Thinner at edges and its thick in the middle
Answer:
The second law of a vibrating string states that for a transverse vibration in a stretched string, the frequency is directly proportional to the square root of the string's tension, when the vibrating string's mass per unit length and the vibrating length are kept constant
The law can be expressed mathematically as follows;
The second law of the vibrating string can be verified directly, however, the third law of the vibrating string states that frequency is inversely proportional to the square root of the mass per unit length cannot be directly verified due to the lack of continuous variation in both the frequency, 'f', and the mass, 'm', simultaneously
Therefore, the law is verified indirectly, by rearranging the above equation as follows;
From which it can be shown that the following relation holds with the limits of error in the experiment
m₁·l₁² = m₂·l₂² = m₃·l₃² = m₄·l₄² = m₅·l₅²
Explanation:
Answer:
energy decreases
Explanation:
The energy of a wave is directly proportional to the square of amplitude of a wave.
Amplitude is defined as the maximum displacement of the wave particle from its mean position.
So, as the amplitude goes on decreasing, the energy of the wave is also decreasing.
Answer:
θ = 28°
Explanation:
For this exercise We will use the second law and Newton, let's set a System of horizontal and vertical.
X axis
Fₓ = m a
Nₓ = m a
Where the acceleration is centripetal
a = v² / r
The only force that we must decompose is normal, let's use trigonometry
sin θ = Nₓ / N
cos θ = / N
Nₓ = N sin θ
= N cos θ
Let's replace
N sin θ = m v² / r
Y Axis
- W = 0
N cos θ = mg
Let's divide the two equations of Newton's second law
Sin θ / cos θ = v² / g r
tan θ = v² / g r
θ = tan⁻¹ (v² / g r)
We reduce the speed to the SI system
v = 61 km / h (1000 m / 1 km) (1h / 3600 s) = 16.94 m / s
Let's calculate
θ = tan⁻¹ (16.94 2 / (9.8 55.1)
θ = tan⁻¹ (0.5317)
θ = 28°