Question

because of them theories on the subiect, and he lost his post at the University of from these experiments were contrary to the An experiment with an incline and three balls produced the following results has a mass of 40 grams, and ball C has a mass of 80 grams.
How much time is required for one of the balls to travel 1.5 inches?​

Answer 1

Explanation:

Hello here for the solution for the party. Calculate the distance traveled by the marble in the inclined plane between one second and a two second. Using the constant acceleration equation that is delta X equal toe we know t plus half 80 square Here we noticed the initial spear off the marble Delta X is the distance off the marble and the A is the acceleration off the marble. Now, the initially speed off the marble in the inclined plane is equal toe the zero. So the V note is equal to 0 m per second squared. Now substitute X for delta X 0 m per second four we note and one second 40 in the equation Delta X equal toe. We know t plus half 80 square and solve for X here. By substituting the value in the equation and solve this, we get X equal to a by two. But now the distance moved by the marble in our time t death is that X Delta X delta equal toe. We know t displace Half 80 Death Square. Therefore, the distance moved by the seconds between the time leaders and T is Delta X dx minus delta X equal toe we know t death plus half, 80 deaths square minus back it we know t plus half 80 square. It will be equal toe We note t ds minus d plus half 80 bracket Tedeschi Square minus T square. Now substitute 0 m per second for we note to 2nd 40 days and one second 40 and x for one by two A. Here in the equation By substituting the value we get it Terry X. Therefore, the distance covered by the marble between one second and two seconds on the inclined plane is kitty X. Now for the party. Calculate the distance traveled by the marble in the inclined plane between two seconds and three seconds, using the drive equation in the part a The Delta excess minus delta X equal toe we know teeters minus D plus half a packet. Peters's square minus T square. Mhm now substitute 0 m per second for we note three second 40 days to 2nd 40 and export half a here in the equation. By substituting the value in solving this vigor. Delta X X minus. Delta X is equal to five x Therefore, the distance covered by the marble between us and three years on. The Inclined plane is five x now for the Parsi. Calculate the distance table by the marble in the inclined plane between and minus one and and seconds using the drive equation in the part A that is Delta X X minus delta X equal toe. We note back it tedious minus T plus half a bracket. Tedious, square minus T square. Now substitute 0 m per second for we note and 40 days and and management 2nd 40 and X and for half Hey here. By substituting the value we get Delta X X minus Delta X equal to two and minus one whole multiple X. Therefore the distance covered by the marble in the N earth. Second off the motion is two and minus one x So this is a complete solution, step by step in detail.