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2 years ago

In a coiled spring, the particles of the medium vibrate to and fro about their

Physics

1 answer:

Kitty [74]2 years ago

8 0

Answer:

In a coiled spring, the particles of the medium vibrate to and fro about their mean positions at an angle of

A. 0° to the direction of propagation of wave

Explanation:

The waveform of a coiled spring is a longitudinal wave, which is made up of vibrations of the spring which are in the same direction as the direction of the wave's advancement

As the coiled spring experiences a compression force and is then released, it experiences a sequential movement of the wave of the compression that extends the length of the coiled spring which is then followed by a stretched section of the coiled spring in a repeatedly such that the direction of vibration of particles of the coiled is parallel to direction of motion of the wave

From which we have that the angle between the direction of vibration of the particles of the coiled spring and the direction of propagation of the wave is 0°.

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2 years ago

Mark creates a graphic organizer to review his notes about electrical force. Which labels belong in the regions marked X and Y?

sveta [45]

Answer:

The correct answer is A

Explanation:

The question requires as well the attached image, so please see that below.

Coulomb's Law.

The electrical force can be understood by remembering Coulomb's Law, that describes the electrostatic force between two charged particles. If the particles have charges $q_1$ and $q_2$ , are separated by a distance r and are at rest relative to each other, then its electrostatic force magnitude on particle 1 due particle 2 is given by:

$|F|=k \cfrac{q_1 q_2}{r^2}$

Thus if we decrease the distance by half we have

$r_1 =\cfrac r2$

So we get

$|F|=k \cfrac{q_1 q_2}{r_1^2}$

Replacing we get

$|F|=k \cfrac{q_1 q_2}{(r/2)^2}\\|F|=k \cfrac{q_1 q_2}{r^2/4}$

We can then multiply both numerator and denominator by 4 to get

$|F|=k \cfrac{4q_1 q_2}{r^2}$

So we have

$|F|=4 \left(k \cfrac{q_1 q_2}{r^2}\right)$

Thus if we decrease the distance by half we get four times the force.

Then we can replace the second condition

$q_{2new} =2q_2$

So we get

$|F|=k \cfrac{q_1 q_{2new}}{r_1^2}$

which give us

$|F|=k \cfrac{q_1 2q_2}{r_1^2}\\|F|=2\left(k \cfrac{q_1 q_2}{r_1^2}\right)$

Thus doubling one of the charges doubles the force.

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3 years ago

Read 2 more answers

Which accounts for an increase in the temperature of a gas that is kept a constant volume

oksian1 [2.3K]

Answer:

An increase in pressure

Explanation:

The ideal gas law states that:

$pV=nRT$

where

p is the gas pressure

V is the volume

n is the number of moles

R is the gas constant

T is the temperature of the gas

in the equation, n and R are constant. For a gas kept at constant volume, V is constant as well. Therefore, from the formula we see that if the temperature (T) is increase, the pressure (p) must increase as well.

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2 years ago

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