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.
The hawk’s centripetal acceleration is 2.23 m/s²
The magnitude of the acceleration under new conditions is 2.316 m/s²
radius of the horizontal arc = 10.3 m
the initial constant speed = 4.8 m/s
we know that the centripetal acceleration is given by
=
= 23.04/10.3
= 2.23 m/s²
It continues to fly but now with some tangential acceleration
= 0.63 m/s²
therefore the net value of acceleration is given by the resultant of the centripetal acceleration and the tangential acceleration
so
=
=
= 2.316 m/s²
So the magnitude of net acceleration will become 2.316 m/s².
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