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Andrew [12]
4 years ago
4

Does the resulting wave demonstrate destructive interference? Explain your answer.

Physics
2 answers:
chubhunter [2.5K]4 years ago
5 0
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.<span>No wave formation is represented by a horizontal line.

From Ingenuity  Wave interactions. </span>
Tcecarenko [31]4 years ago
4 0

Destructive interference happens when one wave is oscillating the opposite way as the other one, so it compensates.

For example:

Let’s say one wave can be described with this formula:

f(x)=sinx

And the other one, with this formula:

g(x)=-sinx

So when you add them, you get 0 for any x value:

h(s)=f(x)+g(x)=sinx-sinx=0

(I recommend to put them in a graph software such as GeoGebra, to see how they look like).

In other cases, the sum of both can be bigger than each one of them. That is called constructive interference. For example:

Wave 1: f(x)=sinx

Wave 2: g(x)=2sinx

Total: h(x)=f(x)+g(x)=sinx+2sinx=3sinx

which is bigger than sinx, and also higher to 2sinx, for any x value.

If you have the graph instead of the formula, take 2 points from the graph which have the same x value, one in the first wave and the other one in the second wave, and add them (y1+y2). If both numbers are the same but one is negative and the other one is positive, the sum will be 0, and that is a destructive interference.

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A jogger moves from a position x =
vlada-n [284]

Answer:

3 m/s

Explanation:

We'll begin by calculating the change in displacement of the jogger. This can be obtained as follow:

Initial displacement (d₁) = 4 m

Final displacement (d₂) = 16 m

Change in displacement (Δd) =?

Δd = d₂ – d₁

Δd = 16 – 4

Δd = 12 m

Finally, we shall determine the determine the average velocity. This can be obtained as follow:

Change in displacement (Δd) = 12 m

Time (t) = 4 s

Velocity (v) =?

v = Δd / t

v = 12 / 4

v = 3 m/s

Thus, the average velocity of the jogger is 3 m/s

5 0
3 years ago
This picture clicked by me<br>How's it​
galben [10]

Answer:

Hey dear

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8 0
3 years ago
Read 2 more answers
a body of mass 5kg falls from height of 10m above the ground what kinetic energy of the body before it strike the ground
Snowcat [4.5K]
Gravitational Potential Energy (GPE) before fall = Kinetic energy on impact
GPE = mgh
GPE = 5kg x 9.8m/s^2 x 10m
GPE = 490 J
Kinetic Energy = 490 J

(This is assuming that gravitation field strength (g) is 9.8m/s^2, sometimes       you may round this to 10m/s^2, therefore making the final result: 
  Kinetic energy = 500 J)
4 0
4 years ago
Cody hits up food king and uses a scale to weigh the mass of an apple. if the spring potential energy in the scale is .09 j and
Lostsunrise [7]

Answer:

oK so  here's  <u>what you should do is  add .09 and 0.6</u>

Explanation:

7 0
3 years ago
Introduction: The specific heat capacity of a substance is the amount of energy needed to change the temperature of that substan
ki77a [65]

Answer:

A) 8,368 J

B) ) 0.893 J/gºC

Explanation:

A)

  • The heat gained by the water can be obtained solving the following equation:

       q_{g} = c_{w} * m *  \Delta T (1)

  • where cw = specific heat of water = 4.184 J/gºC
  • m= mass of water = 1,000 g
  • ΔT = 2ºC
  • Replacing these values in (1) we get:

       q_{g} = c_{w} * m *  \Delta T = 4.184 J/gºC*1,000 g* 2ºC = 8,368 J (2)

B)

  • Assuming that the heat energy gained by the water is equal to the one lost by the aluminum, we can use the same equation, taking into account that the energy is lost by the aluminum, so the sign is negative:  -8,368 J.
  • Replacing by the mass of aluminum (125 g), and the change in temperature (-74.95ºC), in (1), we can solve for the specific heat of aluminum, as follows:

       q_{l} = c_{Al} * m_{Al} *  \Delta T  (3)

⇒    -8,368 J = c_{Al}* 125 g * (-74.95ºC) (4)

       c_{Al} = \frac{-8,368J}{125g*(-74.95ºC} = 0.893 J/gºC (5)

  • which is pretty close to the Aluminum's accepted specific heat value of 0.900 J/gºC.

8 0
3 years ago
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