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zzz [600]
3 years ago
13

A small sphere with a mass of 441 g is moving upward along the vertical +y-axis when it encounters an electric field of 5.00 N/C

. If, due to this field, the sphere suddenly acquires a horizontal acceleration of 13.0 m/s2 , what is the charge that it carries?
Physics
1 answer:
sweet-ann [11.9K]3 years ago
3 0
In your question where the ask is to calculate the charge that the small sphere carries which is the mass of it is 441g moving at an acceleration of 13m/s^2 nad having and electric field of 5N/C. So the formula in getting the charge is mutliply the mass and the quotients of Acceleration and the Electric Field so the answer is 1,146.6
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Which formula represents final velocity of an object with average acceleration?
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Answer:

you would be better off if the car bounced backwards

Explanation:

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0.035 N

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A thin, uniform rod is bent into a square of side length a. If the total mass is M, find the moment of inertia about an axis thr
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Answer:

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    I_s =  \frac{Ma^2}{3}

Explanation:

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    m_1 = \frac{M}{4}

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        I_g =  \frac{1}{12}  *  m_1 * a^2

Generally given that m_1 = m_2 = m_3 = m_4 = m it means that this moment inertia evaluated above apply to every side of the square  

Now substituting for  m_1

  So

       I _g=  \frac{1}{12}  *  \frac{M}{4} * a^2

Now according to  parallel-axis theorem the moment of inertia of one side of the square about an axis through the center and perpendicular to the plane of the square is mathematically represented as

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=>    I_a =  I_g + {\frac{M}{4} }* [\frac{q}{2} ]^2

substituting for I_g

=>    I_a =  \frac{1}{12}  *  \frac{M}{4} * a^2 + {\frac{M}{4} }* [\frac{q}{2} ]^2

=>    I_a = \frac{Ma^2}{48} + \frac{Ma^2}{16}

=>    I_a = \frac{Ma^2}{12}

Generally the moment of inertia of the square about an axis through the center and perpendicular to the plane of the square is mathematically represented as

      I_s = 4 * I_a

=>   I_s = 4 * \frac{Ma^2}{12}

=>   I_s =  \frac{Ma^2}{3}

8 0
3 years ago
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