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babymother [125]
3 years ago
7

4. What is the best estimate of the frequency of the wave shown below?

Physics
1 answer:
tatyana61 [14]3 years ago
7 0

Answer:

The answer is "Option D"

Explanation:

Please find the graph file in the attached file.

Using formula:

\to v= \rho \lambda \\\\\ \to \lambda= 130 \cm \\\\ \to v= 335 \frac{cm}{s}\\\\ \to \rho= \frac{v}{\lambda} \\\\

      =\frac{335}{130} \\\\ = 2.57 \approx 2.50 \ Hz

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Tha atmosphere has no effect on gravity.
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A car travels 10 meters east in 4 seconds. What is the car's velocity? *
MrMuchimi

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your velocity is 2.5 m/sec

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A plastic rod of length d = 1.5 m lies along the x-axis with its midpoint at the origin. The rod carries a uniform linear charge
Serga [27]

Answer:

Explanation:

Let the plastic rod extends from - L to + L .

consider a small length of dx on the rod on the positive x axis at distance x . charge on it =  λ dx where  λ is linear charge density .

It will create a field at point P on y -axis . Distance of point P

= √ x² + .15²

electric field at P due to small charged length

dE = k λ dx x  / (x² + .15² )

Its component along Y - axis

= dE cosθ where θ is angle between direction of field dE and y axis

= dE x .15 / √ x² + .15²

=  k λ dx  .15 / (x² + .15² )³/²

If we consider the same strip along the x axis at the same position  on negative x axis , same result will be found . It is to be noted that the component of field in perpendicular to y axis will cancel out each other . Now for electric field due to whole rod at point p , we shall have to integrate the above expression from - L to + L

E = ∫  k λ  .15  / (x² + .15² )³/² dx

=  k λ  x L / .15 √( L² / 4 + .15² )

6 0
4 years ago
A spring with spring constant 11.5 N/m hangs from the ceiling. A 490 g ball is attached to the spring and allowed to come to res
Natalija [7]

Answer:

The time constant is \tau = 17.5 \ s    

Explanation:

From the question we are told that

   The spring constant is  k = 11.5 \  N/m

   The mass  of the ball is  m_b  = 490 \ g  = 0.49 \ kg

   The amplitude of the  oscillation t the beginning is x =  6.70 cm = 0.067 \  m

    The amplitude after time t is  x_t = 2.20 cm = 0.022 \  m

    The number of oscillation is N  = 30

Generally the time taken to attain the second amplitude is mathematically represented as

       t  = N  *  T                                            Here  T is the period of oscillation

         t = N * 2\pi \sqrt{\frac{m}{k} }

=>     t = 30 * 2 * 3.142 *  \sqrt{\frac{ 0.490}{11.5} }

=>     t = 38.88 \  s

Generally the amplitude at time t is mathematically represented as

         x(t) = x e^{-\frac{at}{2m} }

Here a is the damping  constant so

 at  t = 38.88 \  s ,  x_t = 2.20 cm = 0.022 \  m

So  

     0.022 = 0.067 e^{-\frac{a * 38.88}{2 * 0.490} }

=>  0.3284 = e^{-\frac{a * 38.88}{2 * 0.490} }

taking natural log of both sides

=>  ln(0.3284 ) = -\frac{a * 38.88}{2 * 0.490} }    

=>   a = 0.028

Generally the time constant is mathematically represented as

    \tau = \frac{m}{a}      

=> \tau = \frac{0.490}{  0.028}    

=> \tau = 17.5 \ s    

4 0
3 years ago
Help Please
densk [106]

Answer:

Energy=3.1times 10^-17 J

Rest mass: 6.2 kg

Speed: 47.5 m/s

Wavelength: 2.659 times 10^-6

Momentum: 67.3 kg(m/s)

Explanation:

3 0
3 years ago
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