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anastassius [24]

3 years ago

13

A parallel-plate air capacitor with a capacitance of 260 pF has a charge of magnitude 0.155 Î¼C on each plate. The plates have a

separation of 0.313 mm. What is the potential difference between the plates?

1 answer:

butalik [34]3 years ago

7 0

Answer:

The potential difference between the plates is 596.2 volts.

Explanation:

Given that,

Capacitance $C=260\ pF$

Charge $q=0.155\ \mu\ C$

Separation of plates = 0.313 mm

We need to calculate the potential difference between the plates

Using formula of potential difference

$V= \dfrac{Q}{C}$

Where, Q = charge

C = capacitance

Put the value into the formula

$V=\dfrac{0.155\times10^{-6}}{260\times10^{-12}}$

$V=596.2\ volts$

Hence,The potential difference between the plates is 596.2 volts.

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<h2>Answer 1: a. weight</h2>

Mass and weight are very different concepts.

Mass is the amount of matter that exists in a body, which only depends on the quantity and type of particles within it. This means mass is an intrinsic property of each body and remains the same regardless of where the body is located.

On the other hand, weight is a measure of the gravitational force acting on an object and is directly proportional to the product of the mass $m$ of the body by the acceleration of gravity $g$ :

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<h2>Answer 2: b. Force changes by 2/9</h2>

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$G$ is the universal gravitation constant

$m_{1}$ and $m_{2}$ are the masses of both bodies.

$r$ is the distance between both bodies

If we double the mass of one object (for example $2m_{1}$ ) and triple the distance between both (for example $3r$ ). The equation (1) will be rewritten as:

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$F=\frac{2}{9}G\frac{m_{1}m_{2}}{r^2}$ (3)

If we compare (1) and (2) we will be able to see the force changes by 2/9.

<h2>Answer 3: b. movement</h2>

The Work $W$ done by a Force $F$ refers to the release of potential energy from a body that is <u>moved</u> by the application of that force to overcome a resistance along a path.

When the applied force is constant and <u>the direction of the force and the direction of the movement are parallel,</u> the equation to calculate it is:

$W=(F)(d)$

Now, <u>when they are not parallel, both directions form an angle</u>, let's call it $\alpha$ . In that case the expression to calculate the Work is:

$W=Fdcos{\alpha}$

Therefore, pushing on a rock accomplishes no work unless there is movement (independently of the fact that movement is parallel to the applied force or not).

<h2>Answer 4: a. 40,000 Joules</h2>

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$K=\frac{1}{2}mV^{2}$ (4)

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For the second case (unknown kinetic energy $K_{2}$ for a car with the same mass at $V_{2}=60 mph=26.8224m/s$ ):

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$K_{2}=\frac{1}{2}(111.197kg)(26.8224m/s)^{2}$ (10)

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<h2>Answer 5: c. the soil will be 5°C</h2>

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$Q=m. c. \Delta T$ (12)

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$\Delta T_{s}=5\°C)}$ (19)

Hence the correct option is c.

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