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lisabon 2012 [21]

3 years ago

14

How much work does the electric field do in moving a proton from a point with a potential of +130 v to a point where it is -70 v

? express your answer in joules?

1 answer:

NikAS [45]3 years ago

6 0

Hi Gonnah,
W=qΔV
W= q(-70-130)
q of proton look it up and plug in and find out
it should be negative since you are doing work form greater potential to lower

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In an LC circuit at one time the charge stored by the capacitor is 10 mC and the current is 3.0 A. If the frequency of the circu

Ronch [10]

Answer:

$i_2=3.61\ A$

Explanation:

<u>LC Circuit</u>

It's a special circuit made of three basic elements: The AC source, a capacitor, and an inductor. The charge, current, and voltage are oscillating when there is an interaction between the electric and magnetic fields of the elements. The following variables will be used for the formulas:

$q, q_1, q_2$ = charge of the capacitor in any time $t, t_1, t_2$

$q_o$ = initial charge of the capacitor

$\omega$ =angular frequency of the circuit

$i, i_1, i_2$ = current through the circuit in any time $t, t_1, t_2$

The charge in an LC circuit is given by

$q(t) = q_0 \, cos (\omega t )$

The current is the derivative of the charge

$\displaystyle i(t) = \frac{dq(t)}{dt} = - \omega q_0 \, sin(\omega t).$

We are given

$q_1=10\ mc=0.01\ c, i_1=3\ A,\ q_2=6\ mc=0.006\ c\ ,\ f=\frac{1000}{4\pi}$

It means that

$q(t_1) = q_0 \, cos (\omega t_1 )=q_1\ .......[eq 1]$

$i(t_1) = - \omega q_0 \, sin(\omega t_1)=i_1.........[eq 2]$

From eq 1:

$\displaystyle cos (\omega t_1 )=\frac{q_1}{q_0}$

From eq 2:

$\displaystyle sin(\omega t_1)=-\frac{i_1}{\omega q_0}$

Squaring and adding the last two equations, and knowing that

$sin^2x+cos^2x=1$

$\displaystyle \left ( \frac{q_1}{q_0} \right )^2+\left ( \frac{i_1}{\omega q_0} \right )^2=1$

Operating

$\displaystyle \omega^2q_1^2+i_1^2=\omega^2q_o^2$

Solving for $q_o$

$\displaystyle q_o=\frac{\sqrt{\omega^2q_1^2+i_1^2}}{\omega}$

Now we know the value of $q_0$ , we repeat the procedure of eq 1 and eq 2, but now at the second time $t_2$ , and solve for $i_2$

$\displaystyle \omega^2q_2^2+i_2^2=\omega^2q_o^2$

Solving for $i_2$

$\displaystyle i_2=w\sqrt{q_o^2-q_2^2}$

Now we replace the given values. We'll assume that the placeholder is a pi for the frequency, i.e.

$\displaystyle f=\frac{1}{4\pi}\ KHz$

$w=2\pi f=500\ rad/s$

$\displaystyle q_o=\frac{\sqrt{(500)^2(0.01)^2+3^2}}{500}$

$q_0=0.01166\ c$

Finally

$\displaystyle i_2=500\sqrt{0.01166^2-.006^2}$

$i_2=5\ A$

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3 years ago

Accelerating charges radiate electromagnetic waves. Calculate the wavelength of radiation produced by a proton of mass mp moving

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Explanation:

Let $m_p$ is the mass of proton. It is moving in a circular path perpendicular to a magnetic field of magnitude B.

The magnetic force is balanced by the centripetal force acting on the proton as :

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r is the radius of path,

$r=\dfrac{mv}{qB}$

Time period is given by :

$T=\dfrac{2\pi r}{v}$

$T=\dfrac{2\pi m_p}{qB}$

Frequency of proton is given by :

$f=\dfrac{1}{T}=\dfrac{qB}{2\pi m_p}$

The wavelength of radiation is given by :

$\lambda=\dfrac{c}{f}$

$\lambda=\dfrac{2\pi m_pc}{qB}$

So, the wavelength of radiation produced by a proton is $\dfrac{2\pi m_pc}{qB}$ . Hence, this is the required solution.

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A swimming pool is 4.0 m in depth; a swimmer at this depth feels discomfort in the ear. Calculate the net force on a 0.50-cm-dia

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The net force on a 0.50-cm-diameter eardrum is mathematically given as

F= 0.76969 N

<h3>What is the net force on a 0.50-cm-diameter eardrum?</h3>

Generally, the equation for Pressure is mathematically given as

P = ρgh

Therefore

P= 1000*9.8*4

P= 39200 Pa

Where

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F = PA

F= 39200 *( pi*(0.005/2)^2)

F= 0.76969 N

In conclusion, The net force is

F= 0.76969 N

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A speed does not involve the element of direction.

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olchik [2.2K]

Answer:

27 Joules.

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use the formula for kinetic energy:

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3 years ago