Physics help asap pretty please! An automobile with 0.210 m radius tires travels 85,000 km before wearing...?
Physics help asap pretty please! An automobile with 0.210 m radius tires travels 85,000 km before wearing...?
1) An automobile with 0.210 m radius tires travels 85,000 km before wearing them out. How many revolutions do the tires make, neglecting any backing up and any change in radius due to wear?
2) An ordinary workshop grindstone has a radius of 8.50 cm and rotates at 6500 rpm.
(a) Calculate the centripetal acceleration at its edge in m/s2 and convert it to multiples of g.
(b) What is the linear speed of a point on its edge?
3) Determine the maximum speed that can be reached by a 20.11 N (as measured on Earth) steel ball tied to the end of a thin thread. The ball, which is essentially weightless, is being whirled in a circle of 4.00 m radius by an astronaut out in space. The thread has a breaking strength of 18.0 N.
Physics help asap pretty please! An automobile with 0.210 m radius tires travels 85,000 km before wearing...?
1) An automobile with 0.210 m radius tires travels 85,000 km before wearing them out. How many revolutions do the tires make, neglecting any backing up and any change in radius due to wear?
2) An ordinary workshop grindstone has a radius of 8.50 cm and rotates at 6500 rpm.
(a) Calculate the centripetal acceleration at its edge in m/s2 and convert it to multiples of g.
(b) What is the linear speed of a point on its edge?
3) Determine the maximum speed that can be reached by a 20.11 N (as measured on Earth) steel ball tied to the end of a thin thread. The ball, which is essentially weightless, is being whirled in a circle of 4.00 m radius by an astronaut out in space. The thread has a breaking strength of 18.0 N.
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1) number of revolutions n = S / 2 pi r = 85x10^6 / 6.28 x0.21= 6.445 x10^7
2) a) centripetal acceleration a = r w^2 = 0.085 x (6500 x6.28/60)^2 = 39342 m/sec^2.= 4015 g (nearly)
b) linear speed v = r w = 0.085 x6500x6.28/60 = 57.83 m/sec.
3) Tmax = m Vmax^2 / r so that Vmax = square root of ( Tmax r / m) = square root of ( 18 x 4/ 2)
ie., Vmax = 6 m/sec(nearly)
2) a) centripetal acceleration a = r w^2 = 0.085 x (6500 x6.28/60)^2 = 39342 m/sec^2.= 4015 g (nearly)
b) linear speed v = r w = 0.085 x6500x6.28/60 = 57.83 m/sec.
3) Tmax = m Vmax^2 / r so that Vmax = square root of ( Tmax r / m) = square root of ( 18 x 4/ 2)
ie., Vmax = 6 m/sec(nearly)