Answer:
a)v= 675V
b) v=-675V
c)v=-450V
Explanation:
Formula: V = k*sum(qi/ri)
Given:
q1 = +6.00nC --> R1 = 2.00cm
q2 = -9.00nC --> R2 = 4.00cm
(a) r = 0. So, inside both shells
![V=k( \frac{q1}{R1}+\frac{q2}{R2} )](https://tex.z-dn.net/?f=V%3Dk%28%20%5Cfrac%7Bq1%7D%7BR1%7D%2B%5Cfrac%7Bq2%7D%7BR2%7D%20%29)
![V=(9x10^{9}[\frac{6x10^{-9} }{2x10^{-2} \\} + \frac{-9x10^{-9} }{4x10^{-2} } ]](https://tex.z-dn.net/?f=V%3D%289x10%5E%7B9%7D%5B%5Cfrac%7B6x10%5E%7B-9%7D%20%7D%7B2x10%5E%7B-2%7D%20%5C%5C%7D%20%20%2B%20%5Cfrac%7B-9x10%5E%7B-9%7D%20%7D%7B4x10%5E%7B-2%7D%20%7D%20%5D)
![V= 9x10^{9} * 7.5x10^{-8}](https://tex.z-dn.net/?f=V%3D%209x10%5E%7B9%7D%20%2A%207.5x10%5E%7B-8%7D)
V=675V
(b) r = 4.00cm. So, This point is outside of shell 1 but inside shell 2.
![V= k(\frac{q1}{r} + \frac{q2}{R2} )](https://tex.z-dn.net/?f=V%3D%20k%28%5Cfrac%7Bq1%7D%7Br%7D%20%2B%20%5Cfrac%7Bq2%7D%7BR2%7D%20%29)
![V=(9x10^{9}) [\frac{6x10^{-9} }{4x10^{-2} } +\frac{-9x10^{-9} }{4x10^{-2} } ]](https://tex.z-dn.net/?f=V%3D%289x10%5E%7B9%7D%29%20%5B%5Cfrac%7B6x10%5E%7B-9%7D%20%7D%7B4x10%5E%7B-2%7D%20%7D%20%20%2B%5Cfrac%7B-9x10%5E%7B-9%7D%20%7D%7B4x10%5E%7B-2%7D%20%7D%20%5D)
![V= 9x10^{9} * -7.5x10^{-8}](https://tex.z-dn.net/?f=V%3D%209x10%5E%7B9%7D%20%2A%20-7.5x10%5E%7B-8%7D)
V=-675V
(c) r = 6.00cm. So, This point out of both shells.
![V = k(q1/r + q2/r) = \frac{k}{r} (q1 + q2)](https://tex.z-dn.net/?f=V%20%3D%20k%28q1%2Fr%20%2B%20q2%2Fr%29%20%3D%20%5Cfrac%7Bk%7D%7Br%7D%20%20%28q1%20%2B%20q2%29)
![V = \frac{9x10^{-9} }{6x10^{-2} } [6x10^{-9} + (-9x10^{-9}) ]\\V = -450 V](https://tex.z-dn.net/?f=V%20%3D%20%5Cfrac%7B9x10%5E%7B-9%7D%20%7D%7B6x10%5E%7B-2%7D%20%7D%20%20%5B6x10%5E%7B-9%7D%20%20%2B%20%28-9x10%5E%7B-9%7D%29%20%5D%5C%5CV%20%3D%20-450%20V)
Answer:
4E
Explanation:
From the question given above, the following data were obtained:
Initial elongation (e₁) = 4 cm = 4/100 = 0.04 m
Initial energy (E₁) = E
Final elongation (e₂) = 0.04 + 0.04 = 0.08 m
Final energy (E₂) =?
The energy stored in a s spring is given by:
E = ½Ke²
Where
E => is the energy
K => is the spring constant
e => is the elongation
From:
E = ½Ke²
Energy is directly proportional to the elongation. Thus,
E₁/e₁² = E₂/e₂²
With the above formula, we can obtain the final energy as follow:
Initial elongation (e₁) = 0.04 m
Initial energy (E₁) = E
Final elongation (e₂) = 0.08 m
Final energy (E₂) =?
E₁/e₁² = E₂/e₂²
E / 0.04² = E₂ / 0.08²
E / 0.0016 = E₂ / 0.0064
Cross multiply
0.0016 × E₂ = 0.0064E
Divide both side by 0.0016
E₂ = 0.0064E / 0.0016
E₂ = 4E
Therefore, the final energy is 4 times the initial energy i.e 4E
Answer:
the potential energy is zero, and the kinetic energy must be maximum
F = 0
Explanation:
In this exercise you are asked to complete the sentences of a simple harmonic movement of a mass-spring system.
In this system mechanical energy is conserved
at the most extreme point the carousel potential energy is
K_e = ½ k x²
the kinetic energy is zero for that stopped.
At the equilibrium point
the spring elongation is x = 0 so the potential energy is zero
and the kinetic energy must be maximum since total energy of the system is conserved
the spring force is
F =- k x
as in the equilibrium position x = 0 this implies that the force is also zero
F = 0
Answer:Direct them straight across
Explanation: