Uhhhh...you should have paid attention in class, just saying...
As we know,
so, let's solve for charge (q) :
time = 5 minutes = 5 × 60 seconds = 300 seconds.
hence, the charge = 60 coulombs (C)
Answer:
Explanation:
a) Energy stored in spring = 1/2 k x² = .5 x k 0.1²
500 = 5 x 10⁻³ k ,
k = (500/5) x 10³ = 10⁵ N/m
b )
k = 4.5 x 10¹ = 45 N/m
Stored energy = 1/2 k x² = .5 x 45 x 8² x 10⁻⁴ =1440 x 10⁻⁴ J
This energy gets dissipated by friction .
work done by friction = μ mg d
d is the distance traveled under friction
so 1440 x 10⁻⁴ = μ x 3 x 9.8 x 2
μ = 245 x 10⁻⁴ or 0.00245 which appears to be very small. .
Answer:
Vb = k Q / r r <R
Vb = k q / R³ (R² - r²) r >R
Explanation:
The electic potential is defined by
ΔV = - ∫ E .ds
We calculate the potential in the line of the electric pipe, therefore the scalar product reduces the algebraic product
VB - VA = - ∫ E dr
Let's substitute every equation they give us and we find out
r> R
Va = - ∫ (k Q / r²) dr
-Va = - k Q (- 1 / r)
We evaluate with it Va = 0 for r = infinity
Vb = k Q / r r <R
We perform the calculation of the power with the expression of the electric field that they give us
Vb = - int (kQ / R3 r) dr
We integrate and evaluate from the starting point r = R to the final point r <R
Vb = ∫kq / R³ r dr
Vb = k q / R³ (R² - r²)
This is the electric field in the whole space, the places of interest are r = 0, r = R and r = infinity
