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erma4kov [3.2K]

3 years ago

11

A 400 µF capacitor is constructed out of two parallel plates of very large area which are separated by 1 mm. A battery is used t

o charge the capacitor to a potential of 100 V . (a) If the capacitor is disconnected from the battery, how much work (in Joules) must be done to pull the plates apart to a separation of 2 mm?

1 answer:

Kay [80]3 years ago

6 0

Answer:

W=2 J

Explanation:

Given that

C= 400 µF

V= 100 V

d= 1 mm

Energy before disconnected from the battery

U= 1/2 CV²=Q²/(2C)

Energy after disconnected from the battery

U'=1/2 C'V²=Q²/(2C')

The work done W

W= U' - U

W= Q²/(2C') - Q²/(2C)

$W=\dfrac{Q^2}{2C}\left ( \dfrac{C}{C'} -1\right )$

1/2 CV²=Q²/(2C)

$W=\dfrac{CV^2}{2}\left ( \dfrac{C}{C'} -\right )$

We know that

$C=\dfrac{\varepsilon _oA}{d}$

$C'=\dfrac{\varepsilon _oA}{d'}$

Given that

d'=2mm ,d= 1mm

C/C'= d'/d= 2

$W=\dfrac{CV^2}{2}\left ( \dfrac{C}{C'} -\right )$

By putting the values

$W=\dfrac{400\times 10^{-6}\times 100^2}{2}( 2 -1 )$

W=2 J

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3. What is the velocity of a wave that has a frequency of 750 Hz and a wavelength of 45.7 cm?

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3 years ago

A 2.93 kg particle has a velocity of (2.98 i hat - 3.98 j) m/s.

cupoosta [38]

Answer:

a) The x and y components of the momentum are $8.731\,\frac{kg\cdot m}{s}$ and $-11.661\,\frac{kg\cdot m}{s}$ , respectively.

b) The magnitude and direction of its momentum are approximately 14.567 kilogram-meters per second and 306.823º.

Explanation:

a) The vectorial equation of momentum is represented by the following expression:

$\vec p = m\cdot \vec v$ (1)

Where:

$\vec p$ - Vector momentum, measured in kilogram-meters per second.

$m$ - Mass of the particle, measured in kilograms.

$\vec v$ - Vector velocity, measured in meters per second.

If we know that $m = 2.93\,kg$ and $\vec v = 2.98\,\hat{i}-3.98\,\hat{j}\,\,\,\left[\frac{m}{s} \right]$ , then the momentum is:

$\vec p = (2.93)\cdot (2.98\,\hat{i}-3.98\,\hat{j})\,\,\,\left[\frac{kg\cdot m}{s} \right]$

$\vec p = 8.731\,\hat{i}-11.661\,\hat{j}\,\,\,\left[\frac{kg\cdot m}{s} \right]$

The x and y components of the momentum are $8.731\,\frac{kg\cdot m}{s}$ and $-11.661\,\frac{kg\cdot m}{s}$ , respectively.

b) The magnitude and direction of momentum are represented by the following expressions:

$\|\vec p \| = \sqrt{p_{x}^{2}+p_{y}^{2}}$ (2)

$\theta = \tan^{-1}\left(\frac{p_{y}}{p_{x}} \right)$ (3)

Where:

$\|\vec p\|$ - Magnitude of momentum, measured in kilogram-meters per second.

$\theta$ - Direction of momentum, measured in sexagesimal degrees.

If we know that $p_{x} = 8.731\,\frac{kg\cdot m}{s}$ and $p_{y} = -11.661\,\frac{kg\cdot m}{s}$ , then the magnitude and direction of momentum are, respectively:

$\|\vec p\| = \sqrt{\left(8.731\,\frac{kg\cdot m}{s} \right)^{2}+\left(-11.661\,\frac{kg\cdot m}{s} \right)^{2}}$

$\|\vec p\| \approx 14.567\,\frac{kg\cdot m}{s}$

$\theta =\tan^{-1}\left(\frac{-11.661\,\frac{kg\cdot m}{s} }{8.731\,\frac{kg\cdot m}{s} } \right)$

$\theta \approx 306.823^{\circ}$

The magnitude and direction of its momentum are approximately 14.567 kilogram-meters per second and 306.823º.

6 0

3 years ago