An object falling near the surface of the earth has a constant acceleration of 9

Question

An object falling near the surface of the earth has a constant acceleration of 9.8 m/s2. If the object starts from rest, this means that the A box on a horizontal table is given an initial velocity of 5 m/s. It slides across a rough surface coming to rest in 2 m. The coefficient of kinetic friction is closest to A 15 kg block sitting on a frictlonless table Is connected to a free-hanging 5 kg mass by a stretchless. massles cord that passes over a small frictionless pulley at the edge of the table. The acceleration of the two-bock system is In simple harmonic motion, the speed is greatest at the point in the cycle when A figure skater is spinning slowly with arms outstretched. She brings her arms In close to her body and her moment of inertia decreases by 1/2. Her angular speed changes by a factor of

Explanation / Answer

1)

here

acceleration = (vfinal - vinitial ) / t

9.8 m/s^2 = (vfinal - 0 ) / 1 sec

vfinal = 9.8 m/s

so after 2 sec the speed will be double

so

e.speed of the object increase by 9.8m/s during each second

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