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:

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:

hence, the frequencies for before and after tha train has past are
b) 472HZ, 408HZ
Answer:
25.08m/s
Explanation:
mgh1 + 0.5mv1² = mgh2 + 0.5mv2²
h1 = 0m
v1 = u
h2 = 5m
v2 = 23m/s
putting the values into the formula above;
m(10)(0) + 0.5m(u²) = m(10)(5) + 0.5m(23²)
0 + 0.5mu² = 50m + 264.5m
0.5mu² = 314.5m
dividing through by m
0.5u² = 314.5
u² = 629
u = <u>2</u><u>5</u><u>.</u><u>0</u><u>8</u><u>m</u><u>/</u><u>s</u>
<u>Theref</u><u>ore</u><u>,</u><u> </u><u>the</u><u> </u><u>init</u><u>ial</u><u> </u><u>speed</u><u> </u><u>"</u><u>u</u><u>"</u><u> </u><u>=</u><u> </u><u>2</u><u>5</u><u>m</u><u>/</u><u>s</u>
Option A looks like the best definition
A group of cells together