Two isolated, concentric, conducting spherical shells have radii R1 = 0.520 m and R2 = 1.40 m, uniform charges q1 = +2.50 μC and q2 = +2.70 μC, and negligible thicknesses. What is the magnitude of the electric field E at radial distance (a) r = 4.80 m, (b) r = 0.710 m, and (c) r = 0.210 m? With V = 0 at infinity, what is V at (d) r = 4.80 m, (e) r = 1.40 m, (f) r = 0.710 m, (g) r = 0.520 m, (h) r = 0.210 m, and (i) r = 0?

A) E = 2031.7 N/C

B) E = 44,643.8 N/C

C) E = 0 N/C

D) V = 9752.14 N/C

E) V = 33,435.91 N/C

F) V = 49,058 N/C

G) V = 60,639.65 N/C

H) V = 60,639.65 N/C

I) V = 60,639.65 N/C

Explanation:

We shall make use of the following;

At r < R1; E_in = 0 - - - - - (eq1)

At R1 < r < R2 ;

E_bet = q1/(4πε_o(r²)) -- - - - - (eq2)

At R1 < r < R2 ;

E_out = (q1 + q2)/(4πε_o(r²)) - (eq3)

At R1 < r < R2 ;

V_in = [1/(4πε_o)]•[(q1/R1) + (q2/R2)] - - - - - (eq4)

At R1 < r < R2 ;

V_bet = [1/(4πε_o]•[(q1/r) + (q2/R2)] - - - - - (eq5)

At r > R2 ;

V_out= (q1 + q2)/(4πε_o(r)) - - - - - (eq6)

A) r = 4.80 m. This falls into r > R2. But since we are looking for electric field, E, we will use eq (3)

Thus, E = (q1 + q2)/(4πε_o(r)²)

q1 = +2.50 μC = 2.5 x 10^(-6) C

q2 = 2.7 μC = 2.7 x 10^(-6) C

ε_o = 8.84 x 10^(-12) F/m

E = [(2.5 x 10^(-6))] + (2.7 x 10^(-6))] /(4πx8.84 x 10^(-12)x(4.8)²)

= 2031.7 N/C

B) r = 0.710 m, we'll use eq (2)

E_bet = q1/(4πε_o(r²))

E_bet = [2.5 x 10^(-6)]/(4πx8.84 x 10^(-12)x(0.71²)) = 44,643.8 N/C

C) r = 0.210 m

This corresponds with condition for eq 1 where enclosed charge is zero.

Thus, E_in = 0 N/C

D) V at r = 4.80 m

Corresponds to eq(6) where;

V_out= (q1 + q2)/(4πε_o(r))

Thus,

V_out = [(2.5 x 10^(-6))] + (2.7 x 10^(-6))] /(4πx8.84 x 10^(-12)x(4.8)) = 9752.14 N/C

E) V at r = 1.40 m

Corresponds with condition for eq(6)

V_out= (q1 + q2)/(4πε_o(r))

Thus,

V_out = [(2.5 x 10^(-6))] + (2.7 x 10^(-6))] /(4πx8.84 x 10^(-12)x(1.4)) 33,435.91 N/C

F) r = 0.710 Corresponds with condition for eq(5)

Thus,

V_bet = [1/(4πx8.84 x 10^(-12))]•[(2.5 x 10^(-6)/0.71) + (2.7 x 10^(-6)/1.4)] = 97318.13 N/C = 49,058 N/C

G) V at r = 0.520 m corresponds with condition for eq(5)

Thus,

V_bet = [1/(4π x 8.84 x 10^(-12))]•[(2.5 x 10^(-6)/0.52) + (2.7 x 10^(-6)/1.4)] = 60,639.65 N/C

H) V at r = 0.210 m corresponds with condition for eq(4)

V_in = [1/4πε_o]•[(q1/R1) + (q2/R2)] - - - - - (eq4)

Thus,

V_in = [1/(4π x 8.84 x 10^(-12))]•[(2.5 x 10^(-6)/0.52) + (2.7 x 10^(-6)/1.4)]

= 60,639.65 N/C

I) V at r = 0 m corresponds with condition for eq(4)

V_in = [1/4πε_o]•[(q1/R1) + (q2/R2)] - - - - - (eq4)

Thus,

V_in = [1/(4π x 8.84 x 10^(-12))]•[(2.5 x 10^(-6)/0.52) + (2.7 x 10^(-6)/1.4)]

= 60,639.65 N/C

muscle endurance

explanation:

i can do 50 in a row ( but i get so tired after that)

russian is not my first language, so i'm afraid i can't write out an answer in russian, but hopefully you can find a way to do so. here's my best translation of the problem: you're given a plot of a particle's position in one-dimensional space over time, and you're asked to find either the total distance traveled by the particle after 6 seconds had elapsed, or the total displacement of the particle after 6 seconds.

if you want the total distance traveled, you would first need to determine the particle's velocity as a function of time, then integrate that over the 6-second interval. we know the position function is quadratic with roots at 0 and 6, so we have

where is unknown. because the position is parabolic, there is symmetry about the midpoint at , and we know that

so that the position function is

which means the velocity is

then the total distance traveled is

so the total distance traveled is 20 m.

on the other hand, if you want to find displacement: the particle starts at and ends at , so its displacement is 0.