Sound waves....................................
Answer:
The “terminal speed” of the ball bearing is 5.609 m/s
Explanation:
Radius of the steel ball R = 2.40 mm
Viscosity of honey η = 6.0 Pa/s



While calculating the terminal speed in liquids where density is high the stokes law is used for viscous force and buoyant force is taken into consideration for effective weight of the object. So the expression for terminal speed (Vt)

Substitute the given values to find "terminal speed"




The “terminal speed” of the ball bearing is 5.609 m/s
Answer
given,
y(x,t)= 2.20 mm cos[( 7.02 rad/m )x+( 743 rad/s )t]
length of the rope = 1.33 m
mass of the rope = 3.31 g
comparing the given equation from the general wave equation
y(x,t)= A cos[k x+ω t]
A is amplitude
now on comparing
a) Amplitude = 2.20 mm
b) frequency =


f = 118.25 Hz
c) wavelength




d) speed


v = 105.84 m/s
e) direction of the motion will be in negative x-direction
f) tension


T = 27.87 N
g) Power transmitted by the wave


P = 0.438 W
The angular speed is defined as:
<h2> ω=

</h2>
where



Complete question:
A 200 g load attached to a horizontal spring moves in simple harmonic motion with a period of 0.410 s. The total mechanical energy of the spring–load system is 2.00 J. Find
(a) the force constant of the spring and (b) the amplitude of the motion.
Answer:
(a) the force constant of the spring = 47 N/m
(b) the amplitude of the motion = 0.292 m
Explanation:
Given;
mass of the spring, m = 200g = 0.2 kg
period of oscillation, T = 0.410 s
total mechanical energy of the spring, E = 2 J
The angular speed is calculated as follows;

(a) the force constant of the spring

(b) the amplitude of the motion
E = ¹/₂kA²
2E = kA²
A² = 2E/k
