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Savatey [412]
3 years ago
12

A positively charged light metal ball is suspended between two oppositely charged metal plates on an insulating thread as shown

below. After being charged once, the plates are disconnected from the battery. Describe the behavior of the ball. Please use 3 content related sentences. (
Physics
1 answer:
Tcecarenko [31]3 years ago
6 0

Answer:

The positive ball would go first to the negatively charged plate.

Explanation:

After which, it would hold a more negative charge. Due to the negative charge, it would travel towards the positive plate. Thereby, it would transfer negative electrons to the positive plate once more. In doing so, it would transfer positive protons to the negative plate. After which, it would hold more  negative electrons and be drawn towards the positive plate once more. The Process would continue until the once-positive and once-negative became neutral ( and were discharged.) Additionally, the ball hanging on the insulator thread would also be neutral and discharged.

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Suppose that a 1000 kg car is traveling at 25 m/s (aprox 55mph). Its brakes can apply a force of 5000N. What is the minimum dist
Brrunno [24]

Assuming the the car is travelling to the right (→):

R(\rightarrow), F = ma

\implies a =  \frac{F}{m}

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= -5 \ ms^{-2}

The information we know:

u = 25 \ ms^{-1}

v = 0

a = -5 \ ms^{-2}

Using one of the equations of motion:

v^2 = u^2 + 2as

\implies s = \frac{v^2 -u^2}{2a}

=  \frac{0^2 - 25^2}{2\times (-5)}

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8 0
3 years ago
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A wire 50.0 m long and 2.00 mm in diameter is connected to a source with a potential difference of 9.11 V, and the current is fo
Alex787 [66]

Answer:

ρ = 1.6*10⁻⁸ Ω/m.

Explanation:

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       R = \frac{V}{I} = \frac{9.11V}{36.0A} = 0.253  \Omega (1)

  • At the same time, we know that there exists a relationship between the resistance, the resistivity ρ, the length L and the area A of the wire, that is given for the following expression:

       R = \rho* \frac{L}{A} (2)

  • The area of the circular section of the wire, can be expressed as a function of the diameter d, as follows:

      A = \frac{\pi*d^{2} }{4} = \frac{\pi*(0.002m)^{2}}{4} = \pi*10e-6 (3)

  • Replacing  the left side of (2) by (1), and (3) on the right side, we can solve for the resistivity ρ as follows:

       \rho = \frac{R*A}{L} = \frac{0.253\Omega*\pi*10e-6}{50.0m} = 1.6e-8 \Omega/m

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3 years ago
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LUCKY_DIMON [66]
At some time during her drive she backed up with a substantial negative. ( backwards) acceleration. Since the pocket book is not physically connected to the seat it is free to move. Upon rapid negative acceleration the pocket book remains in its position while the car accelerates backwards away from it. this demonstrates Newtons 1st law of motion. The first law is the law of inertia. Which states, an object at rest. ( pocketbook) will remain at rest and an object in motion will continue in motion at constant velocity, unless acted upon by some outside force to change its motion.
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3 years ago
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mr_godi [17]

Answer:

Solve it

Explanation:

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