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Arada [10]
2 years ago
9

2 boxes connected by a plus sign hold Wave 1 on top and Wave 2 on bottom. The crests of Wave 1 line up with the troughs of Wave

2. 2 curved lines connect these boxes with a third box showing the 2 waves superimposed. An arrow leads from this box to another box labeled Resulting Wave with a wave having twice the amplitude of Waves 1 or 2.
Does the resulting wave demonstrate destructive interference? Explain your answer.
Physics
2 answers:
loris [4]2 years ago
6 0

Answer:

No, the resulting wave in the diagram does not demonstrate destructive interference. The resulting wave in the diagram shows a bigger wave than Wave 1 or Wave 2. If it demonstrated destructive interference, it would be a smaller wave or a horizontal line. With destructive interference, waves break down to form a smaller wave, or cancel each other out, resulting in no wave formation.

Zanzabum2 years ago
3 0

Answer:

Sample Response: No, the resulting wave in the diagram does not demonstrate destructive interference. The resulting wave in the diagram shows a bigger wave than Wave 1 or Wave 2. If it demonstrated destructive interference, it would be a smaller wave or a horizontal line. With destructive interference, waves break down to form a smaller wave, or cancel each other out, resulting in no wave formation.

Explanation:

that's the sample answer on edge. here are the questions they ask you if you want to make your own

What did you include in your response? Check all that apply.

The diagram of the resulting wave does not demonstrate destructive interference.

With destructive interference, waves break each other down to form a smaller wave, or cancel each other out, resulting in no wave formation.

No wave formation is represented by a horizontal line.

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An aluminum "12 gauge" wire has a diameter d of 0.205 centimeters. The resistivity ρ of aluminum is 2.75×10−8 ohm-meters. The el
Tresset [83]

Complete Question

An aluminum "12 gauge" wire has a diameter d of 0.205 centimeters. The resistivity ρ of aluminum is 2.75×10−8 ohm-meters. The electric field in the wire changes with time as E(t)=0.0004t2−0.0001t+0.0004 newtons per coulomb, where time is measured in seconds.

I = 1.2 A at time 5 secs.

Find the charge Q passing through a cross-section of the conductor between time 0 seconds and time 5 seconds.

Answer:

The charge is  Q =2.094 C

Explanation:

From the question we are told that

    The diameter of the wire is  d =  0.205cm = 0.00205 \ m

     The radius of  the wire is  r =  \frac{0.00205}{2} = 0.001025  \ m

     The resistivity of aluminum is 2.75*10^{-8} \ ohm-meters.

       The electric field change is mathematically defied as

         E (t) =  0.0004t^2 - 0.0001 +0.0004

     

Generally the charge is  mathematically represented as

       Q = \int\limits^{t}_{0} {\frac{A}{\rho} E(t) } \, dt

Where A is the area which is mathematically represented as

       A =  \pi r^2 =  (3.142 * (0.001025^2)) = 3.30*10^{-6} \ m^2

 So

       \frac{A}{\rho} =  \frac{3.3 *10^{-6}}{2.75 *10^{-8}} =  120.03 \ m / \Omega

Therefore

      Q = 120 \int\limits^{t}_{0} { E(t) } \, dt

substituting values

      Q = 120 \int\limits^{t}_{0} { [ 0.0004t^2 - 0.0001t +0.0004] } \, dt

     Q = 120 [ \frac{0.0004t^3 }{3} - \frac{0.0001 t^2}{2} +0.0004t] }  \left | t} \atop {0}} \right.

From the question we are told that t =  5 sec

           Q = 120 [ \frac{0.0004t^3 }{3} - \frac{0.0001 t^2}{2} +0.0004t] }  \left | 5} \atop {0}} \right.

          Q = 120 [ \frac{0.0004(5)^3 }{3} - \frac{0.0001 (5)^2}{2} +0.0004(5)] }

         Q =2.094 C

     

5 0
2 years ago
The model of the atom has changed as scientists have gathered new evidence. Four models of the atom are shown below, but one imp
nexus9112 [7]

Answer: Dalton’s model

Explanation:

In the attached image we can see four atomic models labeled with four letters:

W represents the current and accepeted atomic model: a nucleus with an electron cloud, where the orbit and position of the electrons around the nucleus is defined by specific regions (associated with specific energy levels) where there is a greater probability of finding the electron at any given moment. It is important to note this model was improved by the works in quantum physics done by Louis de Broglie and Erwin Schrodinger.

X represents Rutherford's model (This model was proposed after Thomson's model). Ernest Rutherford conducted a series of experiments in order to corroborate Thomson's atomic model. However the results of the experiment led him to find out there is a concentration of charge in the atom's core (which was later called nucleus) surrounded by electrons.  This lead to a new atomic model, in which the atom has a positive charged nucleus surrounded by negative charged particles that move similar to the orbit of the planet around the Sun.

Y represents Thomson's model, also called  the <em>plum pudding</em> model. This scientific found out that atoms contain small subatomic particles with a negative charge (later called electrons). However, taking into consideration that at that time there was still no evidence of the atom nucleus, Thomson thought the electrons were immersed in the atom of positive charge that counteracted the negative charge of the electrons. Just like the raisins embedded in a pudding or bread.

Z represents Bohr's model. This model was proposed by the danish physicist Niels Bohr after Rutherford's model. In fact, this model was Rutherford's model with the following addition: electrons orbit the nucleus (like planets around the sun) in specific orbits at different energy levels around the nucleus.

So, the only missing model is <u>Dalton's model</u>, which was the first atomic model:  the atom represented as a solid, indestructible and indivisible mass. An idea that was already accepted by that time since the ancient Greeks.

4 0
2 years ago
Read 2 more answers
an ice skater, standing at rest, uses her hands to push off against a wall. she exerts an average force on the wall of 120 N and
natulia [17]

Answer:

The skater's speed after she stops pushing on the wall is 1.745 m/s.

Explanation:

Given that,

The average force exerted on the wall by an ice skater, F = 120 N

Time, t = 0.8 seconds

Mass of the skater, m = 55 kg

It is mentioned that the initial sped of the skater is 0 as it was at rest. The change in momentum of skater is :

\Delta p=m(v-u)\\\\\Delta p=mv

The change in momentum is equal to the impulse delivered. So,

J=\Delta p=F\times t\\\\mv=F\times t\\\\v=\dfrac{Ft}{m}\\\\v=\dfrac{120\times 0.8}{55}\\\\v=1.745\ m/s

So, the skater's speed after she stops pushing on the wall is 1.745 m/s.                      

4 0
1 year ago
The image shows positions of the earth and the moon in which region would an astronaut feel the lightest
trapecia [35]

Answer:

The moon region

Explanation:

This is because there is little to no gravity on the moon. That is where the astronaut would feel the lightest.

5 0
2 years ago
Read 2 more answers
It requires 49 J of work to stretch an ideal very light spring from a length of 1.4 m to a length of 2.9 m. What is the value of
nadya68 [22]

Answer:

44 N/m

Explanation:

The extension, e, of the spring = 2.9 m - 1.4 m = 1.5 m

The work needed to stretch a spring by <em>e</em> is given by

W = \frac{1}{2} ke^2

where <em>k</em> is spring constant.

k = \dfrac{2W}{e^2}

Using the appropriate values,

k = \dfrac{2\times 49\text{ J}}{1.5^2\text{ m}^2} = 43.55\ldots\text{ N/m} \approx 44\text{ N/m}

3 0
2 years ago
Read 2 more answers
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