A rock falls for 1.43 seconds how far did it fall? The falls's velocity is an acceleration of -9.81 m/s2

Explanation:

Generally from kinematics equations we have that

Here u is the initial velocity of the rock and the value is  0 m/s given that there was no information that it was in motion before the fall

substituting 1.43 s for t and for g we have that

=>    

The negative sign show that the displacement is in the direction of the negative y-axis

The answer is -10.03 meters

The first is to write out the parameters given in the question

time= 1.43 seconds

distance(s)= ?

acceleration(g)= - 9.81

The formula is given as follows

s= ut +  1/2gt²

= (0)(1.43) + 1/2(-9.81)(1.43²)

= 0 + (-4.905)(2.0449)

= - 10.03

Hence the distance is -10.03 meters

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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.

unlike poles attract