Answer:
Check the explanation
Explanation:
(a) When the electric field is applied to the ends of the wire, electrons move through the wire because the resistance of the wire is less than the resistance of the surrounding air. Resistance is the obstruction to the flow of electrons hence electrons flow through the path having minimum resistance.
(b) As the thermal energy increases, Kinetic energy of electrons as well as that of positive ions also increase. Hence the mean time between collisions of electrons with ions is reduced due to movement of ions also. That is why Resistance of the wire increases.
This is the period in a simple harmonic motion which is 2 seconds in this question.
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What is Period ?</h3>
The period of an oscillatory object can be defined as the total time taken by a vibrating body to make one complete revolution about a reference point.
We are given the below question
2×3.14√(1.0m/(9.8〖ms〗^(2) )= T
This question can as well be expressed as
2π√(L/g) which is equal to period T.
In a nut shell, Period T = 2×3.14√(1.0m/9.8)
T = 6.28√0.102
T = 6.28 × 0.32
T = 2.006 s
Therefore, the period T of the oscillation is 2 seconds approximately.
Learn more about Period here: brainly.com/question/12588483
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The electric force between the two particles are calculated through the equation,
F = kQ₁Q₂ / d²
where F is the force, k is a constant called Coulomb's law constant, Q₁ and Q₂ are the charges, and d is the distance. This equation is called the Coulomb's law.
It can be seen from the equation above that the electric forces between the objects are majorly affected by the substance's charges and distance.
The answer to this item is therefore letter A.
Let us start from considering monochromatic light as an incidence on the film of a thickness t whose material has an index of refraction n determined by their respective properties.
From this point of view part of the light will be reflated and the other will be transmitted to the thin film. That additional distance traveled by the ray that was reflected from the bottom will be twice the thickness of the thin film at the point where the light strikes. Therefore, this relation of phase differences and additional distance can be expressed mathematically as
We are given the second smallest nonzero thickness at which destructive interference occurs.
This corresponds to, m = 2, therefore
The index of refraction of soap is given, then
Combining the results of all steps we get
Rearranging, we find
