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Jet001 [13]
3 years ago
12

An electron has a charge of 1.602 X 10-19.coulomb. When two electrons are separated by 1.2 X 10-9m, what force will they exert o

n each other?(-19 & -9 are exponents)
6.2 X 10-6N
1.6 X 10-10N
3.4 X 10-9N
5.3 X 10-7 N​
Physics
1 answer:
amid [387]3 years ago
3 0

Answer:

The force they will exert on each other is 1.6*10⁻¹⁰ N

Explanation:

The electromagnetic force is the interaction that occurs between bodies that have an electric charge. When the charges are at rest, the interaction between them is called the electrostatic force. Depending on the sign of the interacting charges, the electrostatic force can be attractive or repulsive. The electrostatic interaction between charges of the same sign is repulsive, while the interaction between charges of the opposite sign is attractive.

Coulomb's law is used to calculate the electric force acting between two charges at rest. This force depends on the distance "r" between the electrons and the charge of both.

Coulomb's law is represented by:

F=k*\frac{q1*q2}{r^{2} }

where:

  • F = electric force of attraction or repulsion in Newtons (N). Like charges repel and opposite charges attract.
  • k = is the Coulomb constant or electrical constant of proportionality.
  • q = value of the electric charges measured in Coulomb (C).
  • r = distance that separates the charges and that is measured in meters (m).

In this case:

  • k= 9*10⁹ \frac{N*m^{2} }{C^{2} }
  • q1= 1.602*10⁻¹⁹ C
  • q2= 1.602*10⁻¹⁹ C
  • r= 1.2*10⁻⁹ m

Replacing:

F=9*10^{9} \frac{N*m^{2} }{C^{2} }*\frac{1.602*10^{-19} C*1.602*10^{-19} C}{(1.2*10^{-9} )^{2} }

and solving you get:

F=1.6*10⁻¹⁰ N

<u><em>The force they will exert on each other is 1.6*10⁻¹⁰ N</em></u>

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The density of the woman is calculate as;

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A 1.0 kg football is given an initial velocity at ground level of 20.0 m/s [37 above horizontal]. It gets blocked just after re
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Refer to the diagram shown below.

Neglect air resistance.
The horizontal component of the launch velocity is
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The vertical component of the launch velocity is
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The acceleration due to gravity is g =9.8 m/s².
The time, t s, for the ball to reach a height of 3 m is given by 
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t² - 2.4563t + 0.6122 = 0
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The ball reaches a height of 3 m twice.
The first time it reaches 3 m height is 0.2815 s.

Part a.
The vertical velocity when t = 0.2815 s is
Vy  = 12.036 - 9.8*0.2815
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The horizontal component of velocity is Vx = 15.973 m/s
The resultant velocity is 
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Answer:
The velocity at a height of 3.0 m  is 18.5 m/s (nearest tenth)

Part b.
The horizontal distance traveled is 
d = (15.973 m/s)*(0.2815 s) = 4.4964 m
Answer:
The horizontal distance traveled is 4.5 m (nearest tenth)

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