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Paraphin [41]
2 years ago
10

A small metal ball is given a negative charge, then brought near to end a of the rod (figure 1). What happens to end a of the ro

d when the ball approaches it closely this first time?.
Physics
1 answer:
erma4kov [3.2K]2 years ago
7 0

What happens to end a of the rod when the ball approaches it closely this first time is; It is strongly attracted.

<h3>Electrostatics</h3>

I have attached the image of the rod.

We are told that the ball is much closer to the end of the rod than the length of the rod. Thus, if we point down the rod several times, the distance of approach will experience no electric field and as such the charge on end point A of the rod must be comparable in magnitude to the charge on the ball.

This means that their fields will cancel.

Finally, we can conclude that when a charge is brought close to a conductor, the opposite charges will all navigate to the point that is closest to the charge and as a result, a strong attraction will be created.

This also applies to a strong conducting rod and therefore it is strongly attracted.

Read more about Electrostatics at; brainly.com/question/18108470

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Electromagnetic radiation that has a short wavelength will have a ________ frequency.
ArbitrLikvidat [17]
High frequency , it is because wavelength is inversely proportional to frequency
8 0
3 years ago
One of the world's largest Ferris wheels, the Cosmo Clock 21 with a radius of 50.0 m is located in Yokohama City, Japan. Each of
STatiana [176]

Answer:

a = 0.55 m / s²

Explanation:

The centripetal acceleration is given by the relation

         a = v² / r

angular and linear velocities are related

         v = w r

we substitute

          a = w² r

In the exercise they indicate the angular velocity w = 1 rev/min, let's reduce to the SI system

          w = 1 rev / min (2pi rad / 1rev) (1min / 60s) = 0.105 rad/ s

let's calculate

          a = 0.105² 50.0

          a = 0.55 m / s²

4 0
3 years ago
A point charge with a charge q1 = 4.00 μC is held stationary at the origin. A second point charge with a charge q2 = -4.10 μC mo
GrogVix [38]

Answer:

W = -0.480 J

Explanation:

given,

q₁ = 4 μC

q₂ = -4.10 μC

W = kq_1q_2(\dfrac{1}{a}+\dfrac{1}{b})

b = \sqrt{(0.27-0)^2+(0.27-0)^2}

b = 0.381

k = 8.99 × 10⁹ Nm²/C²

W = 8.99\times 10^9\times 4\times 10^{-6}\times (-4.1 \times 10^{-6})(\dfrac{1}{0.17}+\dfrac{1}{0.381})

W = [-147.436\times (5.88-2.62)\times 10^{-3}]J

W = -0.480 J

Work done by the electric force W = -0.480 J

4 0
2 years ago
If the given wave has a frequency of 100 Hz, what frequency will the sixth harmonic have?
alukav5142 [94]

Answer:

600Hz

Explanation:

In electrical systems of alternating current, the harmonics are, as in acoustics, frequencies multiples of the fundamental working frequency of the system and whose amplitude decreases as the multiple increases. For example, if we have systems fed by the 50 Hz network, harmonics of 100, 150, 200, etc. may appear.

In our case having a fundamental wave of 100Hz, I can have harmonics of 200,300,400, ..., 600Hz

4 0
3 years ago
One way to measure g on another planet or moon by remote sensing is to measure how long it takes an object to fall a given dista
Jlenok [28]

Answer:

(a) 0.94 m/s²

(b) g (planet) = 0.096g

Explanation:

(a)

From Newton's equation of motion,

S = ut + 1/2gt²......................... equation 1

Making g the subject of equation 1

g =( S - ut)/t² ........................ equation 2

Where  s = distance ( m), u = initial velocity (m/s), t = time (s), g = acceleration due to gravity (m/s²)

From the question, S = 12.02 m, t = 3.58 s, u= 0 ( at rest),

Substituting these values in equation 2

g = {12.02 -(0×3.58)}/3.58²

g = (12.02)/12.82

g = 0.94 m/s²

∴ The acceleration due to gravity on the planet = 0.94 m/s²

(b) g (planet)/g (earth) = 0.94/9.80

     g (planet) = 0.096 g (earth).

The acceleration due to gravity of the planet in terms of the earth g  is

g (planet) = 0.096g

5 0
2 years ago
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