2) The position of a 0.30-kg object attached to a spring is described by x = (0.

Question

2) The position of a 0.30-kg object attached to a spring is described by

x = (0.26 m) cos(0.5?t).

(c) Find the position of the object at t = 0.27 s.
m

(d) Find the object's speed at t = 0.27 s.

1) Energy from the Sun arrives at the top of the Earth's atmosphere with an intensity of 1.30 kW/m^2. How long does it take for 1.35 x 10^9 J of energy to arrive on an area of 1.00 m2? This is the average monthly electrical energy consumption of a family of four in the United States. How is intensity related to power and area? How is power related to energy and time ? hr 2) The position of a 0.30-kg object attached to a spring is described by (a) Find the amplitude of the motion. (b) Find the spring constant (c) Find the position of the object at t = 0.27 s. (d) Find the object's speed at t = 0.27 s.

Explanation / Answer

Use the below equation to find the required time.

                  t = E/IA     

                    = [1.35x109 J ] / (1.30x103 W/m2)(1.00 m2)

                    = 1.038x106 s

                    = 12 days

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(a)

The position of the object is defined as,

      x = (0.26 m) cos(0.5(pi)t).

At t = 0.27 s, the position is,

      x = (0.26 m) cos(0.5(pi)(0.27))

         = 0.237 m

(b)

The speed of the object is,

      dx/dt = d/dt [(0.26 m) cos(0.5(pi)t)]

               = (0.26 m) [-(0.5)(pi)]sin(0.5(pi)t)

               = (0.26 m) [-(0.5)(pi)]sin(0.5(pi)(0.27 s))

               = -0.168 m/s

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