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3 years ago

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A parallel plate capacitor has two plates has a separation d. If I charge it with a +Q and a ????Q on the two plates, respective

ly, what is the electrostatic energy? What is it if I move to 1/2 d?

1 answer:

EleoNora [17]3 years ago

5 0

Answer:

Explanation:

Electric energy per unit volume

= (1 / 2) x ε₀ E²

Total energy

= energy per unit volume x volume

= (1 / 2) x ε₀ E² x A d , A is plate area

= (1 / 2A) x (ε₀ (σ/ε₀)² x A² d) [ E is electric field ]

(1 / 2A) x (ε₀ (σA/ε₀)² X d)

(1 / 2A) x (ε₀ (Q/ε₀)² X d

= (1 / 2A) x (Q²/ε₀) x d

So total energy is proportional to d , distance between plates

if d is changed to d/2

total energy also becomes half.

(1 / 2A) x (Q²/ε₀) x d / 2

(1 / 4A) x (Q²/ε₀) x d

=

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Answer:

The potential energy of the more massive one is twice that of the other.

Explanation:

Potential energy is given by

<em>PE</em> = <em>mgh</em>

where <em>m</em> = mass of body, <em>g</em> = acceleration of gravity and <em>h</em> = height or elevation.

For the less massive car, let the mass be $m_1$ . Then its <em>PE</em> is

$PE_1 = m_1gh$

For the massive car, let the mass be $m_2$ . Its <em>PE</em> is

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3 years ago

Two workers are sliding 300 kg crate across the floor. One worker pushes forward on the crate with a force of 400 N while the ot

almond37 [142]

Answer:

The kinetic coefficient of friction of the crate is 0.235.

Explanation:

As a first step, we need to construct a free body diagram for the crate, which is included below as attachment. Let supposed that forces exerted on the crate by both workers are in the positive direction. According to the Newton's First Law, a body is unable to change its state of motion when it is at rest or moves uniformly (at constant velocity). In consequence, magnitud of friction force must be equal to the sum of the two external forces. The equations of equilibrium of the crate are:

$\Sigma F_{x} = P+T-\mu_{k}\cdot N = 0$ (Ec. 1)

$\Sigma F_{y} = N - W = 0$ (Ec. 2)

Where:

$P$ - Pushing force, measured in newtons.

$T$ - Tension, measured in newtons.

$\mu_{k}$ - Coefficient of kinetic friction, dimensionless.

$N$ - Normal force, measured in newtons.

$W$ - Weight of the crate, measured in newtons.

The system of equations is now reduced by algebraic means:

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$\mu_{k} =\frac{P+T}{m\cdot g}$

If we know that $P = 400\,N$ , $T = 290\,N$ , $m = 300\,kg$ and $g = 9.807\,\frac{m}{s^{2}}$ , then:

$\mu_{k} = \frac{400\,N+290\,N}{(300\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)}$

$\mu_{k} = 0.235$

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3 years ago

When light of wavelength 240 nm falls on a cobalt surface, electrons having a maximum kinetic energy of 0.17 eV are emitted. Fin

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Answer:

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$\lambda_{cuttoff}= \frac{hc}{\Phi _{0} }$

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