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iogann1982 [59]
3 years ago
13

Problem 1: A positively charged particle Q1 = +35 nC is held fixed at the origin. A second charge Q2 of mass m = 8.5 micrograms

is floating a distance d = 25 cm above charge Q1. The net force on Q2 is equal to zero. You may assume this system is close to the surface of the Earth. A) What is the sign of charge on Q2? B) Calculate the magnitude of Q2 in units of nanocoulombs.
Physics
1 answer:
tigry1 [53]3 years ago
3 0

Answer:This problem has been solved! See the answer. Two charged particles, Q 1 = + 5.10 nC and Q 2 = -3.40 nC, are separated by 40.0 cm. (a) What is the ... has not been graded yet. (b) What is the electric potential at a point midway between the charged particles? (Note: Assume a reference level of potential V = 0 at r = infinity .)

Explanation:

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3 0
1 year ago
A piano tuner strikes a tuning fork at the same time he strikes a piano key with a note of a similar pitch. If he hears 3 beats
DaniilM [7]
The frequency produced by the string could be 437 Hz or it could be 443 Hz.

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7 0
2 years ago
3. Una cuerda de guitarra tiene 60 cm de longitud y una masa de 0.05 kg de masa. Si se tensiona mediante una fuerza de 20 N. La
jok3333 [9.3K]

Answer:

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Explanation:

To calculate the first harmonic frequency you use the following formula for n = 1:

f_n=\frac{n}{2L}\sqrt{\frac{T}{M/L}}

f_1=\frac{1}{2L}\sqrt{\frac{T}{M/L}}    ( 1 )

It is necessary that the unist are in meters, then you have:

L: length of the string = 60cm = 0.6m

M: mass of the string = 0.05kg

T: tension on the string = 20 N

you replace the values of L, M and T in the expression (1) for getting f1:

f_1=\frac{1}{2(0.6m)}\sqrt{\frac{20N}{0.05kg/0.6m}}=12.90\ Hz

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