Answer:

Explanation:
The capacitance of the parallel-plate capacitor is given by

where
ϵ0 = 8.85x10-12 C2/N.m2 is the vacuum permittivity
k = 3.00 is the dielectric constant
is the area of the plates
d = 9.00 mm = 0.009 m is the separation between the plates
Substituting,

Now we can calculate the energy of the capacitor, given by:

where
C is the capacitance
V = 15.0 V is the potential difference
Substituting,

Answer:
La velocidad de la luz en el vacío es una constante universal con el valor de 299 792 458 m/s (186 282,397 mi/s),aunque suele aproximarse a 3·108 m/s. Se simboliza con la letra c, proveniente del latín celéritās (en español, celeridad o rapidez).
¿Cuál es la consecuencia que a velocidad de la luz sea constante?
Respuesta. En modificaciones del vacío más sutiles, como espacios curvos, efecto Casimir, poblaciones térmicas o presencia de campos externos, la velocidad de la luz depende de la densidad de energía de ese vacío.
they are called "cells"
hope this is the answer is what your looking for.
When t=2, the ball has fallen d(2) = 16 (2²) = 64 feet .
When t=5, the ball has fallen d(5) = 16 (5²) = 400 feet .
Distance fallen from t=2 until t=5 is (400 - 64) = 336 feet.
Time period between t=2 until t=5 is (5 - 2) = 3 seconds.
Average speed of the ball from t=2 until t=5 is
(distance covered) / (time to cover the distance)
= 336 feet / 3 seconds = 112 feet per second.
That's what choice-C says.
Answer:
Magnitude of the net force on q₁-
Fn₁=1403 N
Magnitude of the net force on q₂+
Fn₂= 810 N
Magnitude of the net force on q₃+
Fn₃= 810 N
Explanation:
Look at the attached graphic:
The charges of the same sign exert forces of repulsion and the charges of opposite sign exert forces of attraction.
Each of the charges experiences 2 forces and these forces are equal and we calculate them with Coulomb's law:
F= (k*q*q)/(d)²
F= (9*10⁹*3*10⁻⁶*3*10⁻⁶)(0.01)² =810N
Magnitude of the net force on q₁-
Fn₁x= 0
Fn₁y= 2*F*sin60 = 2*810*sin60° = 1403 N
Fn₁=1403 N
Magnitude of the net force on q₃+
Fn₃x= 810- 810 cos 60° = 405 N
Fn₃y= 810*sin 60° = 701.5 N

Fn₃ = 810 N
Magnitude of the net force on q₂+
Fn₂ = Fn₃ = 810 N