Answer:
31.2 m/s
Explanation:
= Frequency of approach = 480 Hz
= Frequency of going away = 400 Hz
= Speed of sound in air = 343 m/s
= Speed of truck
Frequency of approach is given as
eq-1
Frequency of moving awayy is given as
eq-2
Dividing eq-1 by eq-2


= 31.2 m/s
Answer:
120 J
Explanation:
KE = mv²/2 = (0.15 kg * [40 m/s]²)/2 = 120 J
<em>12,25 km/h</em>
<em>≈ 3,4 m/s </em>
<em>v = d/t</em>
<em>= 12250m/h</em>
<em>= 12,25km/h</em>
<em>or</em>
<em>v = d/t</em>
<em>= 12250m/h</em>
<em>1h = 60m×60s = 3600s</em>
<em>= 12250m/3600s</em>
<em>≈ 3,4 m/s </em>
Answer: I have no idea either i need help aswell
Explanation: