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iragen [17]
3 years ago
15

A uniform rod of length L is pivoted at L/4 from one end. It is pulled to one side through a very small angle and allowed to osc

illate in a vertical plane. Determine the period of oscillation of the pendulum if L = 5.95 m.
Physics
1 answer:
ludmilkaskok [199]3 years ago
8 0

Answer:

T= 4.24sec

Explanation:

We are going to use the formula below to calculate.

T=2\pi \sqrt{\frac{L}{g} }

Where T is period

           L is length of rod

       g is acceleration due to gravity =     9.8m/s^{2}

From the problem, the rod is pivoted at 1/4L which means that three quarter of the rod was used for the oscillation. lets call this L_{O}

L_{O} = 3/4 * 5.95m

        = 4.4625m

thus   T=2\pi \sqrt{\frac{L_{O} }{g} }

          T=2\pi \sqrt{\frac{4.4625 }{9.8} }

          T= 4.24sec

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When a  woman walks south at a speed of 2.0mph for 60 minutes. She then turns around and walks north at a distance of 3000m in 25 minutes. then the woman's average speed during her entire motion would be  73.15 meters /minute.

<h3>What is speed?</h3>

The total distance covered by any object per unit of time is known as speed.

the mathematical expression for speed is given by

speed = total; distance /total time

As given in the problem a woman walks south at a speed of 2.0mph for 60 minutes

60 min = 1 hour

1 mile = 1.60934 km

The distance covered by her southwards  =  speed ×time

                                                                      =2 mph × 60 minutes

                                                                      = 3.218 km

She then turns around and walks north at a distance of 3000m in 25 minutes

The distance covered northward is 3000m

speed = total distance /total time

          =(3218 +3000) /(60+25)

          =73.15 meters /minutes

Thus, The average speed of the woman would be 73.15 meters /minute.

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