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Oliga [24]
2 years ago
15

Two conducting spheres, one having twice the diameter of the other, are separated by a distance that is large compared to their

diameters. The smaller sphere has charge q and the larger sphere is uncharged. If the spheres are connected by a long, thin wire, then after a sufficiently long time:
a.The two spheres are at the same potential.
b. The electric field at the surface of the two spheres has the same magnitude.
c. The two spheres have the same charge.
d.The larger sphere is at twice the potential of the smaller sphere.
e.The smaller sphere is at twice the potential of the larger sphere.
Physics
1 answer:
Snezhnost [94]2 years ago
6 0

Answer: Option (a) is the correct answer.

Explanation:

When these two conducting spheres are connected together through a thin wire then charge from the smaller sphere will travel through the wire. And, this charge will continue to travel towards the neutral sphere until the charge on both the spheres will become equal to each other.

For example, charge on small sphere is 5 C then this charge will continue to travel towards the neutral sphere until its charge also becomes equal to 5 C.

Hence, then their potential will also become equal.

Thus, we can conclude that the spheres are connected by a long, thin wire, then after a sufficiently long time the two spheres are at the same potential.

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Biologists think that some spiders "tune" strands of their web to give enhanced response at frequencies corresponding to those a
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T=2.94*10^-10  N/m.

Explanation:

Biologists think that some spiders "tune" strands of their web to give enhanced response at frequencies corresponding to those at which desirable prey might struggle. Orb spider web silk has a typical diameter of 20μm, and spider silk has a density of 1300 kg/m³.

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μ = mass per unit length)

λ/2=l

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and since frequency * wavelength = speed of wave. we have,

                  150 * 0.28 = √(T/μ)                                        ..................(#)

now μ = mass/length = [volume * density]/length = [(length*area) * density] / length = area * density

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now putting this in equation (#) we get

    150 * 0.28 = √(T/[13π * 10^(-8)]).

thus T = [13π * 10^(-8)] * (42)²     =  

2.94*10^-10  N/m.

6 0
1 year ago
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