Answer:
The ratio (U₁/U₂) = 6
Explanation:
U, the potential energy is given as
U = kqQ/r
k = Coulomb's constant
q = charge we're concerned about
Q = charge of the negative plate of the capacitor
r = distance of q from the negative plate of the capacitor.
For charge q₁
U₁ = kq₁Q/s
U₂ = kq₂Q/2s
But q₂ = q₁/3
U₂ becomes U₂ = kq₁Q/6s
U₁ = kq₁Q/s
U₂ = kq₁Q/6s
(U₁/U₂) = 6
Answer:The higher up an object is the greater its gravitational potential energy. The larger the distance something falls through the greater the amount of GPE the object loses as it falls. As most of this GPE gets changed into kinetic energy, the higher up the object starts from the faster it will be falling when it hits the ground. So a change in gravitational potential energy depends on the height an object moves through.
Explanation: Lifting an apple up 1 metre is easier work than lifting an apple tree the same height. This is because a tree has more mass, so it needs to be given more gravitational potential energy to reach the same height.
We know that the measure of an incident ray is: α 1 = 40°.
The index of refraction:
- for the air : n 1 = 1.00,
- for the water: n 2 = 1.33
Snell`s Law of Refraction :
n 1 · sin α 1 = n 2 · sin α 2
sin α 2 = n 1 · sin α 1 / n 2 =
= 1.00 · sin 40° / 1.33 = 0.64278 / 1.33 = 0.4833
α 2 = sin ^(-1) 0.4833
α 2 = 28.9 °
Answer: The angle relative to the water`s surface of the rays when beneath the surface is 28.9°.
Answer:
When the blood and the dialysate are flowing in the same direction, as the the dialysate and the blood move away from the region of higher concentration of the urea, to a region distant from the source, the concentration of urea in the blood stream and in the dialysis reach equilibrium and diffusion across the semipermeable membrane stops within the higher filter regions such as II, III, IV or V
However, for counter current flow, as the concentration of the urea in the blood stream becomes increasingly lesser the, it encounters increasingly unadulterated dialysate coming from the dialysate source, such that diffusion takes place in all regions of the filter
Explanation: