Calculation, add them. 0.85 x 5 = 4.25 Ω
Answer:
a) 1.082 × 10⁻¹⁹C ( e = 1.6 × 10⁻¹⁹C)
b) 3.466 × 10¹¹ N/C
Explanation:
a)
p(r) = -A exp ( - 2r/a₀)
Q = ₀∫^∞ ₀∫^π ₀∫^2xπ p(r)dV = -A ₀∫^∞ ₀∫^π ₀∫^2π exp ( - 2r/a₀)r² sinθdrdθd∅
Q = -4πA ₀∫^∞ exp ( - 2r/a₀)r²dr = -e
now using integration by parts;
A = e / πa₀³
p(r) = - (e / πa₀³) exp (-2r/a₀)
Now Net charge inside a sphere of radius a₀ i.e Qnet is;
= e - (e / πa₀³) ₀∫^a₀ ₀∫^π ₀∫^2π r² exp (-2r/a₀)dr
= e - e + 5e exp (-2) = 1.082 × 10⁻¹⁹C ( e = 1.6 × 10⁻¹⁹C)
b)
Using Gauss's law,
E × 4πa₀ ² = Qnet / ∈₀
E = 4πa₀ ² × Qnet × 1/a₀²
E = 3.466 × 10¹¹ N/C
Answer:
The maximum power density in the reactor is 37.562 KW/L.
Explanation:
Given that,
Height = 10 ft = 3.048 m
Diameter = 10 ft = 3.048 m
Flux = 1.5
Power = 835 MW
We need to calculate the volume of cylinder
Using formula of volume

Put the value into the formula


We need to calculate the maximum power density in the reactor
Using formula of power density

Where, P = power density
E = energy
V = volume
Put the value into the formula


Hence, The maximum power density in the reactor is 37.562 KW/L.
Answer:
I=2 kg.m/s
Explanation:
The impulse is defined as the change of momentum:
![I=p_f-p_o\\I=m*v_f-m*v_o\\I=0.02kg*[(-60m/s)-40m/s]\\I=2kg.m/s](https://tex.z-dn.net/?f=I%3Dp_f-p_o%5C%5CI%3Dm%2Av_f-m%2Av_o%5C%5CI%3D0.02kg%2A%5B%28-60m%2Fs%29-40m%2Fs%5D%5C%5CI%3D2kg.m%2Fs)
We took the final velocity as negative since it is going on the opposite direction of the intial motion of the ball.