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nexus9112 [7]
3 years ago
7

What is the magnitude of the electric field at the origin produced by a semi-circular arc of charge = 7.6 μc, twice the charge o

f the quarter-circle arc?
Physics
1 answer:
Anna [14]3 years ago
6 0

Orient the semi-circle arc such that it is symmetric with respect to the y-axis. Now, by symmetry, the electric field in the x-direction cancels to zero. So the only thing of interest is the electric field in the y-direction.  


dEy=kp/r^2*sin(a) where k is coulombs constant p is the charge density r is the radius of the arc and a is the angular position of each point on the arc (ranging from 0 to pi. Integrating this renders 2kq/(pi*r^3). Where k is 9*10^9, q is 9.8 uC r is .093 m

I answeared your question can you answear my question pleas

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The graph above shows the position x as a function of time for the center of mass of a system of particles of total mass 6. 0 kg
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The resulting change in momentum of the system will be +18.6 Ns. The momentum is conserved.

<h3>What is the law of conservation of momentum?</h3>

According to the law of conservation of momentum, the momentum of the body before the collision is always equal to the momentum of the body after the collision.

The given data in the problem is;

m is the mass =6.0 kg

t is the time interval=2 second

From Newton's second law;

\rm \triangle P =m \triangle V \\\\ \triangle P= m(\frac{\triangle x}{\triangle t} )\\\\

From the graph;

\rm \triangle t = 2sec\\\\ \triangle x = (12-8) m

The change in the momentum is;

\rm \triangle P = m\tr(\frac{v-u}{t}) \\\\ \triangle P =9.3 \times \frac{12-8}{2} \\\\ \triangle P= +18.6 \  N.s

Hence, the resulting change in momentum of the system will be +18.6 Ns.

To learn more about the law of conservation of momentum, refer;

brainly.com/question/1113396

#SPJ1

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