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 mus
t a 0.35 kg mass be placed to balance the 0.24 kg mass? (NEVER include units with the answer to ANY numerical question.)
1 answer:
Answer:
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
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Answer:
the acceleration of the airplane is 5.06 x 10⁻³ m/s²
Explanation:
Given;
initial velocity of the airplane. u = 34.5 m/s
distance traveled by the airplane, s = 46,100 m
final velocity of the airplane, v = 40.7 m/s
The acceleration of the airplane is calculated from the following kinematic equation;
v² = u² + 2as
Therefore, the acceleration of the airplane is 5.06 x 10⁻³ m/s²