Aparticle of mass 2.0 kg is moving in the xy plane at a constant speed of 0.80 m s in the x direction along the line y 4 m as the particle travels from x 3 m to x 3 m the magnitude of its angular momentum with respect to the origin is?

6.4 kgm²/s

Explanation:

the formula for angular momentum is

Angular momentum, L= radius/distance, r * linear momentum, p

i.e L = r * p

and linear momentum = mass * velocity  i.e p = m*v

Hence, L = m*v*r

we are given the following:

mass of particle, m = 2 kg

velocity/speed of particle, v = 0.8 m/s

radius/distance, r = the value in y axis = 4m

substituting the values, we have

L = 2 * 0.8 * 4 = 6.4 kgm²/s

your answer would be fusion

fgravity=10x10=100n case 1

fgravity=30x20=600n case 2

6 times more