Answer:
a. 1nC
b. 0C
Explanation:
Net charge q stored on plate of capacitor is
q = CV
Where C = 2uF = 2 x 10^-6F
V= 50v
q = 20 x 10^-6 x 50 = 1000 x 10^-6 = 1000uF = 10^-9 = 1nC
b. the total net charge on another plate is equal in absolute value to the first one, but it is charged with opposite Pole so always is valid that total net charge on both plates are equal to zero.
That's the other charge on the plate is -1nc
1 nC + -1nC = 0C
Answer:
p = 59.11 dollars
Explanation:
Given
Price: p(x) = 8eˣ (0 ≤ x ≤ 2)
Revenue; R = x*p = 8xeˣ
p = ? when R be at maximum
We can apply
dR/dx = d(x*p)/dx = 0
⇒ d(8xeˣ)/dx = 8*(1*eˣ + x*eˣ) = 0
⇒ eˣ*(1 + x) = 0 ⇒ x = - 1
as x = - 1 ∉ [0, 2]
then, we have
p(0) = 8e⁰ = 8
R = 0*8 = 0
If x = 1
p(1) = 8e¹ ≈ 21.74
R = 1*21.74 = 21.74
If x = 2
p(2) = 8e² ≈ 59.11
R = 2*59.11 = 118.22
Implies that, R(x) is maximum at x = 2.
Thus, the price that maximize the revenue of the company is 59.11 dollars.
I would have to say D. all of the above
