Speed of particle B is 2v₀/3 m/s to the left. Particle A and particle B will always have equal speed since they experience equal forces.
<h3>Conservation of energy</h3>
The speed and direction of the particle B is determined by applying the principle of conservation of energy as follows;
K.E₁ + P.E₁ = K.E₂ + P.E₂


At any given position, the speed of particle A and particle B will be equal, since they experience equal force and they have equal masses.
The complete question is below:
Particle A and particle B, each of mass M, move along the x-axis exerting a force on each other. The potential energy of the system of two particles assosicated with the force is given by the equation U=G/r 2, where r is the distance between the two particles and G is a positive constant. At time t=T1 particle A is observed to be traveling with speed 2vo/3 to the left. The speed and direction of motion of particle B is ?
Learn more about conservation of energy here: brainly.com/question/166559
Answer:
3a, 2b,4c,1d
Explanation:
what do I need to explain just something you know
Answer:
a = 9.94 m/s²
Explanation:
given,
density at center= 1.6 x 10⁴ kg/m³
density at the surface = 2100 Kg/m³
volume mass density as function of distance

r is the radius of the spherical shell
dr is the thickness
volume of shell

mass of shell


now,

integrating both side



we know,




a = 9.94 m/s²
Answer:
<em>The frequency changes by a factor of 0.27.</em>
<em></em>
Explanation:
The frequency of an object with mass m attached to a spring is given as
= 
where
is the frequency
k is the spring constant of the spring
m is the mass of the substance on the spring.
If the mass of the system is increased by 14 means the new frequency becomes
= 
simplifying, we have
= 
= 
if we divide this final frequency by the original frequency, we'll have
==>
÷
==>
x
==> 1/3.742 = <em>0.27</em>