Answer:
the frequency of the second harmonic of the pipe is 425 Hz
Explanation:
Given;
length of the open pipe, L = 0.8 m
velocity of sound, v = 340 m/s
The wavelength of the second harmonic is calculated as follows;
L = A ---> N + N--->N + N--->A
where;
L is the length of the pipe in the second harmonic
A represents antinode of the wave
N represents the node of the wave
The frequency is calculated as follows;
Therefore, the frequency of the second harmonic of the pipe is 425 Hz.
Answer:
V=15.3 m/s
Explanation:
To solve this problem, we have to use the energy conservation theorem:
the elastic potencial energy is given by:
The work is defined as:
this work is negative because is opposite to the movement.
The gravitational potencial energy at 2.5 m aboves is given by:
the gravitational potential energy at the ground and the kinetic energy at the begining are 0.
Answer:
Explanation:
Given data
Force F=2 N
Length L=17 cm = 0.17 m
Spring Constant k=42 N/m
To find
Relaxed length of the spring
Solution
From Hooke's Law we know that
Answer:
5 is the tripoid stand
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