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
You might be interested in
Answer:
Part a)
Part b)
T = 4.68 s
Explanation:
Part a)
Shell is fired at speed of 40 m/s at angle of 35 degree
so here we have
since gravity act opposite to vertical speed of the shell so at the highest point of its trajectory the vertical component of the speed will become zero
so at the highest point the speed is given
Part b)
After completing the motion we know that the displacement of the object will be zero in Y direction
so we have