The wavelengths of the constituent travelling waves CANNOT be 400 cm.
The given parameters:
- <em>Length of the string, L = 100 cm</em>
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The wavelengths of the constituent travelling waves is calculated as follows;
for first mode: n = 1
for second mode: n = 2
For the third mode: n = 3
For fourth mode: n = 4
Thus, we can conclude that, the wavelengths of the constituent travelling waves CANNOT be 400 cm.
The complete question is below:
A string of length 100 cm is held fixed at both ends and vibrates in a standing wave pattern. The wavelengths of the constituent travelling waves CANNOT be:
A. 400 cm
B. 200 cm
C. 100 cm
D. 67 cm
E. 50 cm
Learn more about wavelengths of travelling waves here: brainly.com/question/19249186
Answer:
the kinetic energy lost due to friction is 22.5 J
Explanation:
Given;
mass of the block, m = 0.2 kg
initial velocity of the block, u = 25 m/s
final velocity of the block, v = 20 m/s
The kinetic energy lost due to friction is calculated as;
Therefore, the kinetic energy lost due to friction is 22.5 J
Assuming acceleration due to gravity of the moon is constant and there’s no initial velocity in the mans jump we can use one of the kinematic equations. x(final)=x(initial)+(1/2)gt^2. Plug in known values. 0=10-(1.62/2)t^2. The value 1.62 is acceleration of gravity on the moon. Now simply solve for t. t=3.513
Answer:
θ = (7π / 3) rad
Explanation:
given,
displacement of simple harmonic motion along x-axis
equation is given as
x = 5 sin (π t + π/3 )
general equation of simple harmonic motion
x = A sin θ
θ is the phase angle
θ = π t + π/3
at t = 2 s
Phase of the motion at t =2 s is θ = (7π / 3) rad
Answer:
I = M R^2 is the moment of inertia about a point that is a distance R from the center of mass (uniform distributed mass).
The moment of inertia about the center of a sphere is 2 / 5 M R^2.
By the parallel axis theorem the moment of inertia about a point on the rim of the sphere is I = 2/5 M R^2 + M R^2 = 7/5 M R^2
I = 7/5 * 20 kg * .2^2 m = 1.12 kg m^2
