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horsena [70]
3 years ago
10

Three identical metallic conducting spheres carry the following charges: q = +3.8 μC, q = −2.6 μC, and q = −8.8 μC. The spheres

that carry the charges q and q are brought into contact. Then they are separated. After that, one of those two spheres is brought into contact with the third sphere that carries the charge q. What is the final charge on the third sphere?
Physics
1 answer:
Kitty [74]3 years ago
6 0

Answer:

-4.1μC is the final charge on the third sphere

Explanation:

From the given data, q1 and q2 are brought into contact as they are both conductors,  as such there will be evenly distribution of charges.

a) charge on each sphere(Q) = q1 + q2 / 2

= +3.8 μC + (- 2.6 μC) / 2 = 1.2μC/2 = 0.6μC

b) Now, one of those two spheres is brought into contact with the third sphere ; Q is brought into contact with q3 = Q + q3 / 2

= 0.6μC - 8.8 μC /2 = -8.2 μC/2

= -4.1μC is the final charge on the third sphere.

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Answer:

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Now, the potential energy of the block at x = 0.08 m is ½kx²

where;

k is the spring constant given by; k = ω²m

ω is the angular velocity of the oscillation

m is the mass of the block.

Thus, potential energy of the spring at the bottle(x = 0.08 m) is;

U = ½ω²m(0.08m)²

Also, potential energy of the spring at the bottle(x = 0.05 m) is;

U = ½ω²m(0.05m)²

and the kinetic energy of the block at x = 0.05 m is;

K = ½mv₀²

Thus;

½ω²m(0.08)² = ½ω²m(0.05)² + ½mv₀²

Inspecting this, ½m will cancel out to give;

ω²(0.08)² = ω²(0.05)² + v₀²

Making v₀ the subject, we have;

v₀ = ω√((0.08)² - (0.05)²)

So,

v₀ = 8.1√((0.08)² - (0.05)²)

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A 13,000-N vehicle is to be lifted by a 25-cm diameter hydraulic piston. What force needs to be applied to a 5.0 cm diameter pis
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Answer:

Explanation:

We shall apply Pascal's Law in fluid mechanics

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25 cm diameter = 12.5 x 10⁻² m radius

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13000 / 490.625 x 10⁻⁴

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5 cm diameter = 2.5 x 10⁻² radius

area = 3.14 x (2.5 x 10⁻²)²

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F = 19.625 x 26.497

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