Suppose that a meter stick is balanced at its center. A 0.24 kg mass is then positioned at the 6-cm mark. At what cm mark must a 0.35 kg mass be placed to balance the 0.24 kg mass? (NEVER include units with the answer to ANY numerical question.)

80.17 cm

Explanation:

Taking moments of forces about the center, the total clockwise moments is equal to the total counter clockwise moment:

Force * distance (counter clockwise) = force * distance (clockwise)

0.24 * 9.8 * (50 - 6) = 0.35 * 9.8 * (x - 50)

0.24 * 44 = 3.43x - 171.5

103.5 = 3.43x - 171.5

=> 3.43x = 103.5 + 171.5

3.43x = 275

x = 275/3.43 = 80.17 cm

1.195000

explanation:

total mass=0.0117+0.0137+0.0881+0.0111

total mass= 0.01172kg

total velocity = 0.283-0.743+0.871-0.799+0.155+0.637+0.301=0.102

momentum=mv=0.1172*0.102=1.195000//