₀
For this case we create a Gaussian surface that is a sphere. We can see that the two of the sphere and the field lines from the spherical shell grant in the direction whereby the scalar product is reduced to the ordinary product
∫ E dA = / ε₀
The area of a sphere is
A = 4π r²
E 4π r² = / ε₀
E = (1 /4πε₀
) q / r²
Having the solution of the problem let's analyze the points:
A ) r = 3R / 4 = 0.75 R.
In this case there is no charge inside the Gaussian surface therefore the electric field is zero
E = 0
B) r = 5R / 4 = 1.25R
In this case the entire charge is inside the Gaussian surface, the field is
E = (1 /4πε₀
) Q / (1.25R)²
E = (1 /4πε₀
) Q / R2 1 / 1.56²
E₀ = (1 /4π ε₀
) Q / R²
= Eo /1.56
²
= 0.41 Eo
C) r = 2R
All charge inside is inside the Gaussian surface
=(1 /4π ε₀
) Q 1/(2R)²
= (1 /4π ε₀
) q/R² 1/4
= Eo 1/4
= 0.25 Eo
D) False the field changes with distance
The correct answer is B
300N/25 kg= divide them for the answer
