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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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two resistors of resistance 6 ohm and 3 ohm are connected in series and then in parallel .calculate the equivalent series resist
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Explanation:

Given that,

Two resistors of resistance 6 ohm and 3 ohm are connected in series and then in parallel.

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For parallel combination,

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When 6 ohm and 3 ohm are in series,

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When 6 ohm and 3 ohm are in paralle,

\dfrac{1}{R_p}=\dfrac{1}{6}+\dfrac{1}{3}\\\\R_p=2\ \Omega

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You wad up a piece of paper and throw it into the wastebasket. How far will
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The range of the piece of paper is C) 1.4 m

Explanation:

The motion of the piece of paper is the motion of a projectile, which consists of two separate motions:

- A uniform motion along the horizontal direction, with constant velocity

- A uniformly accelerated motion along the vertical direction, with constant acceleration (the acceleration of gravity, g=9.8 m/s^2)

From the equation of motion, it is possible to find an expression for the range (the total horizontal distance covered) of a projectile, which is given by:

d=\frac{u^2 sin 2\theta}{g}

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u is the initial velocity

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g is the acceleration of gravity

For the piece of paper in this problem,

u = 4.3 m/s

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Substituting,

d=\frac{(4.3)^2 sin(2\cdot 65^{\circ})}{9.8}=1.45 m \sim 1.4 m

Learn more about projectile motion:

brainly.com/question/8751410

#LearnwithBrainly

6 0
3 years ago
Read 2 more answers
A proton travels with a speed of 5.02×10⁶ m/s in a direction that makes an angle of 60.0° with the direction of a magnetic ficld
lesantik [10]
Answer: The magnitude of the proton's acceleration is 0.748 ×10^14 m/s²

Explanation:
the velocity ,v, of ththe proton = 5.02×10^6 m/s
Magnitude , B , of the magnetic field = 0.180 T

First , we need to find the magnitude of the Force on the proton. This is given by the relation :
F = q(v x B) = qvBsinθ

where 'q' is the charge of proton , q= 1.6×10^-19 C
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Putting the respective values of v, B ,θ in the above equation, we get:

F = (1.6×10^-19 C)(5.02×10^6 m/s)(0.180T) sin60°
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Now , from Newton's second law we know that ,
F=m×a

∴ a = F/m

Mass of a proton = 1.67×10^27 kg
a= 1.25 × 10^-13 N / 1.67 × 10^27 kg

a= 0.748 × 10^14 m/s² =acceleration of the proton

(To know more about Magnetic Fields : brainly.com/question/9095546)

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