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zhuklara [117]
3 years ago
11

You wiggle a string, that is fixed to a wall at the other end, creating a sinusoidal wave with a frequency of 2.00 Hz and an amp

litude of 0.075 m. The speed of the wave is 12.0 m/s. At t=0 the string has a maximum displacement and is instantaneously at rest. Assume no waves bounce back from the far end of the wall. Find the angular frequency, period, wavelength, and wave number. Write a wave function describing the wave. Write equations for the displacement, as a function of time, of the end of the string that is being wiggled and at a point 3.00 m from that end. Determine the speed of the medium and draw history and snapshot graphs for the waves created.
Physics
1 answer:
allochka39001 [22]3 years ago
4 0

Answer:hhhkkzkxixuhhhdhhdhdhhhdhhhshsbbsbzhhdhhdhhhshhhdhhdusudhggdydyydhshdhgddhsjsudhdhhdh

Explanation:NzjzjhbxxhzhdghsjshshhHjajskajakakakakaiaiaiijhayayayagayahahah

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An astronaut sitting on the launch pad on Earth's surface is 6,400 kilometers from Earth's center and weighs 400 newtons. Calcul
PIT_PIT [208]

Answer:

weight at height = 100 N .

Explanation:

The problem relates to variation of weight  due to change in height .

Let g₀ and g₁ be acceleration due to gravity , m is mass of the object .

At the surface :

Applying Newton's law of gravitation

mg₀ = G Mm / R²

At height h from centre

mg₁ = G Mm /h²

Given mg₀ = 400 N

400 = G Mm / R²

400 = G Mm / (6400 x 10³ )²

G Mm = 400 x (6400 x 10³ )²

At height h from centre

mg₁ =  400 x (6400 x 10³ )²/ ( 2 x 6400 x 10³)²

= 400 / 4

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weight at height = 100 N

5 0
3 years ago
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Amiraneli [1.4K]

Answer:

I believe it's A.)

Explanation:

Although light comes into our atmosphere through refraction, it reaches our eyes only through reflection from objects. So when light rays reflect off an object and enter the eyes through the cornea you can then see that object.

Hope this helps you out : )

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3 years ago
A penny is dropped from the Statue of Liberty
Brrunno [24]

Answer:

a) The velocity is 2.94m/s

b) 0.441

Explanation:

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Use the equation below to solve for the velocity at 0.30 seconds

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vf =unknown velocity  vi= initial velocity vi=0m/s  a= 9.8m/s^2 t=0.30seconds

Step 1: Substitute the variables with the knowns

vf=0m/s+(9.8m/s^2)*0.30seconds

Step 2: Solve

vf=2.94m/s

b)

Use the equation below to solve for the displacement at 0.30 seconds

x=vit+\frac{1}{2} at^{2}

Step 1: Substitute the same variables with the knowns

x=\frac{1}{2}*(9.8m/s^2)*(0.30seconds)^{2}

Note that vi*t=0 as vi=0m/s

Step 2: Solve

x=0.441m

5 0
3 years ago
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2 years ago
Explain the application of pascals law<br>​
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Answer:

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