Answer:
The first harmonic is: 250Hz, second harmonic 500Hz, third harmonic 750Hz.
Explanation:
Use the frequency f, speed v, and wavelentgh L relationship:
We are given the speed v=400 m/s. The base wavelength on a string of length 80cm is twice the length of the string (a "half wave" along the full length of the string), so:
The fundamental frequency (first harmonic) is 250 Hz
The second harmonic is produced by one full wave across the string (adding one node in the middle), so L=80cm in this case, therefore the second harmonic frequency is: f2 = 2*250=500Hz
the third harmonic add another node (and a half wave) to the pattern and the wavelength will be 2/3 of 80cm, so f3=3*250Hz = 750Hz
As seagull drops a shell from rest at a height of 12 m, so we use kinematic equation of motion,
Here, h is the height, u is initial velocity , v is final velocity and g is acceleration due to gravity.
Given, h = 12 m.
We take, 
and 
because seagull drops a shell from rest.
Therefore, the speed of shell when it hits the rocks,
Answer:
Refraction
Explanation:
A pencil bends when it enters the water media due to the phenomenon of refraction.
- Refraction is one of the properties of waves.
- Refraction of light occurs which it crosses from a region having different density values.
- Water is denser than air.
- As light moves from air to wave it becomes refracted
Answer:
(a) 135 kV
(b) The charge chould be moved to infinity
Explanation:
(a)
The potential at a distance of <em>r</em> from a point charge, <em>Q</em>, is given by
where 
Difference in potential between the points is
![kQ\left[-\dfrac{1}{0.2\text{ m}} -\left( -\dfrac{1}{0.1\text{ m}}\right)\right] = \dfrac{kQ}{0.2\text{ m}} = \dfrac{9\times10^9\text{ F/m}\times3\times10^{-6}\text{ C}}{0.2\text{ m}}](https://tex.z-dn.net/?f=kQ%5Cleft%5B-%5Cdfrac%7B1%7D%7B0.2%5Ctext%7B%20m%7D%7D%20-%5Cleft%28%20-%5Cdfrac%7B1%7D%7B0.1%5Ctext%7B%20m%7D%7D%5Cright%29%5Cright%5D%20%3D%20%5Cdfrac%7BkQ%7D%7B0.2%5Ctext%7B%20m%7D%7D%20%3D%20%5Cdfrac%7B9%5Ctimes10%5E9%5Ctext%7B%20F%2Fm%7D%5Ctimes3%5Ctimes10%5E%7B-6%7D%5Ctext%7B%20C%7D%7D%7B0.2%5Ctext%7B%20m%7D%7D)
(b)
If this potential difference is increased by a factor of 2, then the new pd = 135 kV × 2 = 270 kV. Let the distance of the new location be <em>x</em>.
![270\times10^3 = kQ\left[-\dfrac{1}{x}-\left(-\dfrac{1}{0.1\text{ m}}\right)\right]](https://tex.z-dn.net/?f=270%5Ctimes10%5E3%20%3D%20kQ%5Cleft%5B-%5Cdfrac%7B1%7D%7Bx%7D-%5Cleft%28-%5Cdfrac%7B1%7D%7B0.1%5Ctext%7B%20m%7D%7D%5Cright%29%5Cright%5D)
The charge chould be moved to infinity
