A waitperson carrying a tray with a platter on it tips the tray at an angle of 12 degrees below the horizontal. If the gravitati
onal force on the platter is 5.0 N, what is the magnitude of the force parallel to the tray that tends to cause the platter to slide down the tray? (Disregard friction.)
1 answer:
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Answer:
91.87 m/s
Explanation:
<u>Given:</u>
- x = initial distance of the electron from the proton = 6 cm = 0.06 m
- y = initial distance of the electron from the proton = 3 cm = 0.03 m
- u = initial velocity of the electron = 0 m/s
<u>Assume:</u>
- m = mass of an electron =
- v = final velocity of the electron
- e = magnitude of charge on an electron =
- p = magnitude of charge on a proton =
We know that only only electric field due to proton causes to move from a distance of 6 cm from proton to 3 cm distance from it. This means the electric force force does work on the electron to move it from one initial position to the final position which is equal to the change in potential energy of the electron due to proton.
Now, according to the work-energy theorem, the total work done by the electric force on the electron due to proton is equal to the kinetic energy change in it.
Hence, when the electron is at a distance of c cm from the proton, it moves with a velocity of 91.87 m/s.
In a string of length L, the wavelength of the n-th harmonic of the standing wave produced in the string is given by:
The length of the string in this problem is L=3.5 m, therefore the wavelength of the 1st harmonic of the standing wave is:
The wavelength of the 2nd harmonic is:
The wavelength of the 4th harmonic is:
It is not possible to find any integer n such that , therefore the correct options are A, B and D.