Assuming that acceleration due to gravity is g = 9.8 m/s^2.
mass of block = 5 kg
<span>μ = 0.2
</span>applied force = 9 N
The force of gravity, Fg, on this block is given by Fg = m * g --> F = 5 kg * 9.8m/s^2 = 49 N.
The normal force on this block exerted by the ground is Fn = Fg --> 49 N. (It points in direction opposite to Fg, but we don't need to worry about that here.)
The <em>maximal </em>force of friction on the block that resists its movement is Ff = Fn * <span>μ --> Ff = 49 N * 0.2 = 9.8 N
So, the max frictional force is 9.8 N.
Since the applied force is only 9 N, the block does not move.
Since we deduced that the block is not moving, then the block must be in equilibrium (all external forces must equal zero). So, the frictional force must match the applied force of 9 N.</span>
Answer:
Explanation:
The force excerted by a magnetic field on a charged particle is given by the Lorentz force:
Lets consider the z-direction of our coordinate system the same direction of the magnetic field, that is:
Let us consider that the velocity of a given particle is:
Therefore, since k×k = 0
And since i, j , k are a rigth hand system:
i × j = k
j × k = i
k × i = j --> i × k= -j
Threfore, if the particle has charge q and velocity v = (vx,vy,vz), the magnetic force it will feel will be
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