Answer:
The moist filter paper is in charge of preventing evaporation and ensuring the proper saturation of the air of the chromatography chamber.
Explanation:
Thin-layer chromatography (TLC) is a chromatography technique used to separate non-volatile mixtures.
A strip of moist filter paper is put into the chromatography chamber so that its bottom touches the solvent and the paper lies on the chamber wall and reaches almost to the top of the container.
The container is closed and left for a few minutes to let the solvent vapors ascend the moist filter paper and saturate the air in the chamber.
The moist filter paper is in charge of preventing evaporation and ensuring the proper saturation of the air of the chamber.
D: Ultraviolet Rays is the answer hope that helps
The key to solve this problem is the conservation of momentum. The momentum of an object is defined as the product between the mass and the velocity, and it's usually labelled with the letter 
:
The total momentum is the sum of the momentums. The initial situation is the following:
(it's not written explicitly, but I assume that the 5-kg object is still at the beginning).
So, at the beginning, the total momentum is
At the end, we have
(the mass obviously don't change, the new velocity of the 15-kg object is 1, and the velocity of the 5-kg object is unkown)
After the impact, the total momentum is
Since the momentum is preserved, the initial and final momentum must be the same. Set an equation between the initial and final momentum and solve it for 
, and you'll have the final velocity of the 5-kg object.
<span>Plasma is a controllable reactive gas that is used to make small PATTERNS in silica which are used in computers and cell phones.</span>
Answer: 14.16
Explanation:
Given
d = 38cm
r = d/2 = 38/2 = 19cm = 0.19m
K.E = 510J
m = 10kg
I = 1/2mr²
I = 1/2*10*0.19²
I = 0.18kgm²
When it has 510J of Kinetic Energy then,
510J = 1/2Iω²
ω² = 1020/I
ω² = 1020/0.18
ω² = 5666.67
ω = √5666.67 = 75.28 rad/s
Velocity is the block, v = ωr
V = 75.28 * 0.19
V = 14.30m/s
The "effective mass" M of the system is
M = (14.0 + ½*10.0) kg = 19.0 kg
The motive force would be
F = ma
F = 14 * 9.8
F = 137.2N
so that the acceleration would be
a = F/m
a = 137.2/19
a = 7.22m/s²
Finally, using equation of motion.
V² = u² + 2as
14.3² = 0 + 2*7.22*s
204.49 = 14.44s
s = 204.49/14.44
s = 14.16m
