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leva [86]
3 years ago
8

Let's take two balls, each of radius 0.1 meters and charge them each to 3 microCoulombs. Then, let's put them on a flat surface

and push them together until they touch. Each ball has a mass of 4 kg. Think carefully about what to use for the distance between the two balls. And to keep things simpler, let's assume they slide without rolling.Now consider the case where both balls are free to move once you release them. Neglecting friction, calculate the final speed of one of the balls.?
Physics
1 answer:
salantis [7]3 years ago
6 0

Explanation:

First, we will calculate the electric potential energy of two charges at a distance R as follows.

                   R = 2r

                      = 2 \times 0.1 m

                      = 0.2 m

where,    R = separation between center's of both Q's. Hence, the potential energy will be calculated as follows.

                U = \frac{k \times Q \times Q}{R}

                    = \frac{8.98 \times 10^{9} \times (3 \times 10^{-6} C)^{2}}{0.1}

                    = 0.081 J

As, both the charges are coming towards each other with the same energy so there will occur equal sharing of electric potential energy between these two charges.

Therefore, when these charges touch each other then they used to posses maximum kinetic energy, that is, \frac{U}{2}.

Hence,   K.E = \frac{U}{2}

                     = \frac{0.081}{2}

                     = 0.0405 J

Now, we will calculate the speed of balls as follows.

                 V = \sqrt{\sqrt{\frac{2 \times K.E}{m}}

                     = \sqrt{\sqrt{\frac{2 \times 0.0405}{4 kg}}

                     = 0.142 m/s

Therefore, we can conclude that final speed of one of the balls is 0.142 m/s.

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3 years ago
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A ball is thrown upwards from the edge of a cliff. The horizontal velocity and vertical velocity are both 20m/s the distance fro
german

Answer:

20m

6.9s

Explanation:

The vertical velocity of the ball is 20m/s. We can calculate the kinetic energy which gets transferred to potential energy once it gets to the top.

E_k = E_p

0.5mv^2 = mgh

h = \frac{0.5v^2}{g}

we can subtitute v = 20m/s and g = 10m/s2

h = \frac{0.5*20^2}{10} = 20 m

So the ball could go 20m high from the child hand, or 120m fro the bottom of the cliff.

The time it takes for the ball to travels to the top is the time it takes for it to decelerate from 20m/s to 0m/s with gravitational deceleration g = 10m/s2

t = v / g = 20 / 10 = 2s

Then the ball will start accelerating down ward with a constant acceleration of g = 10m/s. In order to cover distance d of 120m from the top to the bottom of the cliff

d = \frac{gt_2^2}{2}

t^2 = \frac{2d}{g} = \frac{2*120}{10} = 24

t = \sqrt{24} = 4.9s

So the total time it takes is 4.9 + 2 = 6.9s

3 0
3 years ago
The electric potential a distance r from a small charge is proportional to what power of the distance from the charge?
Gekata [30.6K]

Answer:

The power of the distance is -1.

Explanation:

The equation for the electric potential of a point charge is given by V= k\frac{q}{r}

where V is the electric potential, k is Coulomb's constant (it has a value of 9.0*10^{9} with units \frac{Nm^{2} }{c^{2} }), q is the electric charge of the small charge and r is the distance from the charge.

Now, the power of a number is how many times we multiply that number by itself; we see r appears only once in the equation. So we know the power is 1. But we can see in the equation that k and q are divided by r, which means r is the denominator. This means the power of r is negative (-).

Therefore, the power of r is -1.

5 0
3 years ago
What are two differences between gravity and air resistance
Natalka [10]

1. Air resistance force usually upwards, but gravity doenwards.

2. Objects are affected by gravity the same, but air resistance can affect the speed of an object's descent.

Sorry if im wrong

4 0
3 years ago
Electromagnetic radiation of 8.12×10¹⁸ Hz frequency is applied on a metal surface and caused electron emission. Determine the wo
klemol [59]

Answer:

The work function ϕ of the metal = 53.4196 x 10⁻¹⁶ J      

Explanation:

When light is incident on a photoelectric material like metal, photoelectrons are emitted from the surface of the metal. This process is called photoelectric effect.

The relationship between the maximum kinetic energy (E_{k}) of the photoelectrons to the frequency of the absorbed photons (f) and the threshold frequency (f₀) of the photoemissive metal surface is:

                                        E_{k} = h(f − f₀)

                                        E_{k} = hf - hf₀

E is the energy of the absorbed photons:  E = hf

ϕ is the work function of the surface:  ϕ = hf₀

                                        E_{k} = E - ϕ

Frequency f = 8.12×10¹⁸ Hz

Maximum kinetic energy E_{k} = 4.16×10⁻¹⁷ J  

Speed of light  c = 3 x 10⁸ m/s

Planck's constant h = 6.63 × 10⁻³⁴ Js                                

                                        E = hf = 6.63 × 10⁻³⁴ x 8.12×10¹⁸

                                        E = 53.8356 x 10⁻¹⁶ J

from E_{k} = E - ϕ ;

                                        ϕ = E - E_{k}

                                        ϕ = 53.8356 x 10⁻¹⁶ - 4.16×10⁻¹⁷

                                        ϕ = 53.4196 x 10⁻¹⁶ J

The work function of the metal ϕ = 53.4196 x 10⁻¹⁶ J      

4 0
3 years ago
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