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Kobotan [32]
3 years ago
7

A wave pulse traveling to the right along a thin cord reaches a discontinuity where the rope becomes thicker and heavier. What i

s the orientation of the reflected and transmitted pulses? A wave pulse traveling to the right along a thin cord reaches a discontinuity where the rope becomes thicker and heavier. What is the orientation of the reflected and transmitted pulses? The reflected pulse is right-side up, and the transmitted pulse is inverted. The reflected pulse is inverted, and the transmitted pulse is right-side up. Both the reflected and transmitted pulses are inverted. Both the reflected and transmitted pulses are right-side up.
Physics
2 answers:
Talja [164]3 years ago
8 0

Answer:

the reflected wave is inverted and the transmitted wave is up

Explanation:

To answer this question we must analyze the physical phenomenon, with an wave reaching a discontinuity, we can analyze it as a shock.

Let's start when the discontinuity is with a fixed, very heavy and rigid obstacle, in this case the reflected wave is inverted, since the contact point cannot move

In the event that it collides with an object that can move, the reflected wave is not inverted, this is because the point can rise, they form a maximum at this point.

In the proposed case the shock is when the thickness changes, in this case we have the above phenomena, a part of the wave is reflected by being inverted and a part of the wave is transmitted without inverting.

The amplitude sum of the amplitudes of the two waves is proportional to the lanería that is distributed between them.

When checking the answers the correct one is the reflected wave is inverted and the transmitted wave is up

balandron [24]3 years ago
5 0

Answer:

The answer is: The reflected pulse is inverted, and the transmitted pulse is right-side up.

Explanation:

A tight end would cause reverse reflection, while an open end would cause right side reflection. Using a heavier rope would be approximately equal to this condition (tight end). It can be said that in a heavy rope, the reflection would be reversed, and the transmitted pulse is up, due to the conservation of vertical momentum.

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What is the vector product of two vectors????
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Answer:

The magnitude of the vector product of two vectors can be constructed by taking the product of the magnitudes of the vectors times the sine of the angle (<180 degrees) between them. The magnitude of the vector product can be expressed in the form: and the direction is given by the right-hand rule.

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Which statement describes a characteristic of a question that can be answered through scientific inquiry? (A.It requires the app
Mrrafil [7]

Answer:

B. It can be answered using measurements.

Explanation:

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5 0
3 years ago
The forces acting on a sailboat are 390 n north and 180 n east. if the boat (including crew) has a mass of 270 kg, what are the
antoniya [11.8K]
(a) magnitude of acceleration:
we will imagine that the two components of the force are the sides of a right-angled triangle.
We can thus use the Pythagorean theorem to find the magnitude of the force which is the hypotenuses of this triangle
Magnitude of force = sqrt(390^2 + 180^2) = 429.53 Newton,
Using Newton's second law (F=ma), we can find the acceleration as follows:
429.53 = 270 * a 
a = 1.59 m/sec^2

(b) direction of acceleration:
we will again consider the right angled triangle formed from the components of the force.
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4 0
2 years ago
550 nm light passes through a diffraction grating with 3000 lines per centimeter. The screen is 115 cm away from the grating. Wh
Sloan [31]

Answer:

69.7 cm

Explanation:

What is the distance (in cm) between the 2nd order bright fringe and the 2nd dark fringe?

For a diffraction grating, dsinθ = mλ where d = grating spacing = 1/(3000 lines per cm) = 1/3000 × 100 m per line = 1/300000 m = 1/3 × 10⁻⁵ m, m = order of fringe and λ = wavelength of light = 550 nm = 550 × 10⁻⁹ m.

Also, tanθ = x/D where x = distance of nth order fringe from central maximum and D = distance of screen from grating = 115 cm = 1.15 m

Now sinθ = d/mλ, Since θ is small, sinθ ≅ tanθ

So, d/mλ = x/D for a second order bright fringe, m = 2.

So, d/2λ = x/D

x = dD/2λ

So, x =

For a dark fringe, we have

d/(m + 1/2)λ = x'/D where x' is the distance of the fringe from the central maximum.

For a second-order dark fringe, m = 2. So,

d/(2 + 1/2)λ = x'/D

d/(5/2)λ = x'/D

2d/5λ = x'/D

x' = 2dD/5λ

So, the distance (in cm) between the 2nd order bright fringe and the 2nd dark fringe is x" = dD/2λ - 2dD/5λ

x" = dD/10λ

Substituting the values of the variables into the equation, we have

x"= 1/3 × 10⁻⁵ m × 1.15 m/(10 × 550 × 10⁻⁹ m)

x" = 1.15/165 × 10² m

x" = 0.00697 × 10² m

x" = 0.697 m

x" = 69.7 cm

6 0
2 years ago
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