Answer:
Explanation:
mass of string = .0125 / 9.8
= 1.275 x 10⁻³ kg
Length of string l = 1.5 m .
m = mass per unit length
= ( .1.275 / 1.5) x 10⁻³ kg/m
m = .85 x 10⁻³ kg/m
wave equation: y(x,t) = (8.50 mm)cos(172 rad/m x − 4830 rad/s t)
compare with equation of wave
y(x,t) = Acos(K x − ω t)
ω ( angular velocity ) = 4830 rad/s
k = 172 rad/m
Velocity = ω / k
= 4830/172 m /s
= 28.08 m /s
velocity of wave = 
28.08 = 
788.48 = W / .85 X 10⁻³
W = 670 x 10⁻³ N .
c ) wave length
wave length =2π / k
= 2 x 3.14 / 172
= .0365 m
no of wave lengths over whole length of string
= 1.5 / .0365
= 41
d )
equation for waves traveling down the string
= (8.50 mm)cos(172 rad/m x + 4830 rad/s t)
Answer:
8) 709.8875 J
9) The object is at 7.24375 m from the ground
10) Kinetic energy increases as the object falls.
Explanation:
We use the expression for the displacement h(t) as a function of time of an object experiencing free fall:
h(t) = hi - (g/2) t^2
hi being the initial position of the object (10m) above ground, g the acceleration of gravity (9.8 m/s^2), and t the time (in our case 0.75 seconds):
h(0.75) = 10 - 4/9 (0.75)^2 = 7.24375 m
This is the position of the 10 kg object after 0.75 seconds (answer for part 9)
Knowing this position we can calculate the potential energy of the object when it is at this height, using the formula:
U = m g h = 10kg * 9.8 (m/s^2) * 7.24375 m = 709.8875 J (answer for part 8)
Part 10)
the kinetic energy of the object increases as it gets closer to ground, since its velocity is increasing in magnitude because is being accelerated in its motion downwards.
In a string of length L, the wavelength of the n-th harmonic of the standing wave produced in the string is given by:
The length of the string in this problem is L=3.5 m, therefore the wavelength of the 1st harmonic of the standing wave is:
The wavelength of the 2nd harmonic is:
The wavelength of the 4th harmonic is:
It is not possible to find any integer n such that 
, therefore the correct options are A, B and D.
Answer:
theroy of plate tectonics
