1- A wheel of diameter of 49 cm slows down uniformly from 36 m/sto rest over a d

Question

1- A wheel of diameter of 49 cm slows down uniformly from 36 m/sto rest over a distance of 95 m. What is the total number ofrevolutions the wheel rotates in coming to rest?
Number of revolutions? 2- A bicycle wheel rotates uniformly through 15 revolutions in3.5 s. What is the angular speed of the wheel?
Angular speed in rad/s? 3- A pulsar (a rotating neutron star) emits pulses at a frequencyof 0.29 kHz. The period of its rotation is
Period in seconds? 1- A wheel of diameter of 49 cm slows down uniformly from 36 m/sto rest over a distance of 95 m. What is the total number ofrevolutions the wheel rotates in coming to rest?
Number of revolutions? A wheel of diameter of 49 cm slows down uniformly from 36 m/sto rest over a distance of 95 m. What is the total number ofrevolutions the wheel rotates in coming to rest?
Number of revolutions? A wheel of diameter of 49 cm slows down uniformly from 36 m/sto rest over a distance of 95 m. What is the total number ofrevolutions the wheel rotates in coming to rest?
Number of revolutions? 2- A bicycle wheel rotates uniformly through 15 revolutions in3.5 s. What is the angular speed of the wheel?
Angular speed in rad/s? A bicycle wheel rotates uniformly through 15 revolutions in3.5 s. What is the angular speed of the wheel?
Angular speed in rad/s? A bicycle wheel rotates uniformly through 15 revolutions in3.5 s. What is the angular speed of the wheel?
Angular speed in rad/s? 3- A pulsar (a rotating neutron star) emits pulses at a frequencyof 0.29 kHz. The period of its rotation is
Period in seconds? A pulsar (a rotating neutron star) emits pulses at a frequencyof 0.29 kHz. The period of its rotation is
Period in seconds? A pulsar (a rotating neutron star) emits pulses at a frequencyof 0.29 kHz. The period of its rotation is
Period in seconds?

Explanation / Answer

Diameter d = 49 cm = 0.49 m

Initial velocity u = 36 m / s

Final speed v = 0

Distance   S= 95 m

Circumference S ‘ = (pi) d

So, No.of revolutions N = S / S ‘

                                       = 61.71

(b). Angular speed w = 15 rev / 3.5 s

                                  = 15 * 2 (pi) rad / 3.5 s

                                   = 26.92 rad / s

©. Frequency f= 0.29 kHz = 290 Hz

Period T = 1/ f = 0.00344 s

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