PLEASE ANSWER I DONT HAVE THAT MUCH TIME
1 answer:
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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:
a = 0.1 s b. 10 s
Explanation:
Given that,
The frequency in circular motion, f = 10 Hz
(a) Let T is the period of itsrotation. We know that,
T = 1/f
So,
T = 1/10
= 0.1 s
(b) Frequency is number of rotations per unit time. So,
Hence, this is the required solution.