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krok68 [10]
3 years ago
6

Cousin Throckmorton is playing with the clothesline. One end of the clothesline is attached to a vertical post. Throcky holds th

e other end loosely in his hand, so that the speed of waves on the clothesline is a relatively slow 0.700 m/s . He finds several frequencies at which he can oscillate his end of the clothesline so that a light clothespin 40.0 cm from the post doesn't move. What are these frequencies?
Physics
1 answer:
Oksi-84 [34.3K]3 years ago
3 0

Answer:

The  frequencies are  f_n  =  n (0.875 )

Explanation:

From the question we are told that

   The speed of the wave is  v  =  0.700 \  m/s

   The  length of vibrating  clothesline is  L  =  40.0 \  cm = 0.4 \ m

Generally the fundamental frequency is  mathematically represented as

        f =  \frac{v}{2 L  }

=>     f =  \frac{ 0.700 }{2 *  0.4   }

=>     f =  0.875 \  Hz

Now  this other frequencies of vibration experience by the clotheslines are know as harmonics and they are obtained by integer multiple of  the fundamental frequency

So  

   The  frequencies are mathematically represented as

       f_n  =  n  * f

=>     f_n  =  n (0.875 )

Where  n  =  1, 2, 3 ....

       

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We are only concerned about what it happens on the horizontal axis, so there are two forces acting on the cart+child system: the force F of the man pushing it, and the frictional force F_f acting in the opposite direction. So Newton's second law can be rewritten as
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Girl who’s mass is 52kg, experienced a net force of 1800N at bottom of a roller coster loop during her school physics field to t
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Two parallel, circular conducting plates 32 cm in diameter are separated by 0.5 cm and have charges of +24 nC and -24 nC, respec
inysia [295]

Answer: E = 33762.39 N/c

Explanation: we calculate the capacitance of the two conducting plates ( this is because, 2 circular disc carrying opposite charges and the same dimension form a capacitor).

C = A/4πkd

Where C = capacitance of capacitor

A = Area of plates = πr² ( where r is radius which is half of the diameter)

K = electric constant = 9×10^9

d = distance between plates = 0.5cm = 0.005 m

Let us get the area, A = πr², where r = D/2 where D = diameter

r = 32/2 = 16cm = 0.16m

A = 22/7 × (0.16)² = 0.0804 m²

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C = 0.0804/4×3.142*9×10^9 × 0.005

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C = 142.17*10^(-12) F.

But C =Q/V where V = Ed

Hence we have that

C = Q/Ed

Where C = capacitance of capacitor = 142.17*10^(-12)F

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E = strength of electric field =?

d = distance between plates = 0.005m

142.17*10^(-12) = 24 ×10^-9 / E × 0.005

By cross multiplying

142.17*10^(-12) × E × 0.005 = 24 ×10^-9

E = 24 ×10^-9 / 142.17*10^(-12) × 0.005

E = 33762.39 N/c

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