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Korvikt [17]
1 year ago
15

What is the answer?

Physics
1 answer:
Andrews [41]1 year ago
3 0

Answer:

6.26 m/s

Explanation:

Pretty slow.... the PE (Potential Energy) at 2m will be converted to KE (Kinetic Energy) at the bottom of the track (neglecting friction)

PE    =   KE

mgh = 1/2 mv^2     divide both sides of the equation by 'm'

gh    = 1/2 v^2             multiply both sides by 2

2 gh = v^2               take sqrt of both sides

v = sqrt ( 2gh) = sqrt ( 2*9.81*2) = 6.26 m/s

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Three equal charge 1.8*10^-8 each are located at the corner of an equilateral triangle ABC side 10cm.calculate the electric pote
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Answer:

If all these three charges are positive with a magnitude of 1.8 \times 10^{-8}\; \rm C each, the electric potential at the midpoint of segment \rm AB would be approximately 8.3 \times 10^{3}\; \rm V.

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Convert the unit of the length of each side of this triangle to meters: 10\; \rm cm = 0.10\; \rm m.

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Electric potential due to the charge at \rm A: \displaystyle \frac{k\, q}{d({\rm A})}.

Electric potential due to the charge at \rm B: \displaystyle \frac{k\, q}{d({\rm B})}.

Electric potential due to the charge at \rm A: \displaystyle \frac{k\, q}{d({\rm C})}.

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\begin{aligned}& \frac{k\, q}{d({\rm A})} + \frac{k\, q}{d({\rm B})} + \frac{k\, q}{d({\rm C})} \\ &= k\, \left(\frac{q}{d({\rm A})} + \frac{q}{d({\rm B})} + \frac{q}{d({\rm C})}\right) \\ &\approx 8.99 \times 10^{9}\; \rm N \cdot m^{2} \cdot C^{-2} \\ & \quad \quad \times \left(\frac{1.8 \times 10^{-8} \; \rm C}{0.050\; \rm m} + \frac{1.8 \times 10^{-8} \; \rm C}{0.050\; \rm m} + \frac{1.8 \times 10^{-8} \; \rm C}{\sqrt{3} \times (0.050\; \rm m)}\right) \\ &\approx 8.3 \times 10^{3}\; \rm V\end{aligned}.

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