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vladimir1956 [14]
2 years ago
15

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. Where is the error in this Venn diagram?
A.“Parallel” and “Perpendicular” are switched.
B.The slinky descriptions of motion are switched.
C. Particles travel along the wave in longitudinal waves.
D.Particles move long distances in transverse waves.

Physics
2 answers:
UkoKoshka [18]2 years ago
6 0

The correct answer is:

B.The slinky descriptions of motion are switched.

The explanation:

when:

A Venn diagram (also called primary diagram, set diagram or logic diagram) is a diagram that shows all possible logical relations between a finite collection of different sets. These diagrams depict elements as points in the plane, and sets as regions inside closed curves.


natulia [17]2 years ago
4 0

Answer:

B.The slinky descriptions of motion are switched.

Explanation:

First of all, let's define the two types of wave:

- Transverse wave: in a transverse wave, the oscillation of the wave occurs in a direction perpendicular to the direction of propagation of the wave. An example of transverse wave are electromagnetic waves.

- Longitudinal wave: in a longitudinal wave, the oscillation of the wave occurs in a direction parallel to the direction of propagation of the wave. An example of longitudinal wave are sound waves.

According to these definitions, we notice that the following descriptions of motion must be switched:

- "Like moving a spring up and down" --> this better describes a transverse wave, because the motion up/down is perpendicular to the direction of the spring

- "Like moving a spring back and forth" --> this better describes a longitudinal wave, because the back/forth motion is parallel to the direction of the spring

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Answer:

θ=108rad

t =10.29seconds

α=-8.17rad/s²

Explanation:

Given that

At t=0, Wo=24rad/sec

Constant angular acceleration =30rad/s²

At t=2, θ=432rad as it try to stop because the circuit break

Angular motion

W=Wo+αt

θ=Wot+1/2αt²

W²=Wo²+2αθ

We need to find θ between 0sec to 2sec when the wheel stop

a. θ=Wot+1/2αt²

θ=24×2+1/2×30×2²

θ=48+60

θ=108rad.

b. W=Wo+αt

W=24+30×2

W=84rad/s

This is the final angular velocity which is the initial angular velocity when the wheel starts to decelerate.

Wo=84rad/sec

W=0rad/s, because the wheel stop at θ=432rad

Using W²=Wo²+2αθ

0²=84²+2×α×432

-84²=864α

α=-8.17rad/s²

It is negative because it is decelerating

Now, time taken for the wheel to stop

W=Wo+αt

0=84-8.17t

-84=-8.17t

Then t =10.29seconds.

a. θ=108rad

b. t =10.29seconds

c. α=-8.17rad/s²

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Two convex thin lenses with focal lengths 10.0 cm and 20.0 cm are aligned on a common axis, running left to right, the 10-cm len
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Answer:

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\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\\\Rightarrow \frac{1}{f}-\frac{1}{u}=\frac{1}{v}\\\Rightarrow \frac{1}{v}=\frac{1}{20}-\frac{1}{-10}\\\Rightarrow \frac{1}{v}=\frac{3}{20}\\\Rightarrow v=6.67\ cm

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In a semiclassical model of the hydrogen atom, the electron orbits the proton at a distance of 0.053 nm. Part A What is the elec
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Answer with Explanation:

We are given that

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Using the formula

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