Answer:
The energy lost by the atoms is given off as an electromagnetic wave. ... even if it's not very intense, will always cause electrons to be emitted.
Explanation:
Solution
distance travelled by Chris
\Delta t=\frac{1}{3600}hr.
X_{c}= [(\frac{21+0}{2})+(\frac{33+21}{2})+(\frac{55+47}{2})+(\frac{63+55}{2})+(\frac{70+63}{2})+(\frac{76+70}{2})+(\frac{82+76}{2})+(\frac{87+82}{2})+(\frac{91+87}{2})]\times\frac{1}{3600}
=\frac{579.5}{3600}=0.161miles
Kelly,
\Delta t=\frac{1}{3600}hr.
X_{k}=[(\frac{24+0}{2})+(\frac{3+24}{2})+(\frac{55+39}{2})+(\frac{62+55}{2})+(\frac{71+62}{2})+(\frac{79+71}{2})+(\frac{85+79}{2})+(\frac{85+92}{2})+(\frac{99+92}{2})+(\frac{103+99}{2})]\times\frac{1}{3600}
=\frac{657.5}{3600}
\Delta X=X_{k}-X_{C}=0.021miles
Point A has the largest magnitude of acceleration as compared to other points on the position verses time graph.
On the graph, A is the point where magnitude of the acceleration of the particle is greatest as compared to other positions on the graph because the height of point A is the largest as compared to other points of the graph.
The graph shows at which point acceleration of an object is higher and lower so we can conclude that point A has the largest magnitude of acceleration as compared to other points on the position verses time graph.
Learn more about acceleration here: brainly.com/question/933224
Learn more: brainly.com/question/25887663
