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3 years ago

Prove that the period of a simple pendulum T=2×22/7×rootl/g

Physics

1 answer:

Rudik [331]3 years ago

4 0

Answer:

Explanation:

Check the attachment.

By the definition of the simple harmonic motion,

The motion of a particle whose acceleration is

1. Always directed towards the center of the motion.

2. Proportional to the distance from center of the motion.

We get, a ∝ -x

where

a = acceleration

x = distance of the particle from center of the motion

- sign says acceleration is in opposite direction of motion

Check the other attachment for proof of a = -ω²x (or a = -ω²y)

x and y are both use to denote distance from center of the motion

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3 years ago

A hollow, uniformly charged sphere has an inner radius of r1 = 0.105 m and an outer radius of r2 = 0.31 m. The sphere has a net

Sliva [168]

Answer:

E = 77532.42N/C

Explanation:

In order to find the magnitude of the electric field for a point that is in between the inner radius and outer radius, you take into account the Gauss' law for the electric flux trough a spherical surface with radius r:

$\int E\cdot dS=\frac{Q}{\epsilon_o}$ (1)

Q: net charge of the hollow sphere = 1.9*10-6C

ε0: dielectric permittivity of vacuum = 8.85*10^-12 C^2/Nm^2

Furthermore, you have that the net charge contained in a sphere of radius r is:

$Q=\rho V=\rho \frac{4\pi (r^3-r_1^3)}{3}$ (2)

with the charge density is:

$\rho=\frac{Q}{\frac{4}{3}\pi(r_2^3-r_1^3)}$ (3)

r2: outer radius = 0.31m

r1: inner radius = 0.105m

The electric field trough the Gaussian surface is parallel to the normal to the surface, the, you have in the integral of the equation (1):

$\int E\cdot dS=E(4\pi r^2)$ (4)

where you have used the expression for a surface of a sphere.

Next, you replace the expressions of equations (2), (3) and (4) in the equation (1) and solve for E:

$E(4\pi r^2)=\frac{1}{\epsilon_o}\frac{Q}{\frac{4}{3}\pi(r_2^3-r_1^3)}(\frac{4\pi (r^3-r_1^3)}{3})\\\\E=\frac{1}{\epsilon_o}\frac{Q(r^3-r_1^3)}{4\pi r^2(r_2^3-r_1^3)}$

you replace the values of all parameters, and with r = 0.17m

$E=\frac{(1.6*10^{-6}C)((0.17m)^3-(0.105m)^3)}{4\pi(8.85*10^{-12}C^2/Nm^2)(0.17m)^2((0.31m)^3-(0.105m)^3)}\\\\E=77532.42\frac{N}{C}$

The magnitude of the electric field at a distance r=0.17m to the center of the hollow sphere is 77532.42N/C

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3 years ago

ArbitrLikvidat [17]

Answer:

D) Number of electrons.

Explanation:

Elements in the same group on the periodic table has the same number of electrons. These electrons are those involved in bonding with other atoms to form compounds.

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3 years ago

Prove that the period of a simple pendulum T=2×22/7×rootl/g​

Prove that the period of a simple pendulum T=2×22/7×rootl/g