Besides the force due to gravity or g, what else determines the period of a pendulum?

1 answer:

7 0

We have the weight acting downwards (W=mg) from the bob.
When the pendulum swings, there is air resistance acting on the bob.
The time taken to complete one oscillation.
The length of the string affects the time taken.
The tension of the string.

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b. $6.816\times10^5N/C$

Explanation:

#Apply surface charge density, electric field, and Gauss law to solve:

a. Surface charge density is defined as charge per area denoted as $\sigma$

$\sigma=\frac{Q}{4\pi r_{out}^2}$ , and the strength of the electric field outside the sphere $E=\frac{\sigma _{new}}{\epsilon _o}$

Using Gauss Law, total electric flux out of a closed surface is equal to the total charge enclosed divided by the permittivity.

$\phi=\frac{Q_{enclosed}}{\epsilon_o}\\\\\sigma=\frac{Q}{4\pi r_{out}^2}\\\\\sigma=\frac{0.370\times 10^{-6}}{4\pi \times (0.295m)^2}\\\\=3.383\times10^{-7}C/m^2$ #surface charge outside sphere.

$\sigma_{new}=\sigma_{s}-\sigma\\\\\sigma_{new}=6.37\times10^{-6}C/m^2-3.383\times10^{-7}C/m^2\\\\\sigma_{new}=6.032\times10^{-6}C/m^2$

Hence, the new charge density on the outside of the sphere is $6.032\times10^{-6}C/m^2$

b. The strength of the electric field just outside the sphere is calculated as:

From a above, we know the new surface charge to be $6.032\times10^{-6}C/m^2$ ,

$E=\frac{\sigma _{new}}{\epsilon _o}\\\\=\frac{6.032\times10^{-6}C/m^2}{\epsilon _o}\\\\\epsilon _o=8.85\times10^{-12}C^2/N.m^2\\\\E=\frac{6.032\times10^{-6}C/m^2}{8.85\times10^{-12}C^2/N.m^2}\\\\E=6.816\times10^5N/C$