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Ad libitum [116K]
4 years ago
7

A positive charge, q1, of 5 µC is 3 × 10–2 m west of a positive charge, q2, of 2 µC. What is the magnitude and direction of the

electrical force, Fe, applied by q1 on q2?
Physics
1 answer:
EastWind [94]4 years ago
4 0
The law you need is :
lets call the constant c = 4pi*epsilon naught*r2.
where r is radius ( or seperation of the charges). epsilon naught is a constant of about 8.85*10^-12 or something like that. and 4 pi is just 4 times pi.
the actual equation is 
FORCE= (q1*q2)/c

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Answer:

Inductance, L = 0.0212 Henries

Explanation:

It is given that,

Number of turns, N = 17

Current through the coil, I = 4 A

The total flux enclosed by the one turn of the coil, \phi=5\times 10^{-3}\ Tm^2

The relation between the self inductance and the magnetic flux is given by :

L=\dfrac{N\phi}{I}

L=\dfrac{17\times 5\times 10^{-3}}{4}

L = 0.0212 Henries

So, the inductance of the coil is 0.0212 Henries. Hence, this is the required solution.

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an ice skater starts with a velocity of 2.25 m/s in a 50.0 direction. after 8.33 m/s in a 120 direction. what is the y-component
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The y-component of the acceleration is 0.28 m/s^2

Explanation:

The y-component of the ice skater acceleration can be calculated with the equation

a_y = \frac{v_y-u_y}{t}

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v_y is the y-component of the final velocity

u_y is the y-component of the initial velocity

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Here we have:

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Learn more about acceleration:

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brainly.com/question/11181826

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