A 2.50 kg ball is attached to a 3.00 m bar and swung in a vertical circle. If the ball does not leave the circular loop, what mi
1 answer:
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To solve this problem it is necessary to apply the concepts related to transformers, that is to say passive electrical device that transfers electrical energy from one electrical circuit to one or more circuits.
From the mathematical definition we have that the relationship between the voltage of the first coil and the second coil is proportional to the number of loops of the first and second loop, that is:
Where
input voltage on the primary coil.
input voltage on the secondary coil.
number of turns of wire on the primary coil.
number of turns of wire on the secondary coil.
Replacing our values we have:
Replacing,
From the same relations of number of turns and the voltage of the first and second coil we also have the relation of electricity and voltage whereby:
Where
= Current Primary Coil
= Current secundary Coil
Therefore:
Therefore the maximum values for the secondary coil of the voltage is 410.56V and Current is 1.87A
Answer:
(a) the fundamental frequency of this string is 65 Hz
(b) the harmonics of the given frequencies are third and fourth respectively.
(c) the length of the string is 2.74 m
Explanation:
Given;
mass density of the string, μ = 3 x 10⁻³ kg/m
tension of the string, T = 380 N
resonating frequencies, 195 Hz and 260 N
For the given resonant frequencies;
(c) From any of the equations, solve for Length of the string (L);
(a) the fundamental frequency is calculated as;
(b) harmonics of the given frequencies;
the first harmonic (n = 1) = f₀ = 65 Hz
the second harmonic (n = 2) = 2f₀ = 130 Hz
the third harmonic (n = 3) = 3f₀ = 195 Hz
the fourth harmonic (n = 4) = 4f₀ = 260 Hz
Thus, the harmonics of the given frequencies are third and fourth respectively.