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pentagon [3]
4 years ago
7

Is it possible for you to take a hike and have the distance you cover be equal to the magnitude of your displacement?

Physics
1 answer:
kow [346]4 years ago
3 0

Distance can be equal to displacement ... if the motion is all in a straight line ... but distance can never be greater than displacement.

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Wavelength = speed of light/frequency

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6 \times  {10}^{ - 7}

4 0
3 years ago
Constructive interference of two coherent waves will occur if the path difference is:___.
Olegator [25]

Constructive interference of two coherent waves will occur if the path difference is λ/2.

<h3>Constructive interference:</h3>

When two waves are in phase and their maxima add, a process known as constructive interference occurs where the combined amplitude of the two waves equals the sum of their individual amplitudes.

The resultant wave is created by adding the amplitudes of two waves that are in phase and traveling in the same direction. The waves in this instance are said to have experienced beneficial interference. The upward displacement of the medium is higher than the displacement of the two interfering pulses because upward displacement occurs when the waves experience constructive interference. When the phase difference between the waves is an even multiple of (180°), constructive interference happens.

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8 0
2 years ago
The speed of light in a vacuum is 2.998 x 108 m/s. Calculate its speed in miles per hour.
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7 0
3 years ago
What is the wavelength of sound wave in air that has a temperature of 24.6°C that has a frequency of 440.8 Hz?
nignag [31]
The answer is c have a good day
3 0
3 years ago
A 10.3 kg weather rocket generates a thrust of 240 N. The rocket, pointing upward, is clamped to the top of a vertical spring. T
jenyasd209 [6]

Answer:

(a) x = 0.25 m

(b) v = 1.46 m/s

(c) v = 2.4 m/s  

Explanation:

mass (m) = 10.3 kg

force from thrust (F) = 240 N

spring constant (k) = 400 N/m

stretch distance from thrust (y) = 30 cm = 0.3 m

acceleration due to gravity (g) = 9.8 m/s^{2}

(A) from mg = kx

compression (x) = mg/ k

x = \frac{10.3 x 9.8}{400}

x = 0.25 m

   

(B) from the conservation of forces

  (Fy) + (0.5kx^{2}) =  (0.5ky^{2}) + mgh + (0.5mv^{2})

v = \sqrt{[tex]\frac{(Fy) + (0.5k[tex]x^{2}) -  (0.5ky^{2}) - mgh }{0.5m}[/tex]}[/tex]

v =  \sqrt{[tex]\frac{(240 x 0.3) + (0.5 x 400 x [tex]0.25^{2}) -  (0.5 x 400 x 0.3^{2}) - (10.3 x 9.8 x (0.25 + 0.3)) }{0.5 x 10.3}[/tex]}[/tex]

v = 1.46 m/s

                 

(C) if the rocket weren't attached to the spring, the conservation of energy equation becomes

 (Fy) + (0.5kx^{2}) = mgh + (0.5mv^{2})

v = \sqrt{[tex]\frac{(Fy) + (0.5k[tex]x^{2})  - mgh }{0.5m}[/tex]}[/tex]

v =  \sqrt{[tex]\frac{(240 x 0.3) + (0.5 x 400 x [tex]0.25^{2})  - (10.3 x 9.8 x (0.25 + 0.3)) }{0.5 x 10.3}[/tex]}[/tex]

v = 2.4 m/s                          

7 0
3 years ago
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