A 2 x 10’kg car travels at a constant speed of 12 m/s around a circular curve of radius 30 m. What is the magnitude of the centripetal acceleration in m/s2 of the car as it goes around the curve?
A 0.4
B 3. 800
C 4.8
D 9600

Sound energy is produced when an object vibrates. examples are 1- a balloon popping2-a whistle3- a waterfall

Initial velocity,  u = 4.9 m/sfinal velocity = v m/sinitial height,  h₁ =  245 m    final height, h₂ = 0a = -9.8  m/s²time taken = t secondss = ut + 0.5at²⇒  (h₂-h₁) = ut + 0.5at²⇒  0-245 = 4.9t + 0.5×(-9.8)×t²⇒ -245 = 4.9t - 4.9t²⇒ 4.9t² -4.9t -245 =0solving it, we get   t = 7.59sv  = u +  at = 4.9 -9.8×7.59 = 4.9 - 74.38 = -69.48 m/svelocity is  69.48 m/s downward

To compare speeds speeds of wagons we will use second newton's law. it states: f = m * ahere we can see that force depends on the acceleration. if we know acceleration for both wagons we can compare them to make conclusion of speeds of wagons.we have: m1 = 10lbm2 = 20lbf1 = 2*f2also: m2 = 2*m1for bill we have equation: f1 = m1 * a1for tom we have equation: f2 = m2 * a2after inserting identities we have: 2*f2 =m1 * a1f2 = 2*m1 * a2in order to make conclusions about speeds we will solve these equations for a and find their ratio: a1 = 2f2 / m1a2 = f2 / 2m1a1 / a2 = (2f2/m1) / (f2/2m1)a1 / a2 = (2f2*2m1) / (m1*f2)a1 / a2 = (4f2m1) / (m1f2)a1 / a2 = 4this ratio is bigger than 1 and this means that a1 is greater than a2. a1 is 4 times greater than a2. speed of bill's wagon is 4 times greater than speed of tom's wagon.