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Inessa [10]
3 years ago
10

A string fixed at both ends is 8.40 m long and has a mass of 0.120 kg. It is subjected to a tension of 96.0 N and set oscillatin

g.
(a) What is the speed of the waves on the string?
(b) What is the longest possible wavelength for a standing wave?
(c) Give the frequency of that wave.
Physics
1 answer:
Luden [163]3 years ago
3 0

Answer:

81.9756 m/s

16.8 m

4.8795 Hz

Explanation:

m = Mass of string = 0.12 kg

L = Length of string = 8.4 m

T = Tension on string = 96 N

Linear density is given by

\mu=\dfrac{m}{L}\\\Rightarrow \mu=\dfrac{0.12}{8.4}

Spee of the wave is given by

v=\sqrt{\dfrac{T}{\mu}}\\\Rightarrow v=\sqrt{\dfrac{96}{\dfrac{0.12}{8.4}}}\\\Rightarrow v=81.9756\ m/s

The speed of the waves on the string is 81.9756 m/s

Wavelength is given by

\lambda=2L\\\Rightarrow \lambda=2\times 8.4\\\Rightarrow \lambda=16.8\ m

The longest possible wavelength is 16.8 m

Frequency is given by

f=\dfrac{v}{\lambda}\\\Rightarrow f=\dfrac{81.9756}{16.8}\\\Rightarrow f=4.8795\ Hz

The frequency of the wave is 4.8795 Hz

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As we have to find the time taken, the formula can be altered as,

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Substituting all the given values,

t =\frac{781000 - 226000} {4* 1014}

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How many butter, which has a usable energy content of 6.0 Cal/g (= 6000 cal/g), would be equivalent to the change in gravitation
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175.96 g

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a certain car travels 20 km east then turns south for 13 km finally the car turns east again for 6 km
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Answer:

The new distance is     d = 0.447 d₀

Explanation:

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         F = k q₁ q₂ / r²

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We reduce the charge of sphere B to 1/5 of its initial value (q_{B}=q₂ = q₂ / 5) than new distance (d = n d₀)

dat

     q₁ = q_{A}

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In order for the deviation to maintain the electric force it should not change, so we apply the Coulomb equation for the two points

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          5 n² = 1

          n = √ 1/5

          n = 0.447

The new distance is

         d = 0.447 d₀

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