SUPER BOWL 2012 All x \\ !!! watch NFL Online | NFL·O PSI: 190 Review -> C filo:
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SUPER BOWL 2012 All x !!! watch NFL Online | NFL·O PSI: 190 Review -> C filo:///C/Users/az ::: Apps Bookmarks D Wel Advisor Main M w WebAssign O user Dashboard x PS1 Review xC Chegg Study Guided Sc S15620Raview.pdf Cuyanaca College- LS eg Sw le list D My boo E The Cook's Roadmap vP Spin for Perfect Skin PS1 Review 12 Two blocks, A (mA-8.0 kg) and B (me-5.0 kg), rest on a frictionless, horizontal tabletop as shown in the picture. Although there is no friction between block A and the table, the coefficient of static friction between block A and block B is 0.750. An incredibly light string connects block A to block C via a frictionless, massless pulley. What is the largest mass that C can have so that block B does not slip whern the system is released from rest? 3. ROTATION & GRAVITY Earth rotates around an axis, so technically it is not an inertial reference frame. This means every object on Earth's surface experiences centripetal acceleration pointing toward its rotation axis. At the Equator, both centripetal acceleration and gravity point down (toward the center of the Earth). This must be taken into account for very precise ballistics calculations (and weather predictions similarly include the fictitious Coriolis force) a. If the circumference of the Earth at the equator is approximately 40,000 km, how large is the centripetal acceleration at the equator, compared to acceleration of free fall g? How large is the centripetal acceleration at the North Pole? Now imagine two people: one at the North Pole, and another at the Equator. Each drops a ball from the same height h-1 m. Which ball falls longer? (HARD: 1 point extra credit for a good explanation) b. C. 5:35 PM Type here to searchExplanation / Answer
a) at the equator,
v = 2*pi*r/T
= 40000*10^3/(24*60*60)
= 463 m/s
centripetal acceleration, a = v^2/r
= 463^2/(40000*10^3/(2*pi))
= 0.0338 m/s^2
= 0.0338*g/9.8
= 0.00344*g
b) zero.
c) we know, time taken for an object to reach ground from height ,
t = sqrt(2*h/g)
at poles, t_poles = sqrt(2*h/g)
at equatr, t_equator = sqrt(2*h/(g - 0.0338))
so, t_equator > t_poles
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