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A pendulum has a 0.35\ \text{kg}0.35 kg0, point, 35, space, start text, k, g, end text mass oscillating at a small angle from a

expeople1 [14]

Answer:

The frequency of oscillation of the simple pendulum is 0.49 Hz.

Explanation:

Given that,

Mass of the simple pendulum, m = 0.35 kg

Length of the string to which it is attached, l = 1 m

We need to find the frequency of oscillation. The frequency of oscillation of the simple pendulum is given by :

$f=\dfrac{1}{2\pi}\sqrt{\dfrac{g}{l}} \\\\f=\dfrac{1}{2\pi}\sqrt{\dfrac{9.8}{1}} \\\\f=0.49\ Hz$

So, the frequency of oscillation of the simple pendulum is 0.49 Hz. Hence, this is the required solution.

4 0

3 years ago

A charge of 7.00 mC is placed at opposite corners corner of a square 0.100 m on a side and a charge of -7.00 mC is placed at oth

andrew-mc [135]

Answer:

4.03\times10^{7}N[/tex], 135°

Explanation:

charge, q = 7 mC = 0.007 C

charge, - q = - 7 mC = - 0.007 C

d = 0.1 m

Let the force on charge placed at C due to charge placed at D is FD.

$F_{D}=\frac{kq^{2}}{DC^{2}}$

$F_{D}=\frac{9 \times10^{9}\times 0.007 \times 0.007}{0.1^{2}}=4.41 \times 10^{7}N$

The direction of FD is along C to D.

Let the force on charge placed at C due to charge placed at B is FB.

$F_{B}=\frac{kq^{2}}{BC^{2}}$

$F_{B}=\frac{9 \times10^{9}\times 0.007 \times 0.007}{0.1^{2}}=4.41 \times 10^{7}N$

The direction of FB is along C to B.

Let the force on charge placed at C due to charge placed at A is FA.

$F_{A}=\frac{kq^{2}}{AC^{2}}$

$F_{D}=\frac{9 \times10^{9}\times 0.007 \times 0.007}{0.1 \times\sqrt{2} \times 0.1 \times\sqrt{2}}=2.205 \times 10^{7}N$

The direction of FA is along A to C.

The net force along +X axis

$F_{x}=F_{A}Cos45-F_{D}$

$F_{x}=2.205\times10^{7}Cos45-4.41\times10^{7}=-2.85\times10^{7}N$

The net force along +Y axis

$F_{y}=F_{B}-F_{A}Sin45$

$F_{y}=4.41\times10^{7}-2.205\times10^{7}Sin45=2.85\times10^{7}N$

The resultant force is given by

$F=\sqrt{F_{x}^{2}+F_{y}^{2}}=\sqrt{(-2.85\times10^{7})^{2}+(2.85\times10^{7})^{2}}$

$F = 4.03\times10^{7}N$

The angle from x axis is Ф

tan Ф = - 1

Ф = -45°

Angle from + X axis is 180° - 45° = 135°

5 0

3 years ago

Which of Newton's laws of motion describes the motion of an object that has a net<br> force of ON?

algol [13]

Newton's first and second laws of motion both do, but I think the one you're looking for is: <em>The First Law of Motion</em>. That description is a little more direct.

It says that if an object is not acted on by a net external force, then it continues in "constant, uniform motion".

3 0

2 years ago

Ball 1, with a mass of 100 g and traveling at 10 m/s, collides head-on with ball 2, which has a mass of 300 g and is initially a

qaws [65]

There is 4000 balls in the earth of the world why is that so hard for you

5 0

2 years ago

An electroscope is a simple device consisting of a metal ball that is attached by a conductor to two thin leaves of metal foil p

baherus [9]

Answer:

the electroscope separate by the presence of charge carriers

Explanation:

Metal bodies are characterized by having free (mobile) electrons. In the electroscope the plates are in balance; when the external metal ball is touched, a charge is introduced into the device, when the body that touched the ball is separated, an excess charge remains. This charge, being a metal, is distributed over the entire surface, giving a uniform density and an electric force of repulsion is created between the two charged sheets, which tends to separate the sheets. This force is counteracted by the tension component as the sheets are separated at a given angle, the separation reaches the point where

Fe - Tx = 0

Fe = Tx

In summary, the electroscope separate its leaves by the presence of charge carriers

3 0

3 years ago