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Feliz [49]
4 years ago
11

The relationship between a cathode and an anode involves

Physics
1 answer:
Vadim26 [7]4 years ago
3 0
<span>The relationship between a cathode and an anode involves:
</span>
A. electrons.

Negatively ions which have excess electrons move to the Anode, while Postively charged ions which have less electrons move to the Cathode.
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On a distance-time graph, time is shown on the y-axis.<br> A) True<br> B) False
Artist 52 [7]

Answer:

false : In distance time graph,time is shown on the x -axis

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3 years ago
A nonuniform, but spherically symmetric, distribution of charge has a charge density ρ(r) given as follows: ρ(r)=ρ0(1−r/r) for r
Nadusha1986 [10]

A)<span>
dQ = ρ(r) * A * dr = ρ0(1 - r/R) (4πr²)dr = 4π * ρ0(r² - r³/R) dr 
which when integrated from 0 to r is 
total charge = 4π * ρ0 (r³/3 + r^4/(4R)) 
and when r = R our total charge is 
total charge = 4π*ρ0(R³/3 + R³/4) = 4π*ρ0*R³/12 = π*ρ0*R³ / 3 
and after substituting ρ0 = 3Q / πR³ we have 
total charge = Q ◄ 

B) E = kQ/d² 
since the distribution is symmetric spherically 

C) dE = k*dq/r² = k*4π*ρ0(r² - r³/R)dr / r² = k*4π*ρ0(1 - r/R)dr 
so 
E(r) = k*4π*ρ0*(r - r²/(2R)) from zero to r is 
and after substituting for ρ0 is 
E(r) = k*4π*3Q(r - r²/(2R)) / πR³ = 12kQ(r/R³ - r²/(2R^4)) 
which could be expressed other ways. 

D) dE/dr = 0 = 12kQ(1/R³ - r/R^4) means that 
r = R for a min/max (and we know it's a max since r = 0 is a min). 

<span>E) E = 12kQ(R/R³ - R²/(2R^4)) = 12kQ / 2R² = 6kQ / R² </span></span>

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b

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Answer:

Maybe A.... I'm not sure

Explanation:

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