CHAPTER 4 - 4 - x=v t+!at ; 0 2 402m=0+!a(6.40s)...

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F gases m g y y m g F R x  =  v 0 t  +  ! at 2 402 m = 0 +  ! a  (6.40 s) 2 , which gives  a  = 19.6 m/s 2  =        2.00 g . We find the required horizontal force from F horizontal  =  ma  = (280 kg)(19.6 m/s 2 ) =       5.49 × 10 3  N . 21. From Newton’s third law, the gases will exert a force on the rocket that is  equal and opposite to the force the rocket exerts on the gases.   ( a ) With up positive, we write ? F  =  m from the force diagram for the rocket: F gases  –  mg  =  ma ;   33 × 6  N – (2.75 × 6  kg)(9.80 m/s 2 ) = (2.75 × 6  kg) a which gives        a  = 2.2 m/s 2 . ( b ) If we ignore the mass of the gas expelled and any change in  g , we can  assume a constant acceleration. We find the velocity from v  =  v 0  +  at   = 0 + (2.2 m/s 2 )(8.0 s) =         18 m/s . ( c ) We find the time to achieve the height from y  =  y 0  +  v 0 t  +  !
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CHAPTER 4 - 4 - x=v t+!at ; 0 2 402m=0+!a(6.40s)...

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