The concept needed to solve this problem is average power dissipated by a wave on a string. This expression ca be defined as

Here,
= Linear mass density of the string
Angular frequency of the wave on the string
A = Amplitude of the wave
v = Speed of the wave
At the same time each of this terms have its own definition, i.e,
Here T is the Period
For the linear mass density we have that

And the angular frequency can be written as

Replacing this terms and the first equation we have that



PART A ) Replacing our values here we have that


PART B) The new amplitude A' that is half ot the wavelength of the wave is


Replacing at the equation of power we have that


Answer:
v = 10.47 m/s
a = 219.32 m/s²
Explanation:
The velocity of the stone can be given by the following formula:
v = r ω
where,
v = linear velocity of the stone = ?
r = radius of tire = 0.5 m
ω = angular velocity = (200 rev/min)(2π rad/1 rev)(1 min/60 s) = 20.94 rad/s
Therefore,
v = (0.5 m)(20.94 rad/s)
<u>v = 10.47 m/s</u>
<u></u>
The acceleration of the stone will be equal to the centripetal acceleration. Therefore,
a = v²/r
a = (10.47 m/s)²/0.5 m
<u>a = 219.32 m/s²</u>
Answer:
10mm
Explanation:
According to Hooke's law which states that "the extension of an elastic material is directly proportional to the applied force provided the elastic limit is not exceeded. Direct proportionality there means, increase/decrease in the force leads to increase/decrease in extension.
Mathematically, F = ke where;
F is the applied force
k is the elastic constant
e is the extension
from the formula k = F/e
k = F1/e1 = F2/e2
Given force of 1N indents the spring inwards by 2mm, this means force of 1N generates extension of 2mm
Let F1 = 1N e1 = 2mm
The extension that will be produced If force of 5N is applied to the string is what we are looking for. Therefore F2 = 5N; e2= ?
Substituting this values in the formula above we have
1/2=5/e2
Cross multiplying;
e2 = 10mm
This shows that we must have dent it by 10mm before it pushes outwards by a 5N force
The velocity of the car after it accelerates for 25 seconds would be 820.2 ft/s or 250 m/s