A150 g particle at x = 0 is moving at 8.00 m/s in the +x-direction. as it moves, it experiences a force given by fx=(0.850n)sin(x/2.00m). what is the particle (s) speed when it reaches x = 3.14 m?

9.98 m/s

Explanation:

The force acting on the particle is defined by the equation:

[N]

where x is the position in metres.

The acceleration can be found by using Newton's second law:

where

m = 150 g = 0.150 kg is the mass of the particle. Substituting into the equation,

[m/s^2]

When x = 3.14 m, the acceleration is:

Now we can find the final speed of the particle by using the suvat equation:

where

u = 8.00 m/s is the initial velocity

v is the final velocity

x = 3.14 m is the displacement

Solving for v,

And the speed is just the magnitude of the velocity, so 9.98 m/s.

two charges are fixed in place on the x-axis as shown. the charge at x = 0 is negative and has magnitude q = 2 μcoulombs (1 μcoulomb = 10-6 coulombs). the charge at x = d, (d = +12 cm), is positive and has magnitude q = 4 μcoulombs. find a point x = xo along the x-axis at which the electric field is zero.