ihe net force applied to a 3 kg mass is 12 N at a 55 degree angle relative to th
ID: 1784445 • Letter: I
Question
ihe net force applied to a 3 kg mass is 12 N at a 55 degree angle relative to the positive x-axis. Find the x and y components of the object's acceleration. 6. x-component a. 2.294 m/s b. 2.668 m/s c. 1.742 m/s d. 2.041 m/s e. 2.083 m/s f. 2.354 m/s 7. y-component a. 2.865 m/s b. 3.653 m/s c. 3.814 m/s d. 3.277 m/s e. 3.652 m/s f. 2.931 m/s NDOFF A planet and its two moons are aligned as shown. Find the total force on Moon A due to the combination of Moon B and the planet. Note: maMoon A-1.63E +20-kg, mtMoon B = 7.04E+ 19.kg, rnPlane.-4.18E +22-kg, d.-463000000-m, and d2 9260000000,m. di Moon A Moon B Planet 8. a. 8.87E+12,N b. 1.723E+13.N 6.492E+12.N d. 1.058E+13 N e. 1.673E+13 N f. 1.264E+13,N A hockey puck with initial speed 18 m/s travels 210 m before coming to rest. What is the coefficient of friction between the puck and the ice? 9. d. 0.09813 e. 0.06352 f. 0.07593 a. 0.06333 b. 0.07872 c. 0.07061Explanation / Answer
6) Acceleration = ax = Fx/ m
= F*cos / m
= 12*cos(55) / 3
= 2.294 m/s^2 (Option A)
7) ay = Fy/ m
= F*sin / m
= 12*sin(55) / 3
= 3.277 m/s^2 (option d)
8) Force = [ G*mplanet*m moon A / (d1+d2)2 ] + [ G*mmoon B*m moon A / (d1)2 ]
Force = [ (6.67*10-11)*(4.18*1022)*(1.63*1020) / (463*106+879.7*107)2 ] +
[ (6.67*10-11)*(7.04*1019)*(1.63*1020) / (463*106)2 ]
Force = 5.299*1012 + 3.57*1012
= 8.87*10^12 N (Option a)
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