The slowest line is the solid line and the fastest is the dotted line that crosses the solid line
for future reference you just need to find the slope or the line which is traveling most vertical
Explanation:
a. Net force is mass times acceleration (Newton's second law).
∑F = ma
∑F = (5.0 kg) (2.0 m/s²)
∑F = 10 N
b. The net force is the sum of the individual forces.
10 N = F − 5 N
F = 15 N
c. Friction force here is mgμ.
mgμ = 5 N
(5.0 kg) (10 m/s) μ = 5 N
μ = 0.1
Because the number of valence electrons of an element determines the properties and in particular the reactivity of that element.
In fact, elements of the first group (i.e. only one valence electron) have high reactivity, because they can easily give away their valence electron to atoms of other elements forming bonds. On the contrary, elements of the 8th group (noble gases) have their outermost shell completely filled with electrons, so they do not have valence electrons, and they have little or no reactivity at all.
As per law of Heisenberg uncertainty law
product of uncertainty in position and uncertainty in momentum will be constant


now plug in all data


So above is the uncertainty in velocity of the object
Answer:

Explanation:
If the stone will reach the top position of flag pole at t = 0.5 s and t = 4.1 s
so here the total time of the motion above the top point of pole is given as

now we have



so this is the speed at the top of flag pole
now we have



now the height of flag pole is given as


