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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.

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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Water with a volume flow rate of 0.001 m3/s, flows inside a horizontal pipe with diameter of 1.2 m. If the pipe length is 10m an

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Please help on this one

Sphinxa [80]

Since this is a projectile motion problem, break down each of the five kinematic quantities into x and y components. To find the range, we need to identify the x component of the displacement of the ball.

Let's break them down into components.

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v₂ 32 cos50 m/s ?

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Let's use the following equation of uniform motion for the Y components to solve for time, which we can then use for the X components to find the range.

Δdy = v₁yΔt + 0.5ay(Δt)²

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Now lets put our known values into the same kinematic equation, but this time for the x components to solve for range.

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Δdₓ = 32cos50m/s(<u>5</u>.00s)

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Therefore, the answer is A, 102.9m. According to significant digit rules, neither would be correct, but 103m is the closest to 102.9m so I guess that is what it is.

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