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3 years ago

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Two charged particles, with charges q1 = q and q2 = 4q are located at a distance d = 2cm apart on the x axis A third charged par

ticle, with charge q3 = q is placed on the x axis such that the magnitude of the force that charge 1 exerts on charge 3 is equal to the force that charge 2 exerts on charge 3. Find the position of charge 3 when q = 10n C

1 answer:

weqwewe [10]3 years ago

3 0

Answer:

0.667 cm

Explanation:

|F_1on3| = |F_2on3|

k×q1×q3/x^2 = k×q2×q3/(d-x)^2

q1/x^2 = q2/(d-x)^2

(d-x)^2/x^2 = q2/q1

(d-x)/x = sqrt(q2/q1)

(2 - x)/x = sqrt(4×q/q)

(2 - x)/x = 2

2 - x = 2x

2 = 3x

x = 2/3

x= 0.667 cm

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Object A has of mass 7.20 kilograms, and object B has a mass of 5.75 kilograms. The two objects move along a straight line towar

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If the collision is elastic, there is no loss in kinetic energies, which means that the total energies before and after impact are the same. So no need to worry about final velocities.

Final energy
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4 years ago

Consider Compton Scattering with visible light.A photon with wavelength 500nm scatters backward(theta=180degree) from a free ele

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Answer: $4.86(10)^{-12}m$

Explanation:

The Compton Shift $\Delta \lambda$ in wavelength when photons are scattered is given by the following equation:

$\Delta \lambda=\lambda' - \lambda_{o}=\lambda_{c}(1-cos\theta)$ (1)

Where:

$\lambda'=500 nm=500(10)^{-9} m$ is the wavelength of the scattered photon

$\lambda_{o}$ is the wavelength of the incident photon

$\lambda_{c}=2.43(10)^{-12} m$ is a constant whose value is given by $\frac{h}{m_{e}.c}$ , being $h=4.136(10)^{-15}eV.s$ the Planck constant, $m_{e}$ the mass of the electron and $c=3(10)^{8}m/s$ the speed of light in vacuum.

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3 years ago

The tape in a videotape cassette has a total

xxMikexx [17]

Answer:

$w=19.76 \ rad/s$

Explanation:

<u>Circular Motion </u>

Suppose there is an object describing a circle of radius r around a fixed point. If the relation between the angle of rotation by the time taken is constant, then the angular speed is also constant. If that relation increases or decreases at a constant rate, the angular speed is given by:

$w=w_o+\alpha \ t$

Where $\alpha$ is the angular acceleration and t is the time. If the object was instantly released from the circular path, it would have a tangent speed of:

$\displaystyle v_t=w.r$

We have two reels: one loaded with the tape to play and the other one empty and starting to fill with tape. They both rotate at different angular speeds, one is increasing and the other is decreasing as the tape goes from one to the other. We'll assume the tangent speed is constant for both (so the tape can play correctly). Let's call $w_1$ the angular speed of the loaded reel and $w_2$ that from the empty reel. We have

$w_1=w_{o1}+\alpha_1 \ t$

$w_2=w_{o2}+\alpha_2 \ t$

If $r_f=35\ mm=0.035\ m$ is the radius of the reel when it's full of tape, the angular speed for the loaded reel is computed by

$\displaystyle w_{01}=\frac{v_t}{r_f}$

The tangent speed is computed by knowing the length of the tape and the time needed to fully play it.

$t=1.8\ h=1.8*3600=6480\ sec$

$\displaystyle v_t=\frac{x}{t}=\frac{249\ m}{6480\ sec}=0.0384 \ m/s$

$\displaystyle w_{01}=\frac{0.0384}{0.035}=1,098 \ rad/s$

If $r_e=10\ mm=0.001\ m$ is the radius of the reel when it's empty, the angular speed for the empty reel is computed by

$\displaystyle w_{02}=\frac{0.0384}{0.001}=38.426 \ rad/s$

The full reel goes from $w_{01}$ to $w_{02}$ in 6480 seconds, so we can compute the angular acceleration:

$\displaystyle \alpha_1=\frac{w_{02}-w_{01}}{6480}$

$\displaystyle \alpha_1=\frac{38.426-1.098}{6480}=0.00576 \ rad/sec^2$

The empty reel goes from $w_{02}$ to $w_{01}$ in 6480 seconds, so we can compute the angular acceleration:

$\displaystyle \alpha_2=\frac{1.098-38.426}{6480}=-0.00576 \ rad/sec^2$

So the equations for both reels are

$w_1=1.098+0.00576 \ t$

$w_2=38.426-0.00576 \ t$

They will be the same when

$1.098+0.00576 \ t=38.426-0.00576 \ t$

Solving for t

$\displaystyle t=\frac{38.426-1.098}{0.0115}$

$t=3240 \ sec$

The common angular speed is

$w_1=1.098+0.00576 \ 3240=19.76 \ rad/s$

$w_2=38.426-0.00576 \ 3240=19.76 \ rad/s$

They both result in the same, as expected

$\boxed{w=19.76 \ rad/s}$

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