Because there is no record of all things. As we have partially unknow information, it can never be held as a fact.
True a proton carries a positive charge, a neutron carries a neutral charge and an electron carries a negative charge.
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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Power P is the rate at which energy is generated or consumed and hence is measured in units that represent energy E per unit time t. This is:
P = E/t
Solving for t:
t = E/P
t = 6007 J / 500 W
t = 12.014 s
<h2>
t ≅ 12 s</h2>
Answer:
23376 days
Explanation:
The problem can be solved using Kepler's third law of planetary motion which states that the square of the period T of a planet round the sun is directly proportional to the cube of its mean distance R from the sun.
where k is a constant.
From equation (1) we can deduce that the ratio of the square of the period of a planet to the cube of its mean distance from the sun is a constant.
Let the orbital period of the earth be 
and its mean distance of from the sun be 
.
Also let the orbital period of the planet be 
and its mean distance from the sun be 
.
Equation (2) therefore implies the following;
We make the period of the planet the subject of formula as follows;
But recall that from the problem stated, the mean distance of the planet from the sun is 16 times that of the earth, so therefore
Substituting equation (5) into (4), we obtain the following;
cancels out and we are left with the following;
Recall that the orbital period of the earth is about 365.25 days, hence;
