Explanation:
Given that,
Frequency of train horn, f = 218 Hz
Speed of train, ![v_t = 31.7 m/s](https://tex.z-dn.net/?f=v_t%20%3D%2031.7%20m%2Fs)
The speed of sound, V = 344 m/s (say)
The speed of the observed person, ![V_o=0\ m/s](https://tex.z-dn.net/?f=V_o%3D0%5C%20m%2Fs)
(a) When the train approaches you, the Doppler's effect gives the frequency as follows :
![f'=f(\dfrac{V}{V-v_t})\\\\f'=218\times (\dfrac{344}{344-31.7})\\\\f'=240.12\ Hz](https://tex.z-dn.net/?f=f%27%3Df%28%5Cdfrac%7BV%7D%7BV-v_t%7D%29%5C%5C%5C%5Cf%27%3D218%5Ctimes%20%28%5Cdfrac%7B344%7D%7B344-31.7%7D%29%5C%5C%5C%5Cf%27%3D240.12%5C%20Hz)
(b) When the train moves away from you, the Doppler's effect gives the frequency as follows :
![f'=f(\dfrac{V}{V+v_t})\\\\f'=218\times (\dfrac{344}{344+31.7})\\\\f'=199.6\ Hz](https://tex.z-dn.net/?f=f%27%3Df%28%5Cdfrac%7BV%7D%7BV%2Bv_t%7D%29%5C%5C%5C%5Cf%27%3D218%5Ctimes%20%28%5Cdfrac%7B344%7D%7B344%2B31.7%7D%29%5C%5C%5C%5Cf%27%3D199.6%5C%20Hz)
Hence, this is the required solution.
The valence electrons increase as you move left to right.
Answer 1) The electric field at distance r from the thread is radial and has magnitude
E = λ / (2 π ε° r)
The electric field from the point charge usually is observed to follow coulomb's law:
E = Q / (4 π ε°
)
Now, adding the two field vectors:
= {2.5 / (22 π ε° X 0.07 ) ; 0}
Answer 2)
= {2.3 / (4 2 π ε°) ( - 7/ (√(84); -12 / (√84))
Adding these two vectors will give the length which is magnitude of the combined field.
The y-component / x-component gives the tangent of the angle with the positive x-axes.
Please refer the graph and the attachment for better understanding.
Answer:
r = 0.05 m = 5 cm
Explanation:
Applying ampere's law to the wire, we get:
![B = \frac{\mu_oI}{2\pi r}\\\\r = \frac{\mu_oI}{2\pi B}](https://tex.z-dn.net/?f=B%20%3D%20%5Cfrac%7B%5Cmu_oI%7D%7B2%5Cpi%20r%7D%5C%5C%5C%5Cr%20%3D%20%20%5Cfrac%7B%5Cmu_oI%7D%7B2%5Cpi%20B%7D)
where,
r = distance of point P from wire = ?
μ₀ = permeability of free space = 4π x 10⁻⁷ N/A²
I = current = 2 A
B = Magnetic Field = 8 μT = 8 x 10⁻⁶ T
Therefore,
![r = \frac{(4\pi\ x\ 10^{-7}\ N/A^2)(2\ A)}{2\pi(8\ x\ 10^{-6}\ T)}\\\\](https://tex.z-dn.net/?f=r%20%3D%20%5Cfrac%7B%284%5Cpi%5C%20x%5C%2010%5E%7B-7%7D%5C%20N%2FA%5E2%29%282%5C%20A%29%7D%7B2%5Cpi%288%5C%20x%5C%2010%5E%7B-6%7D%5C%20T%29%7D%5C%5C%5C%5C)
<u>r = 0.05 m = 5 cm</u>