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umka21 [38]
3 years ago
13

Two identical particles of charge 6 μμC and mass 3 μμg are initially at rest and held 3 cm apart. How fast will the particles mo

ve when they are allowed to repel and separate to very large (essentially infinite) distance? Now suppose that the two particles have the same charges from the previous problem, but their masses are different. One particle has mass 3 μμg as before, but the other one is heavier, with a mass of 30 μμg. Their initial separation is the same as before. How fast are the particles moving when they are very far apart?
Physics
1 answer:
krek1111 [17]3 years ago
3 0

Answer:

Explanation:

The charges will repel each other and go away with increasing velocity , their kinetic energy coming from their potential energy .

Their potential energy at distance d

= kq₁q₂ / d

= 9 x 10⁹ x 36 x 10⁻¹² / 2 x 10⁻² J

= 16.2 J

Their total kinetic energy will be equal to this potential energy.

2 x 1/2 x mv² = 16.2

= 3 x 10⁻⁶ v² = 16.2

v = 5.4 x 10⁶

v = 2.32 x 10³ m/s

When masses are different , total P.E, will be divided between them as follows

K E of 3 μ = (16.2 / 30+3) x 30

= 14.73 J

1/2 X 3 X 10⁻⁶ v₁² = 14.73

v₁ = 3.13 x 10³

K E of 30 μ = (16.2 / 30+3) x 3

= 1.47 J

1/2 x 30 x 10⁻⁶ x v₂² = 1.47

v₂ = .313 x 10³ m/s

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tatuchka [14]

Answer:

7.04 m

Explanation:

t = Time taken

u = Initial velocity

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Accelration going up is considered as negetive

Initial Velocity of the ball

v^2-u^2=2as\\\Rightarrow -u^2=2as-v^2\\\Rightarrow -u^2=2\times -9.81\times 1.2-0^2\\\Rightarrow u=\sqrt{2\times 9.81\times 1.2}\\\Rightarrow u=4.85\ m/s

Assuming that the ball is thrown with the same velocity on the Moon, displacement of the ball is

v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{0^2-4.85^2}{2\times -1.67}\\\Rightarrow s=7.04\ m

The displacement of the ball on the moon is 7.04 m

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4 years ago
Hi guys! :) I really need help on this!! Will def mark brainliest
Travka [436]

Wavelength = (speed) / (frequency)

Speed of radio = speed of light.

3 0
3 years ago
What are the foci of the ellipse given by the equation 100x^2 + 64y^2 = 6 400?
NikAS [45]

The equation of our ellipse is:

100 x^2 + 64 y^2 = 6400 (1)

First, let's reduce the equation of the ellipse to the standard form:

\frac{x^2}{a^2}+ \frac{y^2}{b^2}=1

To do that, we should divide both terms of equation (1) by 6400, and we get:

\frac{x^2}{64}+ \frac{y^2}{100}=1

This is a vertical ellipse (because b^2 \ \textgreater \  a^2) centered in the origin, and so the distance of its foci from the origin (on the y axis) is given by

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Superman is standing 393 m horizontally away from Lois Lane. A villain drops a rock from 4.00 m directly above Lois. If Superman
Sergio039 [100]

Answer:

-963.93 m/s²

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

s=ut+\frac{1}{2}at^2\\\Rightarrow 4=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{4\times 2}{9.81}}\\\Rightarrow t=0.903\ s

s=ut+\frac{1}{2}at^2\\\Rightarrow 393=0\times 0.0903+\frac{1}{2}\times a\times 0.903^2\\\Rightarrow a=\frac{393\times 2}{0.903^2}\\\Rightarrow a=963.93\ m/s^2

The acceleration of Superman would be -963.93 m/s² from Lois' perspective

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3 years ago
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const2013 [10]

Answer:

True

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