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:
It's a type of chemical bonding that rises from the electrostatic attractive force between conduction electrons and positively charged metal bars. It can also be described as the sharing of free electrons among a structure of positively charged ions
Answer:
The electric potential is approximately 5.8 V
The resulting direction of the electric field will lie on the line that joins the charges but since it is calculated in the midpoint and the charges are the same we can directly say that its magnitude is zero
Explanation:
The two protons can be considered as point charges. Therefore, the electric potential is given by the point charge potential:
(1)
where is the charge of the particle, the electric permittivity of the vacuum (I assuming the two protons are in a vacuum) and is the distance from the point charge to the point where the potential is being measured. Because the electric potential is an scalar, we can simply add the contribution of the two potentials in the midpoint between the protons. Thus:
Substituting the values , and we obtain:
The resulting direction of the electric field will lie on the line that joins the charges but since it is calculated in the midpoint and the charges are the same we can directly say that its magnitude is zero.
Answer:
0.79 s
Explanation:
We have to calculate the employee acceleration, in order to know the minimum time. According to Newton's second law:
The frictional force is maximum since the employee has to apply a maximum force to spend the minimum time. In y axis the employee's acceleration is zero, so the net force is zero. Recall that
Now, we find the acceleration:
Finally, using an uniformly accelerated motion formula, we can calculate the minimum time. The employee starts at rest, thus his initial speed is zero: