A ball falls 0.35m to the floor. it lands 3m from the edge of the table. what was the velocity of the ball before it left the table?

The ball rolls off the table with speed v from a height of 0.35 m, so that it covers a horizontal distance x with height y at time t of

x = v t

y = 0.35 m - 1/2 g t ²

where g = 9.80 m/s² is the magnitude of the acceleration due to gravity.

Solve for t when y = 0, i.e. the time it takes for the ball to reach the ground:

0 = 0.35 m - 1/2 g t ²

t ² = (0.70 m) / g

t ≈ 0.267 s

Now solve for v given that the ball falls 3 m away from the table:

3 m = v (0.27 s)

v = (3 m) / (0.27 s)

v ≈ 11.2 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.

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-the future of geothermal energy can pretty much be summed up with a single word: more

-scientists plan in going deeper than 120 meters

-we are hoping that one day, we will get all of our energy from renewable energy sources such as geothermal energy

hope this !