Answer:
push on objects force is required. pull force is resli8
Answer:
b) 472HZ, 408HZ
Explanation:
To find the frequencies perceived when the bus approaches and the train departs, you use the Doppler's effect formula for both cases:
![f_o=f\frac{v_s+v_o}{v_s-v}\\\\f_o=f'\frac{v_s-v_o}{v_s+v}\\\\](https://tex.z-dn.net/?f=f_o%3Df%5Cfrac%7Bv_s%2Bv_o%7D%7Bv_s-v%7D%5C%5C%5C%5Cf_o%3Df%27%5Cfrac%7Bv_s-v_o%7D%7Bv_s%2Bv%7D%5C%5C%5C%5C)
fo: frequency of the source = 440Hz
vs: speed of sound = 343m/s
vo: speed of the observer = 0m/s (at rest)
v: sped of the train
f: frequency perceived when the train leaves us.
f': frequency when the train is getTing closer.
Thus, by doing f and f' the subjects of the formulas and replacing the values of v, vo, vs and fo you obtain:
![f=f_o\frac{v_s-v}{v_s+v_o}=(440Hz)\frac{340m/s-25m/s}{340m/s}=408Hz\\\\f'=f_o\frac{v_s+v}{v_s-v_o}=(440Hz)\frac{340m/s+25m/s}{340m/s}=472Hz](https://tex.z-dn.net/?f=f%3Df_o%5Cfrac%7Bv_s-v%7D%7Bv_s%2Bv_o%7D%3D%28440Hz%29%5Cfrac%7B340m%2Fs-25m%2Fs%7D%7B340m%2Fs%7D%3D408Hz%5C%5C%5C%5Cf%27%3Df_o%5Cfrac%7Bv_s%2Bv%7D%7Bv_s-v_o%7D%3D%28440Hz%29%5Cfrac%7B340m%2Fs%2B25m%2Fs%7D%7B340m%2Fs%7D%3D472Hz)
hence, the frequencies for before and after tha train has past are
b) 472HZ, 408HZ
Answer:
emf induced in the loop, at the instant when 9.0s have passed = 1.576 * 10 ⁻² V.
Direction is counter clockwise.
Explanation:
See attached pictures.