Answer:
a) 20.81 J
b) 8.29 J
Explanation:
V = iR + L di/dt
where
i = a(1-e^-kt)
for large t
i = V/R
i = 24 / 9.4
i = 2.55 A
so
i = 2.55(1-e^-kt)
di/dt = 2.55 k e^-kt
24 = 24-24e^-kt + 6.4(2.55)k e^-kt
24 = 6.4(2.55) k
k = 24 / (6.4 * 2.55)
k = 24 / 16.32
k = 1.47 = R/L
so
i = 2.55(1-e^-(Rt/L))
current is maximum at great t
i max = 2.55 - 0
energy = (1/2) L i^2
E = (1/2)(6.4)2.55^2
E = 20.81 Joules
one time constant T = L/R and e^-(Rt/L) = 1/e = .368
i = 2.55 (1 - 0.368)
i = 2.55 * 0.632
i = 1.61 amps
energy = (1/2)(6.4)1.61^2
E = 8.29 Joules
It is b. Wave. It moved by sound waves, water waves etc.
Answer:
Explanation:
To solve this problem we can use the Gauss' Theorem
Hence, we have:
where QN is the total net charge inside the Gaussian surface, r is the point where we are going to compute E and ε0 is the dielectric permitivity. For each value of r we have to take into account what is the net charge inside the Gaussian surface.
a) r=4.80m (r>R2)
QN=+2.50 μC+2.70 μC = 5.2 μC
b) r=0.70m (R1<r<R2)
QN=+2.50 μC
c) r=0.210 (r<R1)
Inside the spherical shell of radius R1 the net charge is zero. Hence
E=0N/C
- For the calculation of the potential we have
Thus, we compute the potential by using the net charge of the Gaussian surface
d) r=0.210 (r<R1)
Inside the spherical shell the net charge is zero, thus
E=0N/C
e) r=1.40m (R1<r<=R2)
In this case we take the net charge from the first spherical shell
QN=+2.50 μC
f) r=0.70m
QN=+2.50 μC
V=3.164*10^{4}Nm/C
g) r=0.52
QN=0
V=0
h) r=0.2
QN=0
V=0
HOPE THIS HELPS!!