Physx141MPch7sol

Physx141MPch7sol - 7.2. Solve: Since = ( d dt ) we have f =...

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7.2. Solve: Since ( dd t = ) ωθ we have θ f = i + area under the ω -versus- t graph between t i and t f From t = 0 s to t = 2 s, the area is ( )( ) 1 2 20 rad/s 2 s 20 rad = . From t = 2 s to t = 4 s, the area is . Thus, the area under the -versus- t graph during the total time interval of 4 s is 60 rad or (60 rad) × (1 rev/2 π rad) = 9.55 revolutions. () ( ) 20 rad/s 2 s 40 rad = 7.4. Model: The airplane is to be treated as a particle. Visualize: Solve: (a) The angle you turn through is 21 5000 miles 180 1.25 rad 1.25 rad 71.6 4000 miles rad s r θθ ° −== = = × = ° (b) The plane’s angular velocity is 1.25 rad 0.139 rad/hr 9 hr tt == = 5 rad 1 hr 0.139 3.86 10 rad/s hr 3600 s = × Assess: An angular displacement of approximately one-fifth of a complete rotation is reasonable because the separation between Kampala and Singapore is approximately one-fifth of the earth’s circumference. 7.8. Solve: Let R E be the radius of the earth at the equator. This means R E + 300 m is the radius to the top of the tower. Letting T be the period of rotation, we have ( ) E 2 E top bottom 2 300 m 2 300 m 26 0 0 m 2.18 10 m/s 24 hrs 24 3600 s R R vv TT ππ + −= − = = 7.10. Solve: The plane must fly as fast as the earth’s surface moves, but in the opposite direction. That is, the plane must fly from east to west. The speed is 3 2 r a d km 1 m i l e 6.4 10 km 1680 1680 1040 mph 24 hr hr hr 1.609 km vr  × = = × =   7.12.
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Physx141MPch7sol - 7.2. Solve: Since = ( d dt ) we have f =...

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