W = ∫ (x from 0.1 to +oo) F dx
= ∫ (x from 0.1 to +oo) A e^(-kx) dx
= A/k x [ - e^(-kx) ](between 0.1 and +oo)
= A/k x [ 0 + e^(-k * 0.1) ]
<span>
= A/k x e^(-k/10) </span>
a) we can answer the first part of this by recognizing the player rises 0.76m, reaches the apex of motion, and then falls back to the ground we can ask how
long it takes to fall 0.13 m from rest: dist = 1/2 gt^2 or t=sqrt[2d/g] t=0.175
s this is the time to fall from the top; it would take the same time to travel
upward the final 0.13 m, so the total time spent in the upper 0.15 m is 2x0.175
= 0.35s
b) there are a couple of ways of finding thetime it takes to travel the bottom 0.13m first way: we can use d=1/2gt^2 twice
to solve this problem the time it takes to fall the final 0.13 m is: time it
takes to fall 0.76 m - time it takes to fall 0.63 m t = sqrt[2d/g] = 0.399 s to
fall 0.76 m, and this equation yields it takes 0.359 s to fall 0.63 m, so it
takes 0.04 s to fall the final 0.13 m. The total time spent in the lower 0.13 m
is then twice this, or 0.08s
Answer:
Difference threshold or also Just Noticeable Difference
Explanation:
The above mentioned case between room mates, where one room mate was able to detect a minute change in volume shows an instance of the difference threshold.
Difference threshold can be defined as stimulation at its minimum level that can be detected by an individual almost 50 % of the times.
It is the lowest possible level of sound that is detectable by a person.
That is what happened in the mentioned case that when the volume was increased from 14 to 15, Amber was able to detect it.
What happens to end a of the rod when the ball approaches it closely this first time is; It is strongly attracted.
<h3>Electrostatics</h3>
I have attached the image of the rod.
We are told that the ball is much closer to the end of the rod than the length of the rod. Thus, if we point down the rod several times, the distance of approach will experience no electric field and as such the charge on end point A of the rod must be comparable in magnitude to the charge on the ball.
This means that their fields will cancel.
Finally, we can conclude that when a charge is brought close to a conductor, the opposite charges will all navigate to the point that is closest to the charge and as a result, a strong attraction will be created.
This also applies to a strong conducting rod and therefore it is strongly attracted.
Read more about Electrostatics at; brainly.com/question/18108470
