Answer: a) 73.41 10^-12 F; b)4.83* 10^3 N/C; c) 3.66 *10^3 N/C
Explanation: To solve this problem we have to consider the following: The Capacity= Charge/Potential Difference
As we know the capacity is value that depend on the geometry of the capacitor, in our case two concentric spheres.
So Potential Difference between the spheres is given by:
ΔV=-
Where E = k*Q/ r^2
so we have 
then
Vb-Va=k*Q(1/b-1/a)=kQ (ab/b-a)
Finally using C=Q/ΔV=ab/(k(b-a))
To caclulate the electric firld we first obtain the charge
Q=ΔV*C=120 V*73.41 10^-12 F=8.8 10^-9 C
so E=KQ/r^2 for both values of r
r=12.8 cm ( in meters)
r2=14.7 cm
E(r1)=4.83* 10^3 N/C
E(r2)=3.66 *10^3 N/C
Answer:
i) 24.5 m/s
ii) 30,656 m
iii) 89,344 m
Explanation:
Desde una altura de 120 m se deja caer un cuerpo. Calcule a 2.5 s i) la velocidad que toma; ii) cuánto ha disminuido; iii) cuánto queda por hacer
i) Los parámetros dados son;
Altura inicial, s = 120 m
El tiempo en caída libre = 2.5 s
De la ecuación de caída libre, tenemos;
v = u + gt
Dónde:
u = Velocidad inicial = 0 m / s
g = Aceleración debida a la gravedad = 9.81 m / s²
t = Tiempo de caída libre = 2.5 s
Por lo tanto;
v = 0 + 9.8 × 2.5 = 24.5 m / s
ii) El nivel que el cuerpo ha alcanzado en 2.5 segundos está dado por la relación
s = u · t + 1/2 · g · t²
= 0 × 2.5 + 1/2 × 9.81 × 2.5² = 30.656 m
iii) La altura restante = 120 - 30.656 = 89.344 m.
I haven't worked on Part-A, and I don't happen to know the magnitude of the gravitational force that the Sun exerts on the Earth.
But whatever it is, it's exactly, precisely, identical, the same, and equal to the magnitude of the gravitational force that the Earth exerts on the Sun.
I think that's the THIRD choice here, but I'm not sure of that either.
The radio frequencies push one air molecule that then bumps into a different air molecule.....which then hits another and another causing a line of crashing molecules that lead inside your ear and hits your ear drum causing it to vibrate which causes the sounds.
I'm not sure if this is correct but it's what I'll do
This is free-fall problem.
Stone A is thrown upward, at the point it falls down to the place where it was thrown, the velocity is -15m/s.
Now I choose the bridge is the origin. From the bridge, stone A and B fall the same distance which means Ya = Yb ( vertical distance )
Ya = Vo(t + 2) + 1/2a(t+2)^2
= -15(t + 2) + 1/2(9.8)(t^2 + 4t + 4)
= -15t - 30 + 4.5(t^2 + 4t + 4)
= -15t - 30 + 4.5t^2 + 18t + 18
= 4.5t^2 +3t - 12
Yb = Vo(t) + 1/2a(t)^2
= 0 + 4.5t^2
4.5t^2 = 4.5t^2 +3t - 12
0 = 3t - 12
4 = t
Time for Stone B is 4s
Time for Stone A is 6s
