Answer:
Explanation:
Let the charge on bead A be q nC and the charge on bead B be 28nC - qnC
Force F between them
4.8\times10^{-4} =
=120 x 10⁻⁸ = 9 x q(28 - q ) x 10⁻⁹
133.33 = 28q - q²
q²- 28q +133.33 = 0
It is a quadratic equation , which has two solution
q_A = 21.91 x 10⁻⁹C or q_B = 6.09 x 10⁻⁹ C
Answer:
The system will take approximately 0.255 seconds to reach the (new) equilibrium position.
Explanation:
We notice that block-spring system depicts a Simple Harmonic Motion, whose equation of motion is:
(1)
Where:
- Position of the mass as a function of time, measured in meters.
- Amplitude, measured in meters.
- Spring constant, measured in newtons per meter.
- Mass of the block, measured in kilograms.
- Time, measured in seconds.
- Phase, measured in radians.
The spring is now calculated by Hooke's Law, that is:
(2)
Where:
- Gravitational acceleration, measured in meters per square second.
- Deformation of the spring due to gravity, measured in meters.
If we know that ,
and
, then the spring constant is:
If we know that ,
,
,
and
, then (1) is reduced into this form:
(1)
And now we solve for . Given that cosine is a periodic function, we are only interested in the least value of
such that mass reaches equilibrium position. Then:
The system will take approximately 0.255 seconds to reach the (new) equilibrium position.