From what we know, we can confirm that this ratio (turning up the volume by one click relative to the TV's overall volume) can be quantified as the Weber fraction.
<h3>What is the Weber fraction?</h3>
This fraction describes the ratio needed for change to a stimulus in which the change is just barely noticeable. This question is a prime example in that it seeks to find out just how low of a difference is needed in TV volume in order for the difference to be noticeable.
Therefore, we can confirm that this ratio (turning up the volume by one click relative to the TV's overall volume) can be quantified as the Weber fraction.
To learn more about Weber visit:
brainly.com/question/5004433?referrer=searchResults
Answer:
W= 4.4 J
Explanation
Elastic potential energy theory
If we have a spring of constant K to which a force F that produces a Δx deformation is applied, we apply Hooke's law:
F=K*x Formula (1): The force F applied to the spring is proportional to the deformation x of the spring.
As the force is variable to calculate the work we define an average force
Formula (2)
Ff: final force
Fi: initial force
The work done on the spring is :
W = Fa*Δx
Fa : average force
Δx : displacement
:Formula (3)
: final deformation
:initial deformation
Problem development
We calculate Ff and Fi , applying formula (1) :
We calculate average force applying formula (2):
We calculate the work done on the spring applying formula (3) : :
W= 11N*(0.7m-0.3m) = 11N*0.4m=4.4 N*m = 4.4 Joule = 4.4 J
Work done in stages
Work is the change of elastic potential energy (ΔEp)
W=ΔEp
ΔEp= Epf-Epi
Epf= final potential energy
Epi=initial potential energy
W=ΔEp= 5.39 J-0.99 J = 4.4J
:
Well, I guess you can come close, but you can't tell exactly.
It must be presumed that the seagull was flying through the air
when it "let fly" so to speak, so the jettisoned load of ballast
of which the bird unburdened itself had some initial horizontal
velocity.
That impact velocity of 98.5 m/s is actually the resultant of
the horizontal component ... unchanged since the package
was dispatched ... and the vertical component, which grew
all the way down in accordance with the behavior of gravity.
98.5 m/s = √ [ (horizontal component)² + (vertical component)² ].
The vertical component is easy; that's (9.8 m/s²) x (drop time).
Since we're looking for the altitude of launch, we can use the
formula for 'free-fall distance' as a function of acceleration and
time:
Height = (1/2) (acceleration) (time²) .
If the impact velocity were comprised solely of its vertical
component, then the solution to the problem would be a
piece-o-cake.
Time = (98.5 m/s) / (9.81 m/s²) = 10.04 seconds
whence
Height = (1/2) (9.81) (10.04)²
= (4.905 m/s²) x (100.8 sec²) = 494.43 meters.
As noted, this solution applies only if the gull were hovering with
no horizontal velocity, taking careful aim, and with malice in its
primitive brain, launching a remote attack on the rich American.
If the gull was flying at the time ... a reasonable assumption ... then
some part of the impact velocity was a horizontal component. That
implies that the vertical component is something less than 98.5 m/s,
and that the attack was launched from an altitude less than 494 m.
