Answer:
10.93 rad/s
Explanation:
If we treat the student as a point mass, her moment of inertia at the rim is
So the system moment of inertia when she's at the rim is:
Similarly, we can calculate the system moment of inertia when she's at 0.456 m from the center
We can apply the law of angular momentum conservation to calculate the post angular speed when she's 0.456m from the center:
Answer:
F1=26N and F2=09N ..this is from the two simultaneously equations
Given that the mass of the toy cart is 2.0 kg and and the acceleration is unknown, the normal formula would be a=f/m where a is acceleration, f is force and m is mass but the string's breaking strength is 40n so I think the formula in this case will be f is greater than m*a
40 is greater than 2a
40 is greater than 2a
40/2 is greater than 2a/2
20m/s² is greater than a
Therefore the maximum speed the toy cart should have should be less than 20m/s²
Answer:
wen I was in the car toing home from school after a bad day n si si I have crazzyyy
Answer:
(a) x0 = 0m and y0 = 49.0m
(b) Vox = 15.0m/s Voy = 0m/s
(c) Vx = Vo = 15.0m/s and Vy = -gt
(d) X = 15.0t and y = 49.0 - 4.9t²
(e) t = 3.16s
(f) Vf = 34.4m/s
Explanation:
