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guapka [62]
4 years ago
6

A point charge q1 = -2.3 μC is located at the origin of a co-ordinate system. Another point charge q2 = 6.2 μC is located along

the x-axis at a distance x2 = 9.9 cm from q1.
Physics
1 answer:
Thepotemich [5.8K]4 years ago
7 0

Answer:

-13.09 N

Explanation:

Hello!

With Coulomb's law, the force of attraction between two charges can be calculated:

F=(K*q_1*q_2)/r^2

K=9*〖10〗^9  (N*m^2)/C^2

9,9 cm = 0,099 m

-2,3µC= -2,3*〖10〗^-6 C

6,2µC= 6,2*〖10〗^-6 C

By simplifying units the force remains in Newton.

Successes with your homework!

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What is the oxidation number of the element Se
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3 years ago
A car accelerates uniformly from rest and reaches a speed of 24.3 m/s in 9.1 s. The diameter of a tire is 80.2 cm. Find the numb
BaLLatris [955]

By using first and third equation of motion, the number of revolutions the tire makes during this motion is 43 rev.

ANGULAR MOTION

Since the car accelerate from rest, initial velocity will be zero.

Given that a car accelerates uniformly from rest and reaches a speed of 24.3 m/s in 9.1 s. The following are the given parameters

  • Initial velocity U = 0
  • Final velocity V = 24.3 m/s
  • Time t = 9.1 s

If the diameter of a tire is 80.2 cm, to find the number of revolutions the tire makes during this motion, we must first calculate the distance travelled by the car by using first and third equation of motion of the car.

First equation

V = U + at

Substitute all necessary parameters into the equation.

24.3 = 0 + 9.1a

a = 24.3/9.1

a = 2.67 m/s^{2}

Third Equation of motion

V^{2} = U^{2} + 2aS

Substitute all the necessary parameters

24.3^{2} = 0 + 2 x 2.67 x S

590.49 = 5.34S

S = 590.49 / 5.34

S = 110.58 m.

Given that the diameter of a tire is 80.2 cm,

the radius (r) will be 80.2/2 = 40.1 cm

convert it to meter

r = 40.1/100 = 0.401 m

The Circumference of the tire = 2\pir

Circumference = 2 x 3.143 x 0.401

Circumference = 2.52 m

Assuming no slipping, number of revolutions = 110.58/2.52

Number of revolutions = 43.89 rev.

Number of revolutions = 43 rev.

Therefore, the number of revolutions the tire makes during this motion is 43 rev.

Learn more about circular motion here: brainly.com/question/6860269

4 0
2 years ago
When are airplanes in motion
Sever21 [200]

Motion is always relative to something.

-- Relative to the moon, an airplane is ALWAYS in motion.

-- Relative to the ground, an airplane is in motion when it's taxiing or flying.

-- Relative to a bag of peanuts in its own galley, an airplane is NEVER in motion.


4 0
3 years ago
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