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3 years ago

How would I explain part a?

Physics

1 answer:

grandymaker [24]3 years ago

3 0

Electric field on the axis of ring on both sides is along the axial line

This electric field is away from the center on both sides

Now when an electron is placed on the axis of the ring then we will see that it will experience the electrostatic force on the electron.

The direction of force is always towards the center of the ring because here electron is a negatively charged particle and it will move opposite to the direction of electric field.

So as we release the electron from a distance of 30 cm the electron will move towards the center of ring and accelerate to its maximum speed till it will reach to the center of the ring

After that the electron will reach to the center of the ring and then moves out on the opposite side and then decelerate to zero speed as it will reach on other side.

Now after this it will again move towards the center of the ring and continue its motion similar way

So this is an oscillating motion about the center of the ring.

You might be interested in

A mass weighing 4 lb stretches a spring 2 in. Suppose that the mass is given an additional 6-in displacement in the positive dir

givi [52]

Answer:

$\frac{1}{8} y'' + 2y' + 24y=0$

Explanation:

The standard form of the 2nd order differential equation governing the motion of mass-spring system is given by

$my'' + \zeta y' + ky=0$

Where m is the mass, ζ is the damping constant, and k is the spring constant.

The spring constant k can be found by

$w - kL=0$

$mg - kL=0$

$4 - k\frac{1}{6}=0$

$k = 4\times 6 =24$

The damping constant can be found by

$F = -\zeta y'$

$6 = 3\zeta$

$\zeta = \frac{6}{3} = 2$

Finally, the mass m can be found by

$w = 4$

$mg=4$

$m = \frac{4}{g}$

Where g is approximately 32 ft/s²

$m = \frac{4}{32} = \frac{1}{8}$

Therefore, the required differential equation is

$my'' + \zeta y' + ky=0$

$\frac{1}{8} y'' + 2y' + 24y=0$

The initial position is

$y(0) = \frac{1}{2}$

The initial velocity is

$y'(0) = 0$

6 0

2 years ago

An LC circuit consists of a 3.4-µF capacitor and a coil with a self-inductance 0.080 H and no appreciable resistance. At t = 0 t

alexira [117]

Answer

given,

capacitance = C = 3.4-µF

inductance = L = 0.08 H

frequency is expressed as

$f = \dfrac{1}{2\pi\sqrt{LC}}$

time period

$T = \dfrac{1}{f}=2\pi\sqrt{LC}$

after time T/4 current reach maximum

$t = \dfrac{T}{4}$

$t = \dfrac{2\pi\sqrt{LC}}{4}$

$t = \dfrac{2\pi\sqrt{0.08 \times 3.4 \times 10^{-6}}}{4}$

t = 8.2 x 10⁻⁴ s

t = 0.82 ms

b) using law of conservation

$\dfrac{1}{2}CV^2=\dfrac{1}{2}LI^2$

$I^2 = \dfrac{CV^2}{L}$

$I^2 = \dfrac{C}{L}\dfrac{Q^2}{C^2}$

$I =\sqrt{\dfrac{Q^2}{CL}}$

$I =\sqrt{\dfrac{(5.4 \times 10^{-6})^2}{0.08 \times 3.4 \times 10^{-6}}}$

I = 0.010 A

I = 10 mA

4 0

3 years ago

How can working together help scientist achieve their goals?

abruzzese [7]

When scientists work together it helps them achieve their goals when they share different ideas with eachother

7 0

3 years ago

The formula for speed is v = s/t. <br> a. True<br> b. False

crimeas [40]

If you mean S is the distance then it is true
Velocity = Distance / time

3 0

2 years ago

A 0.50-kg mass attached to the end of a string swings in a vertical circle (radius 2.0 m). When the mass is at the highest point

il63 [147K]

Answer:

31.1 N

Explanation:

m = mass attached to string = 0.50 kg

r = radius of the vertical circle = 2.0 m

v = speed of the mass at the highest point = 12 m/s

T = force of the string on the mass attached.

At the highest point, force equation is given as

$T + mg =\frac{mv^{2}}{r}$

Inserting the values

$T + (0.50)(9.8) =\frac{(0.50)(12)^{2}}{2}$

T = 31.1 N

7 0

2 years ago

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