(a) A ball is attached to a string of length L = 20.5 cm and suspended from the ceiling, as shown in the figure. A uniform electric field points to the right in the figure. When e = 13.0°, the ball is in equilibrium. Find the net charge on the ball (in µC). E = 1.00 × 10³ N/C m = 2.00 g 84.87 opposite It seems that you swapped sine and cosine in your calculation. Remember that sin(0) = hypotenuse (b) What If? If the electric field is suddenly turned off, what is the speed of the ball at the bottom of its swing (in m/s)? 0.35 Use energy conservation and solve your resulting equation for the speed v. m/s

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Chapter1: Units, Trigonometry. And Vectors
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(a) A ball is attached to a string of length L = 20.5 cm and suspended from the ceiling, as shown in the figure. A uniform electric field points to the right in the figure. When 0
13.0°, the ball is in
%3D
equilibrium. Find the net charge on the ball (in µC).
E = 1.00 × 103 N/C
L
m = 2.00 g
84.87
opposite
It seems that you swapped sine and cosine in your calculation. Remember that sin(0) =
hypotenuse
(b) What If? If the electric field is suddenly turned off, what is the speed of the ball at the bottom of its swing (in m/s)?
0.35
Use energy conservation and solve your resulting equation for the speed v. m/s
Transcribed Image Text:(a) A ball is attached to a string of length L = 20.5 cm and suspended from the ceiling, as shown in the figure. A uniform electric field points to the right in the figure. When 0 13.0°, the ball is in %3D equilibrium. Find the net charge on the ball (in µC). E = 1.00 × 103 N/C L m = 2.00 g 84.87 opposite It seems that you swapped sine and cosine in your calculation. Remember that sin(0) = hypotenuse (b) What If? If the electric field is suddenly turned off, what is the speed of the ball at the bottom of its swing (in m/s)? 0.35 Use energy conservation and solve your resulting equation for the speed v. m/s
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