A ball whose mass is 0.4 kg hits the floor with a speed of 4 m/s and rebounds upward with a speed of 2 m/s. If the ball was in contact with the floor for 2 ms (2multiply10-3 s), what was the average magnitude of the force exerted on the ball by the floor?Favg = N. Calculate the magnitude of the gravitational force that the Earth exerts on the ball:mg = N

Explanation:

We shall apply concept of impulse

Impulse = Force x time

= Force x 2 x 10⁻³ N.s

impulse = change in momentum

change in momentum

= .4 x 4 - ( - .4 x 2 )

= 2.4

Force x 2 x 10⁻³  = 2.4

Force = 2.4 /  2 x 10⁻³

= 1.2 x 10³ N .

average magnitude of the force exerted by floor = 1.2 x 10³ N

If R be reaction force by earth

R - mg = 1.2 x 10³

R = 1.2 x 10³ + mg

= 1200 + .4 x 9.8

= 1200 +3.92

= 1203 .92 N .

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.

5 w for 1hr = 5x60x60 joules=300x60=3000x6=18,000j as total energy conversion. efficiency of bulb needed for radiant to thermal within 18000