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3 years ago

What is the frequency of a pendulum that is moving at 30 m/s with a wavelength of 0.35

Physics

2 answers:

Kitty [74]3 years ago

5 0

V=f*wavelength
30m/s=f*(0.35m)
f=85.7 Hz

Mashcka [7]3 years ago

4 0

Answer:

85.7Hz

Explanation:

We know that the wave velocity is given by:

v=f*λ

where:

f is the frequency

λ is the wavelength

rearranging the equation we have:

f=v/λ

$f=\frac{30m/s}{0.35m}\\f=85.7s^{-1}=85.7Hz$

so the pendulum has a frequency of 85.7Hz

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The component of the force due to gravity perpendicular and parallel to the slope is 113.4 N and 277.8 N respectively.

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Taking into account the simple trigonometric relations, such as sine, cosine and tangent, the value of their components and the value of the angle of application, then the parallel and perpendicular components will be:

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3 years ago

You are holding a positive charge and there are positive charges of equal magnitude 1 mm to your north and 1 mm to your east. Wh

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If I hold a positive charge in my hand and there are positive charges of equal magnitude 1 mm to your north and 1 mm to your east then the direction of the force on the charge I am holding is towards the north-east direction.

Reasoning:

It is given that there is a positive charge in my hand. There are two more positive charges with the same magnitude. One is 1 mm far towards the east, and the other one is 1 mm far towards the north. It is required to find the direction of the force acting on the charge in my hand.

Let the magnitude of the charge in my hand is Q, and the magnitude of the other charges is q.

Thus the electric force applied on the charge in my hand due to each other is,

$F=\frac{kQq}{r^2}$

Here k is the Coulomb constant, and r is the distance between the charges.

It is also known that the force on a positive charge due to another positive charge is acted outwards.

Thus, the force on the charge due to the charge on the east is,

$\vec{F_1}=\frac{kQq}{( 10^{-3}\text{ m})^2}\hat{i}$

And the force on the charge due to the charge on the north is,

$\vec{F_2}=\frac{kQq}{( 10^{-3}\text{ m})^2}\hat{j}$

As the forces are equal in magnitude and one is perpendicular to the other, thus the net force will be acted at an angle of $45^\circ$ from the north or from the north direction.

Thus the net force is acting in the north-east direction.

Learn more about the direction of the force here,

brainly.com/question/2037071

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