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Bumek [7]
3 years ago
10

A charge of 4.5 × 10-5 C is placed in an electric field with a strength of 2.0 × 104 . What is the electric force acting on the

charge?
Physics
1 answer:
arlik [135]3 years ago
7 0

Answer:

0.9N/C

Explanation:

-The force of a charged particle in an electric field is given by the equation;

F=qE\\\\q-charge\\

#we substitute the values of the charge and field strength:

F=qE\\\\=4.5\times10^{-5}\times2.0\times10^4\\\\=0.9 \ N/C

Hence, the electric force acting on the charge is 0.9N/C

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A particle's trajectory is described by x = (0.5t^3-2t^2) meters and y = (0.5t^2-2t), where time is in seconds. What is the part
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Differentiate the components of position to get the corresponding components of velocity :

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v_y = \dfrac{\mathrm dy}{\mathrm dt} = \left(1\dfrac{\rm m}{\mathrm s^2}\right)t-2\dfrac{\rm m}{\rm s}

At <em>t</em> = 5.0 s, the particle has velocity

v_x = \left(1.5\dfrac{\rm m}{\mathrm s^3}\right) (5.0\,\mathrm s)^2 - \left(4\dfrac{\rm m}{\mathrm s^2}\right)(5.0\,\mathrm s) = 17.5\dfrac{\rm m}{\rm s}

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The speed at this time is the magnitude of the velocity :

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The direction of motion at this time is the angle \theta that the velocity vector makes with the positive <em>x</em>-axis, such that

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