A 3 kg ball rolls at 2m/s does this system favor kinetic energy or potential energy?
1 answer:
Answer:
Kinetic energy
Explanation:
Total energy of any system is always conserved.
E= Total energy
K = kinetic energy
U = Potential energy
Kinetic energy is the energy acquired by the system due to the virtue of it's motion.
Potential energy is the energy acquired by the system due to the virtue of it's configuration.
Also, Kinetic energy
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Answer:
a) E =0, b) E = 1,129 10¹⁰ N / C , c) E = 3.33 10¹⁰ N / C
Explanation:
To solve this exercise we can use Gauss's law
Ф = ∫ E. dA = / ε₀
Where we must define a Gaussian surface that is this case is a sphere; the electric field lines are radial and parallel to the radii of the spheres, so the scalar product is reduced to the algebraic product.
E A = /ε₀
The area of a sphere is
A = 4π r²
E = q_{int} / 4πε₀ r²
k = 1 / 4πε₀
E = k q_{int} / r²
To find the charge inside the surface we can use the concept of density
ρ = q_{int} / V ’
q_{int} = ρ V ’
V ’= 4/3 π r’³
Where V ’is the volume of the sphere inside the Gaussian surface
Let's apply this expression to our problem
a) The electric field in center r = 0
Since there is no charge inside, the field must be zero
E = 0
b) for the radius of r = 6.0 cm
In this case the charge inside corresponds to the inner sphere
q_{int} = 5.0 4/3 π 0.06³
q_{int} = 4.52 10⁻³ C
E = 8.99 10⁹ 4.52 10⁻³ / 0.06²
E = 1,129 10¹⁰ N / C
c) The electric field for r = 12 cm = 0.12 m
In this case the two spheres have the charge inside the Gaussian surface, for which we must calculate the net charge.
The charge of the inner sphere is q₁ = - 4.52 10⁻³ C
The charge for the outermost sphere is
q₂ = ρ 4/3 π r₂³
q₂ = 8.0 4/3 π 0.12³
q₂ = 5.79 10⁻² C
The net charge is
q_{int} = q₁ + q₂
q_{int} = -4.52 10⁻³ + 5.79 10⁻²
q_{int} = 0.05338 C
The electric field is
E = 8.99 10⁹ 0.05338 / 0.12²
E = 3.33 10¹⁰ N / C