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slavikrds [6]
1 year ago
5

Suppose you fill two rubber balloons with air, suspend both of them from the same point, and let them hang down on strings of eq

ual length. You then rub each with wool or on your hair so that the balloons hang apart with a noticeable separation between them. Make order-of-magnitude estimates of (b) the charge on each
Physics
1 answer:
Ulleksa [173]1 year ago
3 0

The charge on each the balloon is 100nC or 1.2 × 10^-7 C

Consider two balloons of diameter 0.200m each with a mass of 1.00g hanging apart with 0.0500m separation on the ends of string making angles of 10.0° with the vertical.

The charge on each balloon can be found from

F_{e} = \frac{k_{e}q^{2} }{r^{2}} \\q = \sqrt{\frac{k_{e}q^{2} }{r^{2}}} \\\\q = \sqrt{\frac{(2\times10^{-3N}(0.25m)^{2}}{8.99\times10^{9}N\cdotm^{2}/C^{2}}\\

q = 1.2\times10^{-7}C or 100nC

An electric charge is the property of matter where it has more or fewer electrons than protons in its atoms. Electrons carry a negative charge and protons carry a positive charge. The matter is positively charged if it contains more protons than electrons and negatively charged if it contains more electrons than protons.

Learn more about the charge here:

brainly.com/question/14713274

#SPJ4

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       F_{frmax} = \mu_{s} *F_{n} (1)

       where  μs is the coefficient of static friction, and Fn is the normal force,

       perpendicular to the wall and aiming to the center of rotation.

  • This force is the only force acting in the horizontal direction, but, at the same time, is the force that keeps the riders rotating, which is the centripetal force.
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     F_{frmax} = m* \mu_{s} * \omega^{2} * r (3)

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       g = \mu_{smin} * \omega^{2} * r (5)

  • Prior to solve (5) we need to convert ω from rev/min to rad/sec, as follows:

      60 rev/min * \frac{2*\pi rad}{1 rev} *\frac{1min}{60 sec} =6.28 rad/sec (6)

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