(02.09 MC) An object starts at rest. Its acceleration over 30 seconds is shown in the graph below: An acceleration versus time g
raph is shown with acceleration on the y axis from zero to seven meters per second per second and time on the x axis from zero to 30 second. The graph starts at the origin and goes to five meters per second in the first five seconds, then goes back to zero at time 10 seconds, then stays at zero until 20 seconds, then goes up to 6 meters per second until 30 seconds. Use the graph above to determine the change in speed of the object between 20 and 30 seconds? (2 points) Select one: a. 6 m/s b. 30 m/s c. 60 m/s d. 180 m/s
As you can see in the graph, or the description of the graph, the change in speed since the acceleration at 20 seconds was 0 will be all the acceleration that we added up after that second, since we accelerated at 6 m/s until 30 seconds, that means that the object remained 10 seconds accelerating at 6m/s, so 10seconds times 6 m/s that is 60 m/s in the change in speed between the 20th second and the 30th second.
It's important to know that diffraction gratings can be identified by the number of lines they have per centimeter. Often, more lines per centimeter is more useful because the images separation is greater when this happens. That is, the distance between lines increases.
