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
the answer is spiral galaxies
Answer:
a)At the mean position
b)At the extremes positions
Explanation:
Given that mass is having oscillation motion.
We know that
1. At the mean position -The velocity of the mass is maximum and the acceleration of the mass is minimum.The net force on the mass will be zero.
2. At the extreme position-The velocity of the mass is minimum and the acceleration of the mass is maximum.The net force on the mass will not be zero.
Therefore
a)At the mean position
b)At the extremes positions
Explanation:
solution: mass m = 5g = 0.005kg; extension e = 7cm = 0.07m; force f = 70 N; velocity = ?; using: work done in elastic material w = 1/2 fe = 1/2 ke2 = 1/2 mv2 - the kinetic energy of the moving stone. 1/2 fe =...