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SIZIF [17.4K]
2 years ago
9

A 2.98-kg object oscillates on a spring with an amplitude of 8.05 cm. Its maximum acceleration is 3.55 m/s2. Calculate the total

energy.
Physics
1 answer:
Aloiza [94]2 years ago
6 0

Answer:

a = ω^2 A      formula for max acceleration (ignoring sign)

V = ω A         formula for max velocity

V^2 = ω^2 A^2 = a A   from first equation

E = 1/2 M V^2 = 1/2 * 2.98 * 3.55 * .0805 = .426 J

(kg * m/sec^2 * m = kg m^2 / sec^2 = Joule

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2 years ago
Determine the magnitude as well as direction of the electric field at point A, shown in the above figure. Given the value of k =
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Answer:

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The equation of the electric field is given by:

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I hope it helps you!

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