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Lemur [1.5K]
3 years ago
7

A 1.5m long string weighs 0.0020 kg. It is tensioned to 100N. A disturbance travels along it with a wavelength of 1.5m, find:a)

the propagation velocity of the wave
Physics
1 answer:
Zigmanuir [339]3 years ago
7 0

Answer:

the propagation velocity of the wave is 274.2 m/s

Explanation:

Given;

length of the string, L = 1.5 m

mass of the string, m = 0.002 kg

Tension of the string, T = 100 N

wavelength, λ = 1.5 m

The propagation velocity of the wave is calculated as;

v = \sqrt{\frac{T}{\mu} } \\\\\mu \ is \ mass \ per \ unit \ length \ of \ the \ string\\\\\mu = \frac{0.002 \ kg}{1.5 \ m} = 0.00133 \ kg/m\\\\v = \sqrt{\frac{100}{0.00133} } \\\\v = 274.2 \ m/s

Therefore, the propagation velocity of the wave is 274.2 m/s

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(4A) The mass of Earth is 5.972 * 10^24 kg, and the radius of Earth is 6,371 km.
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Answer:

x₁ = 345100 km

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The direction of the attraction forces between the earth and the object, and between the moon and the object, are in opposite direction and  (along the straight line between the centers of earth and moon) and as gravity is always attractive, the net force will become zero when both forces are equal. According to this:

Let  call "x₁"  distance between center of the earth and the object, and

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F₁ = F₂               K*Mt*m₀ / x₁²   =  K*Ml*m₀ /x₂²

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Mt/ x₁²  =  Ml/ x₂²

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x₁ =  384000 - x₂

Mt/( 384000 - x₂)²  =  Ml / x₂²

Mt *  x₂²  =  Ml *( 384000 - x₂)²

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Mt *  x₂²  =  Ml *[ ( 384000)² + x₂² - 768000*x₂]

By substitution:

5.972*10∧24*x₂² = 7.348*10∧22 * [ 1.47*10∧11 ] + 7.348*10∧22*x₂² -

                                7.348*10∧22*768000*x₂

Simplifying by 10∧22

5.972*10²*x₂²  = 7.348* [ 1.47*10∧11 ] + 7.348*x₂²- 7.348*768000*x₂

Sorting out

5.972*10²*x₂²- 7.348*x₂² = 10.80*10∧11 - 56,43* 10∧5*x₂

(597,2 - 7,348 )* x₂²  = 10.80*10∧11 - 56.43*10∧5*x₂

590x₂²  + 56.43*10∧5*x₂ - 10.80*10∧11 = 0

Is a second degree equation

x₂  =  -56.43*10∧5 ± √3184*10∧10 + 25488*10∧11  / 1160

x₂ ₁  = -56.43*10∧5 + √3184*10∧10 + 25488*10∧11  / 1160

x₂ ₁  =  -56.43*10∧5 + √3184*10∧10 + 254880*10∧10  / 1160

x₂ ₁  = -56.43*10∧5 + 10∧5 [ √3184 + 254880 ] /1160

x₂ ₁  =  -56.43*10∧5 + 508* 10∧5  / 1160

x₂ ₁  =  451.27*10∧5/1160

x₂ ₁  =  4512.7*10∧4 /1160

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x₂ ₁  = 38900 km

x₂ = 38900 km

We dismiss the other solution because is negative and there is not a negative distance

Then the distance between the earth and the object is:

x₁  = 384000 - x₂

x₁ = 384000 - 38900

x₁ = 345100 km

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