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77julia77 [94]
3 years ago
9

Two identical conducting spheres, A and B, carry equal charge. They are stationary and are separated by a distance much larger t

han their diameters. A third identical conducting sphere, C, is uncharged. Sphere C is first touched to A, then to B, and finally removed (to a far away distance). If before the sphere C is introduced the electrostatic force between A and B is Fi , how does Ff , the electrostatic force between A and B after C is removed, compare to Fi?
Physics
1 answer:
podryga [215]3 years ago
7 0

Answer:

8F_i = 3F_f

Explanation:

When two identical spheres are touched to each other, they equally share the total charge. Therefore, When neutral C is first touch to A, they share the initial charge of A equally.

Let us denote that the initial charge of A and B are Q. Then after C is touched to A, their respective charges are Q/2.

Then, C is touched to B, and they share the total charge of Q + Q/2 = 3Q/2. Their respective charges afterwards is 3Q/4 each.

The electrostatic force, Fi, in the initial configuration can be calculated as follows.

F_i = \frac{1}{4\pi\epsilon_0}\frac{q_Aq_B}{r^2} = \frac{1}{4\pi\epsilon_0}\frac{Q^2}{r^2}[/tex}The electrostatic force, Ff, in the final configuration is [tex]F_f = \frac{1}{4\pi\epsilon_0}\frac{q_Aq_B}{r^2} = \frac{1}{4\pi\epsilon_0}\frac{3Q^2/8}{r^2}[/tex}Therefore, the relation between Fi and Ff is as follows[tex]F_i = F_f\frac{3}{8}\\8F_i = 3F_f

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Answer:

389 kg

Explanation:

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Now, we will take 2 ratios of the time period

\frac{T}{T'} = \sqrt{\frac{m}{m'} }

\frac{1.50}{2.00} = \sqrt{\frac{0.500}{m'} }

0.5625 = \sqrt{\frac{0.500}{m'} }

m' = \frac{0.500}{0.5625}

= 0.889 kg

Since mass to be sum that is

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0.389 kg

or

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3 years ago
A valid hypothesis must have test for proving what?
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You can always calculate this right away, if you know his mass, by multiplying his weight in kg by the gravitational constant
g = 9.81 \frac{m}{s {}^{2} }
let's do it for this case:
f_{g}  = m \times g \\ f _{g}  = 65kg \times 9.81 \frac{m}{s {}^{2} }  = 637.65
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so, first step solved, Fg is 637.65N
Fg is a field force by the way, and at the same time, the elevator is pushing up on him with 637.65N, so you draw another arrow pointing upwards, ending at the tip of the downwards arrow.
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shepuryov [24]

<u>Answer</u>

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Average bond energy = 67 + 79.5 = 146.5kj/ mol

3 0
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