A3.30-kg block starts from rest at the top of a 30.0° incline and slides a distance of 2.10 m down the incline in 1.60 s.(a) find the magnitude of the acceleration of the block.

Mass of the block m = 3.3kg Angle of the slide = 30 degrees Distance the block slides s = 2.10 m Time taken to slide t = 1.6 s Initially in rest condition so initial velocity u = 0. We have an equation for distance s = (u x t) + (1/2) x (a t^2) s = (0 x t) + (1/2) x (a x (1.6) ^2) => 2.10 = (1/2) x (a x2.56) 2.56a = 4.20 => a = 1.64 So the magnitude of the Acceleration a = 1.64 m/s^2

Answer: it will take   4.36 seconds to reach her average velocity 10.69 miles/hour.explanation:   we have equation of motion ,where s is the displacement, a is the acceleration, u is the initial velocity and t is the time taken.average velocity = displacement/ time taken.average velocity = 10.69 miles/hr = 10.69*1.6*1000/(60*60) = 4.75 m/s4.75 = s/ts = 4. 1we also have acceleration = 2.18 m/s², initial velocity = 0 m/sso it will take   4.36 seconds to reach her average velocity 10.69 miles/hour.