A particle with mass 2.59 kg oscillates horizontally at the end of a horizontal spring. A student measures an amplitude of 0.897 m and a duration of 127 s for 71 cycles of oscillation. Find the frequency, f, the speed at the equilibrium position, vmax , the spring constant, k, the potential energy at an endpoint, Umax, the potential energy when the particle is located 45.1% of the amplitude away from the equiliibrium position, U, and the kinetic energy, K, and the speed, v, at the same position. f = 0.559 Hz Umax = m/s

Answer must be in standard form scientific notation with SI it’s that do not have prefixes. Also correct number of sig. fig.

D = m/V for V (density = mass divided by volume)

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Could you explain why 2.5 is the correct answer ?

Use Component and Polygon method

Derive the equation .Cd, g: constants.t, m, V: variables.Cd, g: constants.t, m, V: variables.

bn nas set his GPS to track his route. At the end of his hike, his GPS indicates that he walked 3.50 miles north, 1.00 miles at 45 degrees north of east, and 1.50 miles south. How far away is Bill from his starting position? And in what direction is he? Provide the direction in number of degrees in reference to the cardinal directions (North of East, South of West, etc.).

The symbols for speed, wavelength, and frequency are, respectively, v, 1, and f. The MKS units of speed, wavelength, and frequency are, respectively, meters per second, meters, and hertz. Hertz are equal to inverse seconds. Which of the following equations is incorrect by unit analysis? Of =v/A O 1/f = /v O A = vf O v = Af eTextbook and Media Attempts: 0 of 1 used Submit Answer Save for Later tv MacBook Pro G Search or type URL * 24 8.

Convert 0.20 m²/s into micro m²/min considering significant figures