the control would be A. the bottle with 0% concentration, because you're not changing anything.
Answer:
Explanation:
Let the balls collide after time t .
distance covered by falling ball
s₁ = v₀ t + 1/2 g t²
distance covered by rising ball
s₂ = v₀ t - 1/2 g t²
Given ,
s₁ + s₂ = D
D = v₀ t + 1/2 g t² + v₀ t - 1/2 g t²
= 2v₀ t
t = D / 2v₀
s₂ = v₀ t - 1/2 g t²
= v₀ x D / 2v₀ - (1/2) x g x D² / 4v₀²
= D / 2 - gD² / 8 v₀²
The important point here is that volumetric flow rate in the pump and the pipe is the same.
Q = AV, where Q = Volumetric flow rate, A = Cross sectional area, V = velocity
Q (pump) = (π*15^2)/4*2 = 353.43 cm^3/s
Q (pipe) = (π*(3/10)^2)/4*V = 0.071V
Q (pump) = Q (pipe)
0.071V = 353.43 => V = 5000 cm/s
Therefore, the flow of water in the pipe is 5000 cm/s.
<span>Ohm's law deals with the relation between
voltage and current in an ideal conductor. It states that: Potential difference
across a conductor is proportional to the current that pass through it. It is
expressed as V=IR.
V = IR
200 = 20R
R = 10 ohms</span>
