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Answer:
The possible frequencies for the A string of the other violinist is 457 Hz and 467 Hz.
(3) and (4) is correct option.
Explanation:
Given that,
Beat frequency f = 5.0 Hz
Frequency f'= 462 Hz
We need to calculate the possible frequencies for the A string of the other violinist
Using formula of frequency
...(I)
...(II)
Where, f= beat frequency
f₁ = frequency
Put the value in both equations
Hence, The possible frequencies for the A string of the other violinist is 467 Hz and 457 Hz.
Answer:
(a) The distance of the chicken from the finish line is 62.5 m
(b) The stationary time of the wildebeest is 675 s
Explanation:
Given;
total distance traveled by wildebeest and chicken, d = 2 km = 2000 m
speed of the wildebeest, = 16 m/s
speed of the chicken, = 2.5 m/s
Time for wildebeest to finish the race without stopping, 2000 / 16 = 125 s
Time for chicken to finish the race without stopping, 2000/2.5 = 800 s
(b) for how long in time (in s) was the wildebeest stationary?
t(stationary) = t(chicken) - t(wildebeest)
t = 800s - 125 s
t = 675 s
(a) how far (in m) is the chicken from the Finish Line when the wildebeest resume the race?
The time taken for the wildebeest to run 1.6 km (1600 m) is given by;
t = 1600 / 16 = 100 s
The total time spent by the wildebeest before it resumed the race = stationary time + 100s
t (total) = 675 s + 100 s = 775 s
Distnace traveled by the chicken when the wildebeest resumed the race = 2.5m/s x 775s = 1937.5 m
Thus, the distance of the chicken from the finish line = 2000 m - 1937.5 m
the distance of the chicken from the finish line = 62.5 m