Answer:
Explanation:
a ) F = (-kx + kx³/a²)
intensity of field
I = F / m
= (-kx + kx³/a²) / m
If U be potential function
- dU / dx = (-kx + kx³/a²) / m
U(x) = ∫ (kx - kx³/a²) / m dx
= k/m ( x²/2 - x⁴/4a²)
b )
For equilibrium points , U is either maximum or minimum .
dU / dx = x - 4x³/4a² = 0
x = ± a.
dU / dx = x - x³/a²
Again differentiating
d²U / dx² = 1 - 3x² / a²
Put the value of x = ± a.
we get
d²U / dx² = -2 ( negative )
So at x = ± a , potential energy U is maximum.
c )
U = k/m ( x²/2 - x⁴/4a²)
When x =0 , U = 0
When x= ± a.
U is maximum
So the shape of the U-x curve is like a bowl centered at x = 0
d ) Maximum potential energy
put x = a or -a in
U(max) = k/m ( x²/2 - x⁴/4a²)
= k/m ( a² / 2 - a⁴/4a²)
= k/m ( a² / 2 - a²/4)
a²k / 4m
This is the maximum total energy where kinetic energy is zero.
Multiply (Saturn radii) by (60,268) to get the distance in kilometers.
(This is the radius of the planet, not it's orbit.)
The material that the snails are moving on affects their speed. On a smooth material, (petri dish) the snails moved the fastest. On a rough material (sandpaper) the snails moved the slowest. On dirt, a compromise between smooth and rough, the snails moved at a medium pace. The material and possibly friction affect this.
Answer:
the only effect it has is to create more induced charge at the closest points, but the net face remains zero, so it has no effect on the flow.
Explanation:
We can answer this exercise using Gauss's law
Ф = ∫ e . dA = 
/ ε₀
field flow is directly proportionate to the charge found inside it, therefore if we place a Gaussian surface outside the plastic spherical shell. the flow must be zero since the charge of the sphere is equal induced in the shell, for which the net charge is zero. we see with this analysis that this shell meets the requirement to block the elective field
From the same Gaussian law it follows that if the sphere is not in the center, the only effect it has is to create more induced charge at the closest points, but the net face remains zero, so it has no effect on the flow , so no matter where the sphere is, the total induced charge is always equal to the charge on the sphere.
Kinetic Energy,K.E=1/2MV²
mass,m=16kg
velocity,v=4m/s
K.E=1/2×16×4²
=128kgm²/s²
=128 Joules
