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Mila [183]
3 years ago
13

When conducting objects (such as metals) are connected for a brief time, charge can be made to flow quite easily from one to the

other, even if the objects are in contact for an extremely brief amount of time. To that end, consider two identical metallic ball bearings (i.e., two very tiny metallic solid spheres) having unknown charges 41 and 42. It is found that when they are placed 1 m apart, they experience a 25-N attractive force.
(a) What can you conclude about the charges on these metal marbles from the information given?
(b) The ball bearings are made to touch very briefly and then are separated by 1 m again. What can you conclude about the charges now? Explain.
(c) Suppose that when the charges are 1 m apart, after briefly touching, they again feel a force of 25 N Determine the charges 41 and 42.
Physics
1 answer:
sukhopar [10]3 years ago
7 0

Answer:

a) charges of the opposite,  b)  Δq = q₄₂-q₄₁, force is repulsive

c) q = 5.27 10⁻⁵ C

Explanation:

a) In electrostatic studies it is found that charges of the same sign repel and charges of the opposite sign attract.

On the other hand, as the force between the spheres is attractive, the charges on them are of different sign

b) When the balls touch, the charges are quickly distributed between the two spheres, therefore there are two possibilities:

* if the charges were equal and as they are of the opposite sign, they are neutralized, therefore the spheres remain uncharged

    The force between them is zero

* if the charges are different, a residual charge remains

             Δq = q₄₂-q₄₁

that is distributed between the two spheres and the force between them is repulsive

c) For this case the charges on the two spheres is equal

          q₄₁ = q₄₂ = q

the force is repulsive so the charges are of the same sign.

We can apply Coulomb's law

          F = k \frac{q_1q_2}{r^2}

in this case

          F = k \frac{q^2}{r^2}

          q = \sqrt{ \frac{F \ r^2}{k} }

let's calculate

          q = \sqrt{ \frac{25 \ 1^2}{9 \ 10^{9}} } = \sqrt{27.7778 \ 10^{-10}}

          q = 5.27 10⁻⁵ C

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Mazyrski [523]

Answer:

see from this analysis, the apparent weight of the body is lower due to the push created by the air brujuleas

Explanation:

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The weight of a submerged body is the net force between the weight and the thrust

          F_net = W - B

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we substitute

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4 0
4 years ago
Read 2 more answers
An electron is confined in a harmonic oscillator potential well. What is the longest wavelength of light that the electron can a
kipiarov [429]

Answer:

The longest wavelength of light is 209 nm.

Explanation:

Given that,

Spring constant = 74 N/m

Mass of electron m= 9.11\times10^{-31}\ kg

Speed of light c= 3\times10^{8}\ m/s

We need to calculate the frequency

Using formula of frequency

f =\dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}}

Where, k= spring constant

m = mass of the particle

Put the value into the formula

f=\dfrac{1}{2\pi}\sqrt{\dfrac{74}{9.11\times10^{-31}}}

f=1.434\times10^{15}\ Hz

We need to calculate the longest wavelength that the electron  can absorb

\lambda=\dfrac{c}{f}

Where, c = speed of light

f = frequency

Put the value into the formula

\lambda =\dfrac{3\times10^{8}}{1.434\times10^{15}}

\lambda=2.092\times10^{-7}\ m

\lambda=209\ nm

Hence, The longest wavelength of light is 209 nm.

6 0
3 years ago
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