an 8.25 kgkg point mass and a 13.5 kgkg point mass are held in place 50.0 cmcm apart. a particle of mass mm is released from a point between the two masses 16.0 cmcm from the 8.25 kgkg mass along the line connecting the two fixed masses.Find the magnitude and direction of the acceleration of the particle.

Respuesta :

The particle's acceleration is 3.7*10-9 m/s2, according to the formula. A point mass of 8.25 kg and a point mass of 13.5 kg are kept in site at a distance of 50.0 cm.

Here, the mass m is pulled toward the two particles by gravity. the force generated by the initial mass moving in a -X direction and the second mass moving in a +X direction.

The following is the expression for Newton's law of gravitation:

F =GMm/r^2

The gravitational constant G, the masses M and m, the gravitational force F, and the separation between the mass center's r are all present in this situation. the force acting against the first mass is negative, When we substitute F=GM1m/r1, we obtain,

F1 = -(6.67x10^-11 Nm'/kg)(8.25kg)m/(0.2m)*(0.2m)=-1.37x10^-8 m

F2 also equals 1*10^-8.

F2 = F1+F2

ma = F1+F2

acceleration = F1+F2/m

acceleration = -1.3*10^-8 +1*10^-8m/m

a = -3.7*10^-9 m/s^2

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