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Answer: option 1 : the electric potential will decrease with an increase in y
Explanation: The electric potential (V) is related to distance (in this case y) by the formulae below
V = kq/y
Where k = 1/4πε0
Where V = electric potential,
k = electric constant = 9×10^9,
y = distance of potential relative to a reference point, ε0 = permittivity of free space
q = magnitude of electronic charge = 1.609×10^-19 c
From the formulae, we can see that q and k are constants, only potential (V) and distance (y) are variables.
We have that
V = k/y
We see the potential(V) is inversely proportional to distance (y).
This implies that an increase in distance results to a decreasing potential and a decrease in distance results to an increase in potential.
This fact makes option 1 the correct answer
Do cherry popsicles freeze slower than orange popsicles
Answer:
1.7323
Explanation:
To develop this problem, it is necessary to apply the concepts related to refractive indices and Snell's law.
From the data given we have to:
Where n means the index of refraction.
We need to calculate the index of refraction of the liquid, then applying Snell's law we have:
Replacing the values we have:
Therefore the refractive index for the liquid is 1.7323
