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soldier1979 [14.2K]
3 years ago
5

A thin hoop is supported in a vertical plane by a nail. What should the radius of the hoop be in order for it to have a period o

f oscillation of 1.00 s
Physics
1 answer:
maxonik [38]3 years ago
4 0

Answer:

0.124 m

Explanation:

the period of a simple pendulum with a small amplitude is given as

T = 2π [√(I/mgd)]

From the above stated formula,

I = moment of inertia

m = mass of the pendulum

g = acceleration due to gravity, 9.8 m/s²

d = distance from rotation axis due to center of gravity

Also, moment of Inertia, I = 2mr², if we substitute this in the above formula, we have

T = 2π [√(2mr²/mgd)]

If we assume that our r = d, then we have

T = 2π [√(2r/g)]

If we make r the subject of the formula in the above equation, we get

r = gT² / 8π²

r = (9.8 * 1²) / 8 * π²

r = 9.8 / 78.98

r = 0.124 m

Thus, the radius of the hoop is 0.124 m

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Calculate the electric field at one corner of a square 50 cm on a side if the other corners are occupied by 250x10-7C (charges)
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The electric field at one corner of a square is 1614217 N/C.

Explanation:

The distance between x and y direction diagonals.

As per the given details the distance between diagonals is calculated as

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Similarly as shown in x direction the charge is same for y direction also

Charge below:  For y direction

e = kq/r²

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e = 9 x 10^{5} N/C

Charge diagonal:

e = kq/r²

e = [(9 x 10^{9})(250 x 10^{-7}) / (0.5)²] sin 45

e = 159099 N/C

Y direction sum = 1059099 N/C

Resultant electric field strength:

1218198 ² + 1059099² = e²

e = 1614217 N/C [45 degrees below the horizontal]

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