Question #277625

Two vehicles are approaching an intersection. One is a 2500-š‘˜š‘” pickup traveling at 14.0š‘š/š‘  from east to west (the āˆ’š‘„-direction), and the other is a 1500-š‘˜š‘” sedan going from south to north (the +š‘¦-direction) at 23.0 š‘š/š‘ . (a) Find the š‘„- and š‘¦-components of the net momentum of this system. (b) What are the magnitude and direction of the net momentum?


Expert's answer

px=āˆ’(2500)(14)=āˆ’35000kgmspy=(1500)(23)=34500kgmsp=350002+345002=49100kgmsĪø=arctan⁔345350=44.6°p_x=-(2500)(14)=-35000\frac{kgm}{s}\\ p_y=(1500)(23)=34500\frac{kgm}{s}\\ p=\sqrt{35000^2+34500^2}=49100\frac{kgm}{s}\\\theta=\arctan{\frac{345}{350}}=44.6\degree


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