The block has maximum kinetic energy at the bottom of the curved incline. Since its radius is 3.0 m, this is also the block's starting height. Find the block's potential energy <em>PE</em> :
<em>PE</em> = <em>m g h</em>
<em>PE</em> = (2.0 kg) (9.8 m/s²) (3.0 m)
<em>PE</em> = 58.8 J
Energy is conserved throughout the block's descent, so that <em>PE</em> at the top of the curve is equal to kinetic energy <em>KE</em> at the bottom. Solve for the velocity <em>v</em> :
<em>PE</em> = <em>KE</em>
58.8 J = 1/2 <em>m v</em> ²
117.6 J = (2.0 kg) <em>v</em> ²
<em>v</em> = √((117.6 J) / (2.0 kg))
<em>v</em> ≈ 7.668 m/s ≈ 7.7 m/s
Answer: The fundamental frequency of the slinky = 8Hz
An input frequency of 28 Hz will not create a standing wave
Explanation:
Let Fo = fundamental frequency
At third harmonic,
F = 3Fo
If F = 24Hz
24 = 3Fo
Fo = 24/3 = 8Hz
If an input frequency = 28 Hz at 3rd harmonic
Let find the fundamental frequency
28 = 3Fo
Fo = 28/3
Fo = 9.33333Hz
Since Fo isn't a whole number, it can't create a standing wave
Answer:
1. 
2. 
3. 
Explanation:
Given:
- mass of slinky,
- length of slinky,
- amplitude of wave pulse,
- time taken by the wave pulse to travel down the length,
- frequency of wave pulse,
1.
2.
<em>Now, we find the linear mass density of the slinky.</em>
We have the relation involving the tension force as:
3.
We have the relation for wavelength as:
Answer:
a = 3 m/s^2
Explanation:
Vi = 10 m/s
Vf = 40 m/s
t = 10 s
Plug those values into the following equation:
Vf = Vi + at
40 = 10 + 10a
---> a = 3 m/s^2
