The first three harmonics of the string are 131.8 Hz, 263.6 Hz and 395.4 Hz.
<h3>
Velocity of the wave</h3>
The velocity of the wave is calculated as follows;
v = √T/μ
where;
- T is tension
- μ is mass per unit length = 2 g/m = 0.002 kg/m
v = √(50/0.002)
v = 158.1 m/s
<h3>First harmonic or fundamental frequency of the wave</h3>
f₀ = v/λ
where;
f₀ = v/2L
f₀ = 158.1/(2 x 0.6)
f₀ = 131.8 Hz
<h3>Second harmonic of the wave</h3>
f₁ = 2f₀
f₁ = 2(131.8 Hz)
f₁ = 263.6 Hz
<h3>Third harmonic of the wave</h3>
f₂ = 3f₀
f₂ = 3(131.8 Hz)
f₂ = 395.4 Hz
Thus, the first three harmonics of the string are 131.8 Hz, 263.6 Hz and 395.4 Hz.
Learn more about harmonics here: brainly.com/question/4290297
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<h2>
Answer: 56.718 min</h2>
Explanation:
According to the Third Kepler’s Law of Planetary motion<em> </em><em>“The square of the orbital period of a planet is proportional to the cube of the semi-major axis (size) of its orbit”.
</em>
In other words, this law states a relation between the orbital period 
of a body (moon, planet, satellite) orbiting a greater body in space with the size 
of its orbit.
This Law is originally expressed as follows:
(1)
Where;
is the Gravitational Constant and its value is 
is the mass of Mars
is the semimajor axis of the orbit the spacecraft describes around Mars (assuming it is a <u>circular orbit </u>and a <u>low orbit near the surface </u>as well, the semimajor axis is equal to the radius of the orbit)
If we want to find the period, we have to express equation (1) as written below and substitute all the values:
(2)
(3)
(4)
Finally:
This is the orbital period of a spacecraft in a low orbit near the surface of mars
The first law is about force or push and pull
The image produced is magnified and real.
A large force is required to accelerate the mass of the bicycle and rider. Once the desired constant velocity is reached, a much smaller force is sufficient to overcome the ever-present frictional forces.
