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kotegsom [21]
3 years ago
12

Venn diagrams are used for comparing and contrasting topics. The overlapping sections show characteristics that the topics have

in common, and the sections that are unique to each topic show the characteristics that apply to only that topic. This Venn diagram is comparing longitudinal waves and transverse waves.
“Parallel” and “Perpendicular” are switched.
The slinky descriptions of motion are switched.
Particles travel along the wave in longitudinal waves.
Particles move long distances in transverse waves.

Physics
2 answers:
Kisachek [45]3 years ago
8 0

Answer: The correct answer is "the slinky descriptions of motion are switched".

Explanation:

Longitudinal wave: The particles of the medium move parallel to the direction of the motion of the wave in which the energy from one particle to the another is transported. For example, sound wave. Here, the motion of the wave is like back and forth.

Transverse wave: The particles of the medium move perpendicular to the direction of the motion of the wave in which the energy is transported from one particle to the another.  For example, light wave. Here, the motion of the wave is like up and down.

Therefore, there is an error in the given Venn diagram. The slinky descriptions of motion are switched.

Zanzabum3 years ago
7 0
The second option is the best.
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If an object on a horizontal frictionless surface is attached to a spring, displaced, and then released, it will oscillate. If i
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Explanation:

An object is attached to the spring and then released. It begins to oscillate. If it is displaced a distance 0.125 m from its equilibrium position and released with zero initial speed. The amplitude of a wave is the maximum displacement of the particle. So, its amplitude is 0.125 m.  

After 0.800 seconds, its displacement is found to be a distance 0.125 m on the opposite side. The time period will be, t=2\times 0.8=1.6\ s

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4 years ago
What is the magnitude of the force required to stretch a 20 cm-long spring, with a spring constant of 100 N/m, to a length of 21
GalinKa [24]

The correct answer to the question is: 1 N.

EXPLANATION:

As per the question, the spring constant or the force constant of the spring is given as k = 100 N/m.

The original length of the spring L = 20 cm.

The stretched length of the spring L'= 21 cm.

Hence, the change in length will be-

                              ∆L = L' - L

                                    = 21 cm - 20 cm

                                    = 1 cm

                                    = 0.01 m

We are asked to calculate the magnitude of force acting on the spring .

From Hooke's  law, we know that the restoring force that acts on the spring is proportional to the distance .

Mathematically it can be written as -

                F = - kx.

Here, k is the force constant.

         x is the change in length due to compression or elongation.

The negative sign is due to the fact that it is opposite to the applied force.


Hence, the applied force on the spring is calculated as -

            F = kx

               = k × ∆L

               = 100 N/ m × 0.01 m

               = 1 N.

Hence, the force acting on the spring is 1 N.


                                                   

8 0
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